//textdata.js //copyright Robert M. Hanson, St. Olaf College 9:19 PM 3/17/2003 //use . to go to a specific frame (0-based): CI["13-8:cell potential"]=new Array("_ti13.1") //use addScript(13,"The Chemical Connection","_bat13") to create scripts anywhere in here. //revised 8:56 PM 3/10/2008 atmospheres -- bar //working on chapter 13 5:21 PM 3/19/2003 //_ch01 CI["_q1"]=new Array("
" +"

What makes a reaction \"happen\"? " +"

Is it the release of energy?" +"

Some special \"force\"?" +"

These are the fundamental questions
we wish to answer in this study
of molecular thermodynamics.

The answers might surprise you!

" ,"") CI["1-1:chemical change"]=new Array("_q1") CI["_eqp1"]=new Array("|

You probably already know quite a bit about equilibrium. Now we want to understand both what equilibrium is and how chemical reactions proceed toward it.

The goal, then, is to explore with you two important ideas:

" +"

(1)

" +"The equilibrium state (especially, equilibrium constants) can be predicted based on probability alone.

" +"

(2)

" +"The predictive nature of probability is especially well suited to chemical systems due to the shear magnitude of the number of particles involved." +"

" ,"equil.gif") CI["1-2:equilibrium"]=new Array("_eqp1") CI["_Kq1"]=new Array("No matter what percent H and D we start with, the system adjusts to give the same value for the reaction quotient, Q, which we then call the equilibrium \"constant\" K.","equilh2d2q.gif") CI["equilibrium constant (K), and isotope exchange"]=new Array("_Kq1") CI["1:equilibrium constant (K), and reaction quotient (Q)"]=new Array("_Kq1") CI["isotope exchange, equilibrium constant for"]=new Array("_Kq1") addScript(1,"Equilibrium Ideas","_q1,_eqp1,_Kq1") CI["_pd1"]=new Array(" ","probways.gif") CI["probability, defined"]=new Array("_pd1") CI["_and1"]=new Array("AND-type probability multiplies. In this case we calculate the probability of rolling a 4 ten times in a row using a standard six-sided die.","probandmult.gif") CI["probability, and-type"]=new Array("_and1") CI["_or1"]=new Array("OR-type probability adds. In this case we calculate the probability of rolling either a 2 or a 3 with a single role of a standard six-sided die.","proboradd.gif") CI["probability, or-type"]=new Array("_or1") CI["_add1"]=new Array("AND-type and OR-type probability can be combined. In this case we calculate the probability of rolling an 8 using two standard six-sided dice.","probcomb.gif") CI["probability, combining"]=new Array("_add1") CI["_not1"]=new Array("The probability of \"not x\" is simply 1 - Px. In this case we calculate the probability of NOT drawing two hearts off the top of a standard deck of 52 cards using two different methods.","probnot.gif") CI["probability, not-type"]=new Array("_not1") CI["_big1"]=new Array("For large populations, we can approximate when calculating the most probable distribution.","probapprox.gif") CI["probability, for large populations"]=new Array("_big1") CI["_pp1"]=new Array("Two perspectives are possible:

" +"" +"
Gambler's Pespective\"If I roll this die or pick up that card, my chances of winning are....\"
Casino’s Perspective\"If this game is played 10,000 times, we should expect a profit of....\"
" +"

In chemistry, due to the vast number of particles involved, it is the casino's perspective that relates most to equilibrium. The casino operator’s advantage is precisely why probability works to explain chemical reactions. If you think the odds are pretty likely to work out after 10,000 games are played, try 602,213,670,000,000,000,000,000 (a mole)!" ,"" ) CI["probability, interpretation"]=new Array("_pp1") addScript(1,"Probability","_pd1,_and1,_or1,_add1,_not1,_big1,_pp1") CI["_pc1"]=new Array("There are 24 different ways of distributing four cards, one of each suit, into two \"hands\". But if we only look at suit color, there are only two distinct distributions. One distribution is twice as likely as the other.","cards.gif") CI["distribution, playing cards"]=new Array("_pc1") CI["_iso1"]=new Array("Distributions are distinctly different observable outcomes. For example, if you had 800 H atoms and 200 D atoms, there are 101 distinct distributions of these atoms into molecules.","distribution.gif") CI["distribution, isotope exchange"]=new Array("_iso1") CI["_pi1"]=new Array("To investigate the distributions in the H2/D2/HD system, click here","probh2d2.gif") CI["probability, and isotope exchange"]=new Array("_pi1") addScript(1,"Distributions","_pc1,_iso1,_pi1") CI["_lec1"]=new Array("For relative probability, the most probable distribution is assigned the number 1, and all other probabilities are scaled accordingly.","probrel.gif") CI["probability, and Le Châtelier's principle"]=new Array("_lec1") CI["_eqn1"]=new Array("","equilibrium.gif") CI["1:equilibration"]=new Array("_eqn1") CI["1:equilibrium, and Le Châtelier's principle"]=new Array("_eqn1") CI["1:equilibrium, approach to"]=new Array("_eqn1") CI["_peq1"]=new Array( "Imagine starting with a \"deck\" of 8 moles of H atoms and 2 moles of D atoms already \"stacked\" in the form of 4.0 moles of H2 and 1.0 moles of D2 molecules. Start shuffling. " +"The relative probability of this starting distribution, with absolutely no HD, is essentially 0. Once the collisions and exchanges start, " +"we expect to see some HD formed. HD will build up, but the reverse reaction will also be taking place. " +"At some point, the rate of the forward reaction and the rate of the reverse reaction will become equal. Roughly as many HD will split up in any shuffle as are made. " +"This is equilibrium -- a dynamic fluctuation around the most probable distribution." +"

Initial amounts: 4.0 mol H2 and 1.0 mol D2" +"|

Equilibrium amounts: 3.2 mol H2, 0.2 mol D2, and 1.6 mol HD" ,"equilh2d2.gif" ) CI["1-13:probability, and equilibrium"]=new Array("_peq1") CI["_pr1"]=new Array("The system starts in state A. 300 D2 molecules (600 D atoms) are added, giving state B. The system is no longer at equilibrium. In time, the system finds its way to the a new most probable distribution, state C.","probandlechat.gif" ,"While these states are possible, there are others. States A and C are simply the most probable for their particular number of H and D atoms. Notice that relative to state C, state B is very improbable.","equilibrium.gif" ) CI["probability, relative"]=new Array("_pr1") addScript(1,"Le Chatelier's principle","_lec1,_eqn1,_peq1,_pr1") CI["_pk1"]=new Array("For isotope exchange reactions, we can estimate K based on probability alone.","probandk.gif") CI["probability, and equilibrium constant (K)"]=new Array("_pk1") CI["_pcalc1"]=new Array( "Calculating the amounts of different molecules present at equilibrium
in isotope exchange reactions involves six steps:" +"
The steps include:" +"

1) Determine the number of moles of each of the exchanging atom types." +"
2) Add these to get the total number of exchanging atoms." +"
3) Determine the probability of a randomly selected atom being of each type." +"
4) From this information, determine the approximate probability of each molecule type." +"
5) Determine the total number of molecules." +"
6) Determine the number of each type of molecule at equilibrium." ,"" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" ,"" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

1) Determine the number of moles of each of the exchanging atom types." +"

nH = 8.0 moles
nD = 2.0 moles" ,"" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

nH = 8.0 moles
nD = 2.0 moles
" +"

2) Add these to get the total number of exchanging atoms." +"

natoms = 10.0 moles" ,"" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

nH = 8.0 moles
nD = 2.0 moles
natoms = 10.0 moles
" +"

3) Determine the probability of a randomly selected atom being of each type:|In this case there is an 80% chance of an atom being H and a 20% chance of an atom being D." ,"calcprob3.gif" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

nH = 8.0 moles
nD = 2.0 moles
natoms = 10.0 moles		PH = 0.80
PD = 0.20
" +"

4) From this information, determine the approximate probability of each molecule type:|Note the difference here between atoms (H and D) and molecules (H2, D2, and HD).
Notice, too, that we have to consider the possibilities of both HD and DH in the calculation of PDH." ,"calcprob4.gif" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

nH = 8.0 moles
nD = 2.0 moles
natoms = 10.0 moles		PH = 0.80
PD = 0.20		PH2 = 0.64
PD2 = 0.04
PHD = 0.32
" +"

5) Determine the total number of molecules. In this case we are starting with 4.0 moles of H2 and 1.0 moles of D2, for a total of 5.0 moles of molecules. Note that in isotope exchange reactions the total number of molecules does not change, so in the end we still have 5.0 moles of molecules." ,"" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

nH = 8.0 moles
nD = 2.0 moles
natoms = 10.0 moles		PH = 0.80
PD = 0.20		PH2 = 0.64
PD2 = 0.04
PHD = 0.32		nmolecules = 5.0 moles
" +"

6) Finally, using the idea that nx = Px × nmolecules, determine the number of each type of molecule at equilibrium.|From these numbers we could calculate the equilibrium constant for any valid equilibrium equation for this system." ,"calcprob6.gif" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

nH = 8.0 moles
nD = 2.0 moles
natoms = 10.0 moles		PH = 0.80
PD = 0.20		PH2 = 0.64
PD2 = 0.04
PHD = 0.32		nmolecules = 5.0 moles
nH2 = 3.2 moles
nD2 = 0.2 moles
nHD = 1.6 moles
" +"
The steps include:" +"

1) Determine the number of moles of each of the exchanging atom types." +"
2) Add these to get the total number of exchanging atoms." +"
3) Determine the probability of a randomly selected atom being of each type." +"
4) From this information, determine the approximate probability of each molecule type." +"
5) Determine the total number of molecules." +"
6) Determine the number of each type of molecule at equilibrium." ,"" ,"Assume that we start with 4.0 moles of H2 and 1.0 mole of D2.

The question is,

How much H2, D2, and HD are present at equilibrium?" +"

It's a good idea to check your work:" +"

" +"
nH = 8.0 moles
nD = 2.0 moles		PH = 0.20
PD = 0.80		PH2 = 0.64
PD2 = 0.04
PHD = 0.32		nH2 = 3.2 moles
nD2 = 0.2 moles
nHD = 1.6 moles
natoms = 10.0 moles
Patoms = 1
Pmolecules = 1
nmolecules = 5.0 moles
" ,"" ) CI["equilibrium, determining amounts"]=new Array("_pcalc1") addScript(1,"Probability and Equilibrium","_pk1,_pcalc1") CI["_f1"]=new Array("Fluctuation is the result of there being more than one possible distribution.","fluctuation.gif" ,"The more particles there are in the system, the less pronounced are the fluctuations.","fluctuation2.gif" ,"Given enough particles in the system, the fluctuations are essentially unobservable.|We say that the \"equilibrium state\" and the \"most probable distribution\" are the same, even though there is still a tremendous amount of fluctuation going on.","fluctuation3.gif" ) CI["fluctuation, effect of population size"]=new Array("_f1") addScript(1,"Fluctuation","_f1,_eq1") CI["_seq1"]=new Array("|

There is no magic here. The \"shift\" in the equilibrium due to a chemical reaction is simply the establishment of a new \"most probable distribution\" based entirely on the idea that that distribution, along with those indistinguishable from it, have many more ways of being found. With a few thousand atoms it might be possible to observe slightly different distributions now and then, but with mole quantities, we don’t have to worry about them. It’s just too unlikely that the system will be found in any significantly different distribution. There will be fluctuations around this distribution. We just won’t be able to detect them.","equilh2d2x.gif") CI["1-17:probability, and equilibrium"]=new Array("_seq1") CI["_eq1"]=new Array("What we call the equilibrium state is really just the most probable distribution, along with the generally undetectable set of fluctuations very similar to it. That is, chemical equilibrium is a dynamic, fluctuating state centered around the most probable distribution.","fluctuation2.gif") CI["equilibrium, as most probable distribution"]=new Array("_eq1") CI["distribution most probable, as equilibrium"]=new Array("_eq1") CI["equilibrium, and fluctuation"]=new Array("_eq1") addScript(1,"Summary","_seq1,_eq1") //_ch02 CI["_qt2"]=new Array("Why do you suppose the following statement is fundamentally true:

If heat is absorbed when a reaction occurs, then increasing the temperature will drive the reaction further toward completion.

","") CI["2-2:heat (q), and temperature (T)"]=new Array("_qt2") CI["_tef2"]=new Array("","tempeffect.gif") CI["2:temperature (T), effect on chemical reactions"]=new Array("_tef2") CI["2:chemical reactions, effect of temperature"]=new Array("_tef2") CI["_the2"]=new Array("","tempeffect2.gif") CI["temperature (T), and heat energy"]=new Array("_the2") CI["chemical reactions, endothermic and exothermic"]=new Array("_the2") addScript(2,"Temperature Effect on Chemical Reactions","_qt2,_tef2,_the2") CI["_q2"]=new Array("We start with the simple idea of particles that can have only unit amounts of energy.
We call these units \"quanta.\"
We put the particles in a box, give them some energy, and let them go. What is going to happen? " +"
Click here to watch them go.","quanta.gif") CI["2-3:quanta, in small systems"]=new Array("_q2") CI["2-3:energy, quanta"]=new Array("_q2") CI["_f2"]=new Array("In order to calculate the number of ways the energy in a system can be distributed, we will make use of factorials.","factorial.gif") CI["factorial"]=new Array("_f2") CI["_tww2"]=new Array("One way of calculating the thermodynamic probability of a distribution is to use combinations.","thermoprobways.gif") CI["thermodynamic probability (W), as number of ways"]=new Array("_tww2") CI["thermodynamic probability (W), and combinations"]=new Array("_tww2") CI["_w2"]=new Array(" ","thermoprobdef.gif") CI["thermodynamic probability (W), defined"]=new Array("_w2") addScript(2,"Thermodynamic Probability","_q2,_f2,_tww2,_w2") CI["_d33_2"]=new Array("The 10 ways of distributing 3 units of energy among 3 particles. animate","energysmall.gif") CI["energy distributions, in small systems"]=new Array("_d33_2") CI["_dss2"]=new Array( "Read on to see how to systematically determine all possible distributions of energy in small systems. A web page that does this using random collisions is also available." ,"distrsmall.gif" ,"1) Put all the energy into one particle and call this Distribution A. In this example, there are a total of U units of energy to be distributed among n particles. W for this distributions is just n itself." ,"distrsmall1.gif" ,"2) Next, move this particle down one level and move a particle from Level 0 up one level to get Distribution B, which still has the same total energy, U, but is now more probable:" ,"distrsmall2.gif" ,"3) Repeat Step 2, but consider BOTH possibilities—either the particle from Level 1 could move up or another particle from Level 0 could become energized. This thinking generates Distributions C and D:" ,"distrsmall3.gif" ,"4) Continue in this way, always moving the highest particle down one level while moving all possible particles below it up one level until no new distributions are found. There isn’t any need to actually calculate W at this stage, but be careful not to label the same distribution twice!" ,"distrsmall.gif" ) CI["distribution energy, determining for small systems"]=new Array("_dss2") CI["probability, of energy distributions"]=new Array("_dss2") addScript(2,"Distributions of Energy","_d33_2,_dss2") CI["_b2"]=new Array("The Boltzmann Distribution","boltzmanndist.gif") CI["Boltzmann distribution"]=new Array("_b2") CI["_bdp2"]=new Array("The Boltzmann law relates the ratio of particles in any two levels to the difference in energy between the levels. The ratio will depend upon the temperature. This equation predicts the most probable distribution for a system when there are many particles and many units of energy.","boltzmannlaw.gif") CI["2-11:Boltzmann distribution, as most probable"]=new Array("_bdp2") CI["2-11:distribution most probable, Boltzmann"]=new Array("_bdp2") CI["Boltzmann law, defined"]=new Array("_bdp2") CI["_bk2"]=new Array("","boltzmannk.gif") CI["Boltzmann's constant (k)"]=new Array("_bk2") CI["_bd2"]=new Array("When two levels have the same energy, their populations are equivalent. Each population is independently calculated using the Boltzmann law.","boltzdegeneracy.gif") CI["Boltzmann distribution, and degeneracy"]=new Array("_bd2") CI["_tb2"]=new Array("At low temperature, the ratio of n1/n0 approaches 0, while at high temperature it approaches 1.","tempeffectboltz.gif") CI["temperature (T), and Boltzmann distribution"]=new Array("_tb2") CI["_bmp2"]=new Array("It can be shown without too much difficulty that if you start with a Boltzmann distribution and then make even the smallest possible change, the result is a less probable distribution.","boltzmannprob.gif") CI["2-18:Boltzmann distribution, as most probable"]=new Array("_bmp2") CI["_bkr2"]=new Array("Boltzmann's constant turns out to be the ideal gas constant, R, divided by Avogadro's number. Using R instead, we can talk about energy in units of energy/mole. For this we use a capital E.","boltzmannkr.gif") CI["Boltzmann's constant (k), and ideal gas constant (R)"]=new Array("_bkr2") addScript(2,"The Boltzmann Distribution","_b2,_bdp2,_bk2,_bd2,_tb2,_bmp2,_bkr2") CI["_bg2"]=new Array(" ","boltzmanngeom.gif") CI["Boltzmann distribution, as a geometric progression"]=new Array("_bg2") CI["_tefe2"]=new Array("","tempeffecteven.gif") CI["temperature (T), effect in evenly spaced systems"]=new Array("_tefe2") CI["_sb2"]=new Array(" ","spacingeffect.gif") CI["energy level separation, and Boltzmann distribution"]=new Array("_sb2") CI["_bpl2"]=new Array("For evenly spaced systems only, the population of the lowest level (no) can be calculated as follows, where n is the total number of particles:","boltzmannlowest.gif") CI["Boltzmann distribution, population of the lowest level"]=new Array("_bpl2") CI["_bnp2"]=new Array("For evenly spaced systems only, the highest level containing particles (j) can be approximated as the follows, where n is the total number of particles:","boltzmanntoplevel.gif") CI["Boltzmann distribution, number of populated levels"]=new Array("_bnp2") addScript(2,"Evenly Spaced Systems","_bg2,_tefe2,_sb2,_bpl2,_bnp2") CI["_besc2"]=new Array("|If you have an equation-solving calculator, you will find it very easy to do this calculation. You can very quickly change the value of one variable and see what the result is for any of " +"the other variables in the equation. No algebra!" , "boltzmanncalc.gif" ,"|This really saves time doing homework! If you want to experiment with an on-line equation-solving calculator, click here." , "boltzmanncalc2.gif" ) CI["Boltzmann law, equation-solving calculator"]=new Array("_besc2") addScript(2,"Using an Equation-Solving Calculator","_besc2") CI["_bf2"]=new Array("The Boltzmann distribution is simply the most probable one. There will still be fluctuations. However, the fluctuations are very, very small when " +"any significant number of particles are involved. Consider two distributions, A and B, where A is the Boltzmann distribution, and B is one that differs from A " +"by just a bit. If the populations (n0, ni, etc.) in B differ from A by p percent on average, " +"the relative probability of B vs. A, WB/WA is:|This is very small, even for a tiny fraction of a percent when n is on the order of a mole." ,"boltzfluc.gif") CI["fluctuation, and Boltzmann distribution"]=new Array("_bf2") addScript(2,"Fluctuations and Boltzmann","_bf2") //_ch03 CI["_exe3"]=new Array("","electexcite.gif") CI["excitation, electronic"]=new Array("_exe3") CI["electronic excitation"]=new Array("_exe3") CI["_exv3"]=new Array("","vibexcite.gif") CI["excitation, vibrational"]=new Array("_exv3") CI["vibrational excitation"]=new Array("_exv3") CI["_exr3"]=new Array("","rotexcite.gif") CI["excitation, rotational"]=new Array("_exr3") CI["rotational excitation"]=new Array("_exr3") CI["_ext3"]=new Array("","transexcite.gif") CI["excitation, translational"]=new Array("_ext3") CI["translational excitation"]=new Array("_ext3") addScript(3,"The Four Types of Excitation","_exe3,_exv3,_exr3,_ext3") CI["_pl3"]=new Array("Planck's law and Einstein's famous equation are closely related and allow the connection between mass and wavelength used by de Broglie in 1924. Planck's constant, h, 6.6260755x10−34 Joule-seconds, serves as a conversion factor between frequency (1/sec) and energy (Joules).","planck.gif") CI["Planck's law"]=new Array("_pl3") CI["de Broglie, Louis"]=new Array("_pl3") CI["_mw3"]=new Array("The relationship among mass, wavelength, frequency, and energy for a photon was established by Einstein and Planck well before de Broglie set out to do the same for the electron. It is not easy to see the difference between \"NU\" (frequency) and \"VEE\" (velocity) in these equations.","masswave.gif") CI["mass, and energy"]=new Array("_mw3") CI["frequency, and energy"]=new Array("_mw3") CI["wavelength, and energy"]=new Array("_mw3") CI["mass, and wavelength"]=new Array("_mw3") CI["_pbox3"]=new Array("de Broglie (pronounced \"de BROY-lee\") developed the particle-in-a-box approximation to deal with electrons, but its form is the basis for all of the forms of energy discussed here. The approximation used by de Broglie involves constraints. In particular, the function must have zero amplitude at the walls. This restriction leads to the use of a quantum number--an integer that describes the allowed states.","pinab.gif") CI["particle in a box"]=new Array("_pbox3") CI["quantum number, defined"]=new Array("_pbox3") CI["constraints, leading to quantum numbers"]=new Array("_pbox3") CI["de Broglie, and probability"]=new Array("_pbox3") addScript(3,"Planck, Einstein, and De Broglie","_pl3,_mw3,_pbox3") CI["_ee3"]=new Array("The important thing to see in this equation is that there are both a very small mass (the mass of the electron) and a very small distance (the Bohr radius) in the denominator of this equation. That causes the electronic energy levels of the hydrogen atom to be relatively far apart in energy. Electronic excitation thus typically involves high-energy visible or ultraviolet light.","elec.gif") CI["energy, electronic"]=new Array("_ee3") CI["electronic energy"]=new Array("_ee3") CI["Bohr radius"]=new Array("_ee3") CI["hydrogen atom"]=new Array("_ee3") CI["electronic energy, hydrogen atom"]=new Array("_ee3") CI["_hs3"]=new Array("Each line in the hydrogen emission spectrum corresponds to a difference in energy between two electronic states of the hydrogen atom. Only the Balmer series is visible.","hspec.gif") CI["emission spectrum, of hydrogen atom"]=new Array("_hs3") CI["hydrogen atom, emission spectrum"]=new Array("_hs3") CI["Balmer series"]=new Array("_hs3") CI["Paschen series"]=new Array("_hs3") CI["Lyman series"]=new Array("_hs3") CI["_sch3"]=new Array("Schrödinger's \"orbitals\" can be seen as descriptions of the relative probability of finding the electron at various coordinates around the nucleus. While the equation itself is not terribly hard to write, the solution to the equation is a bit intimidating at first. " +"Nonetheless, hiding in the factorials of that solution are all the quantum rules we associate with the quantum numbers n, l, and m.","elecprob.gif") CI["Schrödinger, Erwin"]=new Array("_sch3") CI["quantum rules"]=new Array("_sch3") addScript(3,"Electronic Energy and the Hydrogen Atom","_ee3,_hs3,_sch3") CI["_ves3"]=new Array("The \"size\" of the box in vibration amounts to the small displacement that the atoms undergo when vibrations occur. These displacement, which amount to a \"shivering\" of the molecule, are dependent upon the strength of the bond (the \"force constant\", kf) and the reduced mass, MU_. They are on the order of one tenth the length of a typical bond for low energy vibrations, but get larger as the energy increases until, ultimately, the bond breaks.","vibsize.gif") CI["vibrational energy, and size"]=new Array("_ves3") CI["_mu3"]=new Array("Reduced mass, MU_, takes into account that fact that it is the less massive particle that does most of the moving when there are two particles involved.","umass.gif") CI["mass, reduced"]=new Array("_mu3") CI["_v3"]=new Array("The important thing to note about this equation is its similarity to the others we find in this chapter. Namely, there is an integer quantum number, a constant including Planck's constant, and a fraction with reduced mass and a distance terms squared in the denominator.","vib1.gif") CI["vibrational energy"]=new Array("_v3") CI["vibrational energy, and reduced mass"]=new Array("_v3") CI["_vib3"]=new Array("While this equation for vibrational energy is more common, it tends to hide the similarity between vibration and the other forms of energy. " +"The vibrational frequency, NU_, is defined such that the difference in energy between two adjacent vibrational energy levels is hNU_.","vib.gif") CI["vibrational energy, alternative formula"]=new Array("_vib3") CI["vibrational energy, and frequency"]=new Array("_vib3") CI["_vhcl3"]=new Array("With care, one can calculate the Boltzmann ratio n1/n0 for HCl vibration at various temperatures. We start with a calculation of the reduced mass. Along with knowledge of the force constant, kf (481 kg/s2), " +"one can calculate the vibrational frequency, NU_. From that, we get the energies of the first two vibrational energy levels, and from THOSE, we get the desired ratio. This calculation shows that just about all HCl molecules are in the ground vibrational state at 298 Kelvin (room temperature)." ,"vibhcl.gif") CI["hydrogen chloride, vibrational energy"]=new Array("_vhcl3") CI["_vo3"]=new Array("The equations we use for vibrational energy are only approximations that consider vibrational energy levels to be evenly spaced. A more realistic picture involves an \"anharmonic oscillator\" for which energy levels get closer and closer together as energy increases, and the bond ultimately breaks.","anharm.gif") CI["vibrational energy, and anharmonic oscillator"]=new Array("_vo3") CI["_vec3"]=new Array("","vibrcomp.gif") CI["vibrational energy, in H2, HCl, and Cl2"]=new Array("_vec3") addScript(3,"Vibrational Energy","_ves3,_mu3,_v3,_vib3,_vhcl3,_vo3,_vec3") CI["_re3"]=new Array("The \"size\" of the box in rotation is on the order of the size of the molecule, which for diatomics is the bond distance. This results in rotational energy levels being much more closely spaced than vibrational energy levels. Thus, larger molecules will have more closely-spaced rotational energy levels.","rotsize.gif") CI["rotational energy, and size"]=new Array("_re3") CI["_rot3"]=new Array("The rotational energy equation, like the one for vibration, involves reduced mass. Again the mass term is in the denominator along with a distance-squared term. R here is about 10 times the magnitude of d in the vibrational energy equation, leading to considerably closer spaced energy levels in rotation relative to vibration.","rot.gif") CI["rotational energy"]=new Array("_rot3") CI["rotational energy, and reduced mass"]=new Array("_rot3") CI["_rhcl3"]=new Array("With care, one can calculate the Boltzmann ratio n1/n0 for HCl rotation at various temperatures. We start again with a calculation of the reduced mass. " +"We calculate the energies of the first two rotational energy levels using i=0 and i=1, then use those values to get the desired ratio. This calculation shows that a substantial number of HCl molecules are rotating at 298 Kelvin (room temperature). In fact, several hundred rotational levels are populated." ,"rothcl.gif") CI["hydrogen chloride, rotational energy"]=new Array("_rhcl3") CI["_rcl2_3"]=new Array("","rotcl2.gif") CI["rotational energy, chlorine"]=new Array("_rcl2_3") addScript(3,"Rotational Energy","_re3,_rot3,_rhcl3,_rcl2_3") CI["_tran3"]=new Array("The key point regarding translation is that the levels are incredibly closely spaced. Primarily this is due to the very large distance-squared term in the denominator. d here is the distance from wall to wall across the container, which is typically millions of times larger than a bond length. The calculation here is for HCl. We will see that, though closely spaced, translational energy levels are critical in explaining pressure effects in chemical reactions.","trans.gif") CI["translational energy"]=new Array("_tran3") CI["translational energy, hydrogen chloride"]=new Array("_tran3") addScript(3,"Translational Energy","_tran3") CI["_ema3"]=new Array("Molecules can absorb a wide range of energies due to the many electronic, vibrational, rotational, and translational energy levels available. In many cases an electronic absorption is accompanied by vibrational and/or rotational excitation as well.","evib.gif") CI["molecule, and energy absorption"]=new Array("_ema3") CI["energy, absorption by molecules"]=new Array("_ema3") CI["_col3"]=new Array("","color.gif") CI["color, from electronic excitation"]=new Array("_col3") CI["_fl3"]=new Array("","fluorescence.gif") CI["fluorescence"]=new Array("_fl3") CI["_ph3"]=new Array("In phosphorescence, the molecule undergoes \"inter-system crossing\" to a " +"different excited state due to an overlap of energy levels. This excited state generally takes the " +"form of a \"triplet\" in which two electrons are unpaired. The fact that the electrons are unpaired " +"makes it more difficult for the molecule to drop back to the ground state, where all electrons must be paired. " +"The emission can be delayed for many seconds--in some cases hours. The slow trickle of molecules down to the " +"ground state releases a slow, steady stream of photons creating the \"glow-in-the-dark\" effect.","phosphor.gif xpos=right") CI["phosphorescence"]=new Array("_ph3") CI["_las3"]=new Array("","laser.gif") CI["laser, and stimulated emission"]=new Array("_las3") CI["stimulated emission, in lasers"]=new Array("_las3") addScript(3,"Molecular Excitation and Spectroscopy","_ema3,_col3,_fl3,_ph3,_las3") CI["_vr3"]=new Array("Molecular vibration is at the heart of most chemical reactions.","vibh2react.gif") CI["chemical reactions, and vibrational energy"]=new Array("_vr3") CI["hydrogen, dissociation reaction"]=new Array("_vr3") CI["_eeq3"]=new Array("Chemical reactions can be seen as changes in the ground-state energy and energy-level spacings for a system. In this case it should not be hard to imagine that increasing the temperature drives the reaction forward, to dissociated H atoms.","levh2react.gif") CI["energy levels, and equilibrium"]=new Array("_eeq3") CI["_gse3"]=new Array("The lower ground-state energy in the product of the combustion of H2 and O2 suggests that at low temperature equilibrium will favor the product. As the temperature is raised, the equilibrium should shift toward the left in this case.","levh2o2react.gif") CI["ground state energy, temperature effect"]=new Array("_gse3") CI["temperature (T), effect of ground state energy"]=new Array("_gse3") addScript(3,"Chemical Reactions and Vibrational Energy","_vr3,_eeq3,_gse3") CI["_con3"]=new Array("All of the types of energy level systems have very similar mass and distance constraints. In each case, though, the mass and distance terms are significantly different, " +"leading to radically different energy level spacings. This in turn leads to significant temperature effects in reactions.","constraint.gif") CI["constraints, effect on energy level separation"]=new Array("_con3") CI["energy levels, effect of mass"]=new Array("_con3") CI["energy levels, effect of size"]=new Array("_con3") CI["mass, effect on energy level separation"]=new Array("_con3") CI["size, effect on energy level separation"]=new Array("_con3") addScript(3,"General Ideas","_con3") //_ch04 CI["_sf4"]=new Array("Certain properties of a system that are intrinsic to that system as it exists in its current state and not based on how it got there." +" Clearly, the energy levels themselves in a system and the number of particles in each level in the most probable distribution are of this sort. Other such properties include the temperature, volume, and" +" pressure of a system, and the masses or concentrations of all of a mixture’s components." +" It is these properties, also called state functions or state variables, that together define" +" the state of a system. If any of these properties change, then we say that the system has" +" changed its state. ","") CI["state function"]=new Array("_sf4") CI["function, state"]=new Array("_sf4") CI["_u4"]=new Array("Internal energy is defined as the sum of all the energies of all the particles in all the levels of a system. The change in internal energy of a system is related to the heat added and work applied to the system by the First Law of thermodynamics.","u.gif") CI["internal energy (U)"]=new Array("_u4") CI["First Law"]=new Array("_u4") CI["thermodynamics, First Law"]=new Array("_u4") addScript(4,"State Functions and Internal Energy","_sf4,_u4") CI["_w4"]=new Array("Particularly in gases, compression (decrease in volume) leads to more widely spaced translational energy levels. If the compression is carried out quickly enough, then the particles do not have enough time to exchange energy, and the net effect is that the populations remain the same, but the energy increases. This is the microscopic action of applying work to the system. In this way a system can be \"heated\" without adding any \"heat.\"|" +"(Check it out, using your browser or wintropy.exe" ,"w.gif") CI["work (w), microscopic"]=new Array("_w4") CI["work (w), and compression"]=new Array("_w4") CI["compression, as work applied to a system"]=new Array("_w4") CI["heating, without adding heat"]=new Array("_w4") CI["_wc4"]=new Array("Expansion of gases (increase in volume) leads to more closely spaced translational energy levels. If the expansion is carried out quickly enough, then the particles do not have enough time to exchange energy, and the net effect is that the populations remain the same, but the energy decreases. This is the microscopic action of removing work from the system.|" +"(Check it out, using your browser or wintropy.exe" ,"w2.gif") CI["work (w), and expansion"]=new Array("_wc4") CI["expansion, as work removed from a system"]=new Array("_wc4") addScript(4,"Microscopic Work","_w4,_wc4") CI["_q4"]=new Array("Adding heat (q) to a system involves moving particles up the energy ladder.|" +"(Check it out, using your browser or wintropy.exe" ,"q.gif") CI["heat (q), microscopic"]=new Array("_q4") CI["_qt4"]=new Array("The heat required to raise the temperature of a system is related to the change in temperature (T) by the heat capacity (C)." ,"qc.gif") CI["heat (q), macroscopic"]=new Array("_qt4") CI["_sph4"]=new Array("Specific heat is defined as the amount of energy required to raise one gram of a substance one degree Kelvin. " +"The specific heat of water is approximately 1 calorie/gram-Kelvin or 4.1868 Joules/gram-Kelvin. (Some standards use 4.184 Joules/gram-Kelvin for the specific heat of water.)

Specific heat is used in calculations calling for heat capacity (C). The heat capacity of a substance, then, is its specific heat times its mass. " +" This is analogous to how heat capacity is molar heat capacity times the number of moles of the substance being heated. " +"

Since a one-degree rise in Kelvin is the same as a one-degree rise in Celcius, either Celcius or Kelvin may be used in calculations involving specific heat or heat capacity." +"

For example, if we wish to know how much energy will be required to raise 25 grams of water 10 oC (10 Kelvin), we calculate:","specificheat.gif") CI["heat (q), specific"]=new Array("_sph4") CI["specific heat"]=new Array("_sph4") CI["_c4"]=new Array("Heat capacity (C) relates a change in temperature to the amount of heat (q) required to effect that change. Systems with more particles or more closely-spaced energy levels (gases, for example) will have higher heat capacities. The heat capacity of 25 grams of liquid water is 25 calories/Kelvin. Thus, to raise the temperature of liquid water from 15 oC to 25 oC (a change of 10 oC, or 10 Kelvin) requires 250 calories (1050 Joules) of heat.|" +"(Check it out, using your browser or wintropy.exe -- Compare two systems with different numbers of particles or different energy level spacings. What is the effect of these changes on the energy required to heat the system 100 K?)" ,"c.gif") CI["heat capacity (C)"]=new Array("_c4") addScript(4,"Heat and Heat Capacity","_q4,_qt4,_sph4,_c4") CI["_scheme4"]=new Array("Heat released by the reaction is transferred to the surroundings. The surroundings absorb the heat and increase in temperature. The heat capacity must account for all of the material in the calorimeter, including the calorimeter device itself.","calschem.gif") CI["calorimeter, schematic of"]=new Array("_scheme4") CI["_qsur4"]=new Array("In the case of an exothermic reaction, while q < 0, qsur is > 0.","heatsurfromfig7_6.gif") CI["4-11:heat (q), and surroundings"]=new Array("_qsur4") CI["_cal4"]=new Array("Consider a reaction carried out in a calorimeter containing 50 mL of water. The temperature is found to be 1.5 oC. The heat capacity of the calorimeter device (not counting the water), Ccal, is known to be 14 Joules/Kelvin. What is the change in internal energy for the system? We start by ignoring work","heatcal.gif") CI["4-12:calorimeter, and internal energy (U)"]=new Array("_cal4") CI["_calb4"]=new Array("Changes in internal energy can be determined using a bomb calorimeter. Since there is no volume change, w=0, and DELTA_U = q. Thus, in a bomb calorimeter, a measure of the heat given off by the reaction is a measure of DELTA_U.","bombcal.gif") CI["calorimeter, bomb"]=new Array("_calb4") CI["_cals4"]=new Array("A simple calorimeter can be constructed from two nested styrofoam cups and a styrofoam lid that has a hole in it to accomodate a thermometer. This simple device is not terribly precise, and it suffers from the fact that if a gas is produced, the work energy lost in the expansion of the gas is not accounted for. " +" Nonetheless, it is amazingly accurate if used carefully." ,"simplecal.gif") CI["calorimeter, simple"]=new Array("_cals4") CI["_rxn4"]=new Array("","eqnrxn.gif") CI["chemical equations, vs. chemical reactions"]=new Array("_rxn4") addScript(4,"Calorimeters","_scheme4,_qsur4,_cal4,_calb4,_cals4,_rxn4") CI["_wm4"]=new Array("The sort of work (w) that we consider here is pressure-volume work. We only consider situations where the external pressure is constant, but the volume changes. (If the pressure changes too, we need to use calculus.) In an expansion, when the volume increases, DELTA_V is positive. The minus sign is necessary because expansion must decrease the energy of the system.","workpv.gif") CI["work (w), and pressure"]=new Array("_wm4") CI["work (w), and volume"]=new Array("_wm4") CI["work (w), macroscopic"]=new Array("_wm4") CI["_ig4"]=new Array("In most chemical reactions work can be ignored because the magnitude of the heat that goes into or out of the reaction is much greater.","ignorework.gif") CI["chemical reactions, ignoring work"]=new Array("_ig4") CI["work (w), ignored in chemical reactions"]=new Array("_ig4") CI["water, vaporization"]=new Array("_ig4") CI["hydrogen, reaction with oxygen"]=new Array("_ig4") addScript(4,"Why Chemists Ignore Their Work","_wm4,_ig4") CI["_eng4"]=new Array("An engine is a device that turns heat into work.","engine.gif") CI["engine, and work (w)"]=new Array("_eng4") CI["work (w), in engines"]=new Array("_eng4") CI["_w!pv4"]=new Array("There are several forms of work besides the pressure-volume work we are dealing with here. Each involves some sort of force and some sort of change.","workother.gif") CI["work (w), non-PV"]=new Array("_w!pv4") CI["_bio4"]=new Array('All work involves movement. So why is holding a heavy book out at arm\'s length for even just 10 minutes hard work?' +' The problem is simply that our muscles are not designed to stay rigid. (If they were, we\'d be clams.)' +'

There are two kinds of muscles in your body: striated and smooth.' +' Striated muscle is what you have in your arms. Smooth muscle, on the other hand, is what' +' your intestines and blood vessels are made of (and what clams have). Smooth muscles are' +' made to hold a position for a long time and move very slowly.

The problem is, when you' +' hold your arm in the air, you are using your striated muscle, not your smooth muscle.' +' Your striated muscle acts by contracting every time a nerve pulse comes along and then' +' immediately relaxing again (so that it can move again to some other position very quickly).' +'

Thus, holding that book out at arm’s length requires innumerable "firing" of the muscle,' +' which requires work to be expended, and your muscles really are moving.

A clam, on the' +' other hand, using its smooth muscle, can hold its shell open indefinitely, expending no work' +' at all.' ,"") CI["4-14:work (w), biological"]=new Array("_bio4") CI["4-14:biological work"]=new Array("_bio4") addScript(4,"Engines and Non-PV Work","_eng4,_w!pv4,_bio4") //_ch05 CI["_hl5"]=new Array("Hess's law states:

","hessslaw.gif") CI["Hess's law"]=new Array("_hl5") CI["Hess's law, vector graph"]=new Array("_hl5") CI["_hlu5"]=new Array("","hessslawu.gif") CI["internal energy (U), and isolated atoms"]=new Array("_hlu5") CI["bonding, and internal energy (U)"]=new Array("_hlu5") CI[""]=new Array("_hlu5") CI[""]=new Array("_hlu5") CI[""]=new Array("_hlu5") CI["_hluc5"]=new Array("","hessslawuc.gif") CI["internal energy (U), calculating DELTA_U using Hess's law"]=new Array("_hluc5") CI["Hess's law, calculating DELTA_U"]=new Array("_hluc5") addScript(5,"Hess's Law","_hl5,_hlu5,_hluc5") CI["_bde5"]=new Array("","tablebde.gif") CI["bond dissociation energy, table of"]=new Array("_bde5") CI["_bdeu5"]=new Array("","bdeu.gif") CI["bond dissociation energy, and internal energy (U)"]=new Array("_bdeu5") addScript(5,"Bond Dissociation Energy","_bde5,_bdeu5") CI["_bdu5"]=new Array("N2 has a very strong triple bond, which is hard to break. If the reaction between N2 and O2 had a negative DELTA_U, the Earth's atmosphere almost certainly would be unstable. If N2 were a little less stable or O2 were a little more reactive, there would be no life on Earth!","bdeun2o2.gif") CI["chemical reactions, and bond dissociation energy"]=new Array("_bdu5") CI["spontaneity, and life on Earth"]=new Array("_bdu5") CI["5-9:atmosphere, combustion of"]=new Array("_bdu5") CI["_phos5"]=new Array("|","highphos.gif") CI["high-energy phosphate bond"]=new Array("_phos5") CI["phosphate bond, high-energy"]=new Array("_phos5") CI["_acid5"]=new Array("","aciddis.gif") CI["acid, dissociation"]=new Array("_acid5") addScript(5,"Using Bond Dissociation Energies","_bdu5,_phos5,_acid5") CI["_ueq5"]=new Array("","equ5.gif") CI["internal energy (U), and equilibrium"]=new Array("_ueq5") addScript(5,"Internal Energy and Equilibrium","_ueq5") //_ch06 CI["_iso6"]=new Array("In an isomerization reaction, a single molecule rearranges to give another with the same molecular formula.","isomer.gif") CI["chemical reactions, isomerization"]=new Array("_iso6") CI["isomerization, reactions"]=new Array("_iso6") CI["_te6"]=new Array("An equilibrium mixture of reactants and products can be seen as two independent Boltzmann distributions.","ab6.gif") CI["temperature (T), effect on isomerization"]=new Array("_te6") CI["isomerization, temperature effect"]=new Array("_te6") CI["6-2:temperature (T), effect on equilibrium constant (K)"]=new Array("_te6") CI["6-2:equilibrium constant (K), effect of temperature"]=new Array("_te6") addScript(6,"Chemical Reactions and Equilibria","_iso6,_te6") CI["_ani6"]=new Array("This simulation involves 50,000,000 particles.|The ground state of A is just a little below the ground state of B, but the energy level spacings in B are about twice as closely spaced as those of A. Click the next radio button to begin." ,"../js/animate/kab100.gif" ,"T = 100 Kelvin|K = [B] / [A] = 0.027/0.973 = 0.028" ,"../js/animate/kab100.gif" ,"T = 200 Kelvin|K = [B] / [A] = 0.160/0.840 = 0.190" ,"../js/animate/kab200.gif" ,"T = 300 Kelvin|K = [B] / [A] = 0.280/0.720 = 0.390" ,"../js/animate/kab300.gif" ,"T = 400 Kelvin|K = [B] / [A] = 0.362/0.638 = 0.567" ,"../js/animate/kab400.gif" ,"T = 500 Kelvin|K = [B] / [A] = 0.418/0.582 = 0.718" ,"../js/animate/kab500.gif" ,"T = 600 Kelvin|K = [B] / [A] = 0.458/0.542 = 0.845" ,"../js/animate/kab600.gif" ,"T = 700 Kelvin|K = [B] / [A] = 0.488/0.512 = 0.953" ,"../js/animate/kab700.gif" ,"T = 800 Kelvin|K = [B] / [A] = 0.510/0.490 = 1.04" ,"../js/animate/kab800.gif" ,"T = 900 Kelvin|K = [B] / [A] = 0.527/0.473 = 1.11" ,"../js/animate/kab900.gif" ,"T = 1000 Kelvin|K = [B] / [A] = 0.541/0.459 = 1.18" ,"../js/animate/kab1000.gif" ,"T = 10000 Kelvin|K = [B] / [A] = 0.655/0.345 = 1.89" ,"../js/animate/kab10000.gif" ) CI["6-5:equilibrium constant (K), effect of temperature"]=new Array("_ani6") CI["6-5:temperature (T), effect on equilibrium constant (K)"]=new Array("_ani6") CI["_els6"]=new Array("","els6.gif") CI["equilibrium constant (K), high-temperature limit"]=new Array("_els6") addScript(6,"The Effect of Temperature on Equilibrium Constants","_ani6,_els6") //_ch07 CI["_1st7"]=new Array("The idea that energy is conserved (the First Law of Thermodynamics) derives from the fact that all the energy going into a system is coming from the surroundings. Overall, the energy change of the \"universe\" is constant. |The First Law of Thermodyamics cannot possibly \"explain\" why heat flows or chemical reactions occur.","energycon.gif") CI["7-1:energy, conservation"]=new Array("_1st7") addScript(7,"The First Law of Thermodynamics","_1st7") CI["_s07"]=new Array("Entropy (S) is defined as Boltzmann's constant times the natural logarithm of thermodynamic probability (W). Thus, an increase in entropy is associated with changes in the direction of the most probable distribution.","entropy.gif") CI["entropy (S)"]=new Array("_s07") CI["entropy (S), defined"]=new Array("_s07") CI["7-3:entropy (S), and thermodynamic probability (W)"]=new Array("_s07") CI["_ds7"]=new Array("Differences in entropy, then are logarithms of relative probability. This means that positive changes in entropy are associated with processes that lead to more probable distributions.|Note that the expansion involves no change in entropy because the populations all stay the same.","deltas.gif") CI["entropy (S), changes"]=new Array("_ds7") CI["7-4:entropy (S), and thermodynamic probability (W)"]=new Array("_ds7") addScript(7,"The Definition of Entropy","_s07,_ds7") CI["_2nd7"]=new Array("The Second Law of thermodynamics states that in any real process the entropy of the universe (which includes system and surroundings) must increase.","2ndlaw.gif") CI["Second Law"]=new Array("_2nd7") CI["Second Law, and entropy of the universe"]=new Array("_2nd7") CI["universe, entropy of"]=new Array("_2nd7") CI["entropy (S), of the universe"]=new Array("_2nd7") CI["thermodynamics, Second Law"]=new Array("_2nd7") addScript(7,"The Second Law of Thermodynamics","_2nd7") CI["_hs7"]=new Array("Two changes occur here: first a compression, then the removal of heat. The overall entropy change is calculated from either SC − SA or (SB − SA) + (SC − SB). For all processes, the overall entropy change can be calculated as Sproducts − Sreactants. In this case, the overall entropy change is −2.35x10−22 Joules/Kelvin.","hessands.gif") CI["7-4:Hess's law, and entropy change"]=new Array("_hs7") CI["7-4:Hess's law, and entropy (S) change"]=new Array("_hs7") CI["_ss7"]=new Array("The amount of heat going into the surroundings, qsur, divided by the temperature of the surroundings turns out to be the entropy change of the surroundings. (This is true for any situation where there is only \"simple heating.\")","ssur.gif") CI["7-6:entropy (S), and surroundings"]=new Array("_ss7") CI["7-6:heat (q), and surroundings"]=new Array("_ss7") CI["_wm7"]=new Array("It is not difficult to calculate the entropy change in the surroundings for the melting of 10 g of ice at 0 oC. " +"Melting requires energy, and the amount of energy required to melt a mole of solid H2O (ice) is 6010 J. " +"When 10.0 g of ice melts at 0 oC (273 K), 3340 Joules of heat energy is required. " +"That means that qsur = −3340 Joules. " +"From this we calculate that the entropy change in the surroundings required to get this job done " +"is −12.2 Joules/Kelvin. In order to melt ice at any temperature, the entropy of the surroundings must decrease." ,"watermelts.gif") CI["7-8:water, melting"]=new Array("_wm7") CI["7-8:entropy (S), and melting"]=new Array("_wm7") addScript(7,"Entropy Changes and the Surroundings","_hs7,_ss7,_wm7") CI["_st7"]=new Array("","sandt.gif") CI["entropy (S), and temperature (T)"]=new Array("_st7") CI["_sms7"]=new Array("

The standard molar entropy (So) for a substance is defined as the entropy of

","") CI["entropy (S), standard molar"]=new Array("_sms7") CI["_sc7"]=new Array("Assuming electronic, vibrational, rotational, and translational excitation are relatively independent ways of increasing the energy of a system, we can multiply their ways to get the overall thermodynamic probability (W) of a system. Since ln(xy) = ln (x) + ln(y), we can then think" +" of the total entropy of a system as being a sum of entropy terms. This means that what we already know about energy level separations and thermodynamic probability can be applied to understanding entropy.|Remember, at a given temperature, systems with more closely spaced energy levels will have larger W and thus also more entropy.","sadds.gif") CI["7-9:entropy (S), comparisons"]=new Array("_sc7") CI["_smt7"]=new Array("","tables.gif") CI["entropy (S) standard molar, table of"]=new Array("_smt7") addScript(7,"Standard Molar Entropy","_st7,_sms7,_smt7") CI["_scg7"]=new Array("Several generalizations regarding entropy can be made based a table of standard molar entropies. You should be able to rationalize all these trends based on what you know about electronic, vibrational, rotational, and translational energy level spacings.

"+ "

","") CI["7-11:entropy (S), comparisons"]=new Array("_scg7") CI["_os7"]=new Array("The standard molar entropy of O2 gas is unusually high. This is because of the special electronic nature of O2, which has two unpaired electrons. This leads to a 3-fold degeneracy in the ground electronic state of O2 and adds about 9 J/K to its molar entropy.","o2s.gif") CI["entropy (S), and degeneracy"]=new Array("_os7") CI["entropy (S), oxygen"]=new Array("_os7") CI["7-12:entropy (S), comparisons"]=new Array("_os7") CI["_stir7"]=new Array("Stirling developed several approximations for calculating the natural logarithms of large numbers. The one that is useful for us is shown here. It is useful for calculating the entropy due to the degeneracy of the ground electronic state in O2.","stirling.gif") CI["Stirling's approximation"]=new Array("_stir7") addScript(7,"Making Sense of Standard Molar Entropies","_smt7,_sc7,_scg7,_os7,_stir7") CI["_dds7"]=new Array("To calculate the standard change in entropy at 298 Kelvin for a chemical reaction, we use Sproducts − Sreactants being careful to include the proper number of moles of reactants and products in the calculation.","no2ds.gif") CI["entropy (S) change, determining for chemical reactions"]=new Array("_dds7") CI["chemical reactions, determining DELTA_So"]=new Array("_dds7") CI["entropy (S) standard, of reaction, determining from tabulated values"]=new Array("_dds7") CI["_h2om7"]=new Array("Consider melting a mole of ice. This is going to require heat from the surroundings. The entropy of the system (ice/water) will increase, but the entropy of the surroundings (which is delivering heat) will decrease. (Particles in the surroundings will drop down to lower energy levels.) " +"A comparison of the entropy change of the system and the entropy change of the surroundings suggests that right at 273 Kelvin (0 oC, the \"melting point\" of water) those two quantities are equal. Below that temperature, the loss of entropy in the surroundings" +" would outweigh the entropy gain in the system. Ice will not melt below its melting point, because that would violate the Second Law of thermodynamics!","h2omelts.gif") CI["melting point, water"]=new Array("_h2om7") CI["melting, water"]=new Array("_h2om7") CI["spontaneity, and entropy (S)"]=new Array("_h2om7") CI["7-17:entropy (S), and melting"]=new Array("_h2om7") CI["_spo7"]=new Array("Heat going into the system by definition comes from the surroundings. The amount of change in entropy in the system depends upon what is happening there," +" but the entropy change in the surroundings can be calculated simply from qsur/T. |For melting, which is endothermic, qsur < 0." +" That means the entropy of the surroundings will decrease. Melting will only occur if the the increase in entropy of the system is greater than the decrease in entropy of the surroundings. For water, that is the case for all temperatures greater than 0oC. ","h2omelts2.gif") CI["spontaneity, and entropy (S)"]=new Array("_spo7") CI["7-16:surroundings, and entropy (S)"]=new Array("_spo7") CI["entropy (S), and heat input"]=new Array("_spo7") CI["7-16:entropy (S), and melting"]=new Array("_spo7") addScript(7,"Entropy Changes and Spontaneity","_dds7,_h2om7,_spo7") //ch08 CI["_imp8"]=new Array("

Imagine you’ve graduated and have a fine job and a house in the " +"suburbs with a swimming pool in the back yard. You’re enjoying a nice summer’s day at the edge of your pool. " +"

Suddenly all the water jumps out and flies over your fence into your neighbor’s yard." +"

Impossible? Is it?

Or is it just improbable?

You sure?

","") CI["spontaneity, and impossibility"]=new Array("_imp8") CI["impossible, vs. improbable"]=new Array("_imp8") addScript(8,"Impossible vs. Improbable","_imp8") CI["_sv8"]=new Array("|When the stopcock is opened, there is an overwhelming probability that some of the xenon atoms will leave Flask A and enter Flask B. " +"Einstein's argument for the increase in entropy is based on solely on basic ideas of probability. " +"\"n\" here is the number of moles. Then nNavk = nR, where R is the ideal gas constant. " +"V2 is the final volume (VA + VB); " +"V1 is the initial volume, VA. Read on to see an sample calculation." ,"voldeltas.gif" ,"|Flask A is 2.0 L; Flask B is 3.0 L. Initially there is 0.50 mol of xenon gas in flask A. What is the entropy change when the valve is opened?" ,"voldeltac1.gif" ,"|Flask A is 2.0 L; Flask B is 3.0 L. Note that the final volume, V2, is 5.0 L, because the xenon ends up in both flasks." ,"voldeltac2.gif" ) CI["entropy (S), and volume"]=new Array("_sv8") CI["_mix8"]=new Array("The entropy of \"mixing\" is simply the sum of the two entropy changes. If the pressure is the same in both flasks, then the change in entropy can be written in terms of mole fraction, \"CHI_\". ","voldeltasab.gif") CI["entropy (S), and mixing"]=new Array("_mix8") CI["expansion, and mixing"]=new Array("_mix8") addScript(8,"Entropy and Expansion","_sv8,_mix8") CI["_gas8"]=new Array("The critical aspect of the ideal-gas equation that we are interested in is its expression for the relationship between pressure and volume. Pressure and volume for an ideal gas are inversely proportional. As one increases, the other decreases. We will use this relationship in calculating the difference in entropy between two states of a gas having the same temperature but different pressure.","ideal8.gif") CI["ideal gases, and entropy (S)"]=new Array("_gas8") CI["_sp8"]=new Array("If we assume that the gases we are working with are ideal, we can express the entropy change due to expansion in term of P1 and P2 rather than V1 and V2. While not particularly useful in and of itself, this will prove to be very useful when we wish to calculate the entropy of a gas at nonstandard pressure.","sp8.gif") CI["8-8:entropy (S), and pressure"]=new Array("_sp8") addScript(8,"Entropy and Pressure","_sp8") CI["_sad8"]=new Array("The molar entropy of a gas depends upon its pressure. The more pressurized a gas is, the less volume a mole occupies, and the lower the molar entropy.

Likewise, for solutes, the higher the concentration, the lower the entropy.|Natural logarithm terms must not contain units; we divide out the units of bar or molarity to properly calculate the adjustment to standard molar entropy." +" Note that the result is in units of the ideal gas constant, R. The value we typically use for R is 8.3145 J/mol-K." ,"sadjust.gif") CI["8-9:entropy (S), adjustment to"]=new Array("_sad8") CI["_esp8"]=new Array("","sandp.gif") CI["8-17:entropy (S), and pressure"]=new Array("_esp8") CI["_q8"]=new Array( "Consider the reaction of dinitrogen tetroxide to give nitrogen dioxide. |A bond is broken. " +"An additional mole of gas is produced. We expect a large increase in entropy. Applying what we know " +"about the adjustment to standard molar entropy based on pressure, we can calculate the entropy change for this " +"reaction at any given partial pressures of reactants and products." ,"sandq.gif" ,"The logarithmic term is commonly called the reaction quotient, Q. |Notice that Q is unitless, " +"because in the calculation it was assumed that the pressures were in bars. We have to be careful " +"if we decide to write this without the units. Overall, the calculation must give units of J/K, not J/mol-K, as " +"the final \"shorthand\" form might imply. Read on for a sample calculation." ,"sandq2.gif" ,"Calculating for P(N2O4) = 0.01 atm and P(NO2) = 3.0 atm:|" +"The result is that at these pressures, the entropy change for the reaction is only about 2/3 its standard value." ,"sandq3.gif" ,"|

The steps for the calculation of the nonstandard change in entropy for a reaction thus include:

" +"

" +"" +"" +"" +"" +"
(1)

Write the balanced chemical equation for the reaction.

(2)

Using values from a table of standard molar entropies, determine the standard change in entropy based on this chemical equation.

(3)

Write Q, and calculate its value.

(4)

Calculate DELTA_S based on DELTA_So and Q.

" ,"sfromso.gif") CI["8-10:reaction quotient (Q), and entropy (S)"]=new Array("_q8") addScript(8,"Calculating DELTA_S for a Reaction","_sad8,_esp8,_q8") CI["_sc8"]=new Array("In a dilution, the entropy change of \"expansion\" is the entropy change due to the solute going from a small volume to a larger one." ,"sc8.gif") CI["8-9:entropy (S), and dilution"]=new Array("_sc8") CI["_sid8"]=new Array("An \"ideal solution\" is one in which dissolved substances are randomly distributed and not attracted " +"or repelled from each other, in analogy to an ideal gas. Basically, we ignore solvent. In the case of ideal solutions, we use " +"concentration instead of pressure, because concentration, like pressure, is inversely proportional to volume." ,"sid8.gif") CI["entropy (S), and ideal solutions"]=new Array("_sid8") CI["concentration, and entropy (S)"]=new Array("_sid8") addScript(8,"Dilution of Ideal Solutions","_sc8,_sid8") CI["_pico8"]=new Array("

A limitation of the calculation of DELTA_S " +"for a reaction is that typically the pressures and concentrations of reactants and products are changing all the time." +"

We either imagine:

  1. A HUGE reaction where a mole or two reacting won't make much of a dent in the overall " +"pressures or concentrations, or " +"

  2. A typical reaction on the scale of a mole, but only a very, very small amount reacting--perhaps a picomole (10−12 mole). " +"In that case, our calculation, which is for a mole of reactants would be \"scaled down\" to give the " +"correct final answer for DELTA_S." +"

In either case, we will see that it is the sign of DELTA_S that is most important, not the magnitude. " +"If DELTA_S > 0, we know the reaction could conceivably happen. If not, forget it!

" ,"") CI["8-11:spontaneity, picomole argument"]=new Array("_pico8") addScript(8,"Picomole Argument","_pico8") CI["_evap8"]=new Array( "|
This sequence illustrates how the entropy of the universe would change if this reaction occurs at different temperatures and pressures. " ,"h2oevaptable1.gif" ,"|
" ,"h2oevaptable2.gif" ,"|
" ,"h2oevaptable3.gif" ,"|
" ,"h2oevaptable4.gif" ,"|
" ,"h2oevaptable5.gif" ) CI["8-15:evaporation, water"]=new Array("_evap8") CI["8-15:water, vapor pressure"]=new Array("_evap8") CI["8-16:spontaneity, water evaporation"]=new Array("_evap8") addScript(8,"The Effect of Temperature and Pressure on Entropy","_evap8") CI["_slq8"]=new Array("Since the molar entropy of a solid or a liquid is always its standard molar entropy, we don't have to consider them when adjusting " +" DELTA_S for a reaction for nonstandard pressure or concentration. We write, for example,","solliqs.gif") CI["8-14:solids, and reaction quotient (Q)"]=new Array("_slq8") CI["8-14:liquids, and reaction quotient (Q)"]=new Array("_slq8") CI["8-14:evaporation, water"]=new Array("_slq8") CI["reaction quotient (Q), rules for writing"]=new Array("_qr8") CI["_qr8"]=new Array( "|The following \"rules\" for writing Q all arise from the above relationships:" +"

" +"" +"" +"" +"" +"" +"
(1)

Always start with a balanced chemical equation.

(2)

Gases are included as partial pressures in atm.

(3)

Solutes are included as concentrations in mol/L.

(4)

Liquids and solids do not appear because their entropy is always standard.

(5)

Products are in the numerator; reactants are in the denominator.

(6)

All substances appear to their stoichiometric power.

" ,"qrules.gif") addScript(8,"Reaction Quotient Rules","_slq8,_qr8") //ch09 CI["_h9"]=new Array("Enthalpy (H) is defined as \"U + PV.\" By itself, this isn't particularly useful. " +"but when we consider the change in enthalpy at constant pressure, we see that DELTA_H = q (because w = P DELTA_V. " +"This turns out to be very useful!","hdef.gif") CI["9-2:enthalpy (H)"]=new Array("_h9") addScript(9,"The Definition of Enthalpy","_h9") CI["_su9"]=new Array("The key idea in this book is that all processes naturally progress in a way that " +"leads to the most probable distribution of energy. We measure that increase in " +"probability as an increase in entropy. Overall, it is not the entropy of the system " +"alone that must increase, but rather the entropy of the universe, which also" +"includes the change in entropy of the surroundings.","deltasu.gif") CI["9-1:entropy (S), of the universe"]=new Array("_su9") CI["_she9"]=new Array("In order to calculate the change in entropy of the universe, we need the change in entropy of the surroundings." +"Since there are no reactions going on in the surroundings (by definition), " +"we can assume a Boltzmann distribution. In that case, the change in entropy is the " +"heat going into the surroundings (coming out of the system) divided by its temperature. " +"By considering only changes at constant pressure, we can equate q with the change in enthalpy for the system.","heatsur.gif") CI["9-2:surroundings, and heat (q)"]=new Array("_she9") CI["9-2:enthalpy (H), and surroundings"]=new Array("_she9") CI["_hq9"]=new Array('It\'s important to realize that the amount of heat that enters (or leaves) ' +'the system generally depends upon how a process occurs. ' +'Therefore q is not a state function. ' +'(That is, q is not a function that only depends upon the initial and final states ' +'of the system.) However, if we define the sort of path we wish to take, ' +'then perhaps q could be a state function. ' +'The path we choose is one of \"constant pressure.\" ' +'This leads to a new state function we call enthalpy, ' +'and allows us to write the change in entropy of the universe solely in terms of ' +'other state functions.',"heatstate.gif") CI["9-1:heat (q), not a state function"]=new Array("_hq9") addScript(9,"Heat, Entropy, and the Universe","_su9,_she9,_hq9") CI["_sh9"]=new Array("","hstand.gif") CI["9-3:enthalpy (H), standard molar, of formation"]=new Array("_sh9") CI["_ht9"]=new Array("|more extensive information is available at http://webbook.nist.gov/chemistry.","tableh.gif") CI["9-4:enthalpy (H) standard molar, of formation, table of"]=new Array("_ht9") CI["_hess9"]=new Array("Each of the values for DELTA_Ho is from the table. Values for reactants in the desired equation must be reversed.|" ,"hessandh.gif" ,"|" ,"hessandhf.gif" ,"|" ,"hesshgraph.gif" ) CI["9-5:enthalpy (H) standard, of reaction, and Hess's law"]=new Array("_hess9") CI["9-5:Hess's law, and standard enthalpy of reaction"]=new Array("_hess9") CI["_ca9"]=new Array("If this value were a just a bit less, enough to make it negative, would Earth have an atmosphere conducive to life?","combust.gif") CI["9-7:combustion, atmosphere"]=new Array("_ca9") CI["9-7:atmosphere, combustion of"]=new Array("_ca9") addScript(9,"Standard Molar Enthalpy of Formation","_sh9,_ht9,_hess9,_ca9") CI["_hu9"]=new Array("What are the differences between DELTA_H and DELTA_U?

    " +"
  • DELTA_H is of particular use when the pressure is constant; DELTA_U is valid anytime." +"
  • DELTA_H takes into account only the energy change of a process associated with making and breaking bonds, while DELTA_U also accounts for changes in volume due to expansion and contraction." +"
  • Since for typical chemical reactions w is small, and DELTA_H = DELTA_U - w, the two are nearly the same." +"
  • Tables of DELTA_Hof take into account the phase of the substance (solid, liquid, or gas), while tables of \"Bond Dissociation Energies\" do not." +"
","deltahandu.gif") CI["9-8:enthalpy (H), vs. internal energy (U)"]=new Array("_hu9") addScript(9,"DELTA_H vs. DELTA_U","_hu9") CI["_tb9"]=new Array("Increasing the temperature makes bond breaking more likely specifically because at higher temperature the loss of entropy in the surroundings is diminished.","tandh.gif") CI["9-9:temperature (T), and bonding"]=new Array("_tb9") CI["_evg9"]=new Array("For many chemical reactions where bonds are broken (even simple melting), DELTA_H and DELTA_S will both be positive. " +"In that case, there exists a temperature above which the reaction is \"favorable\" because at that temperature the entropy gain in the system overcomes the entropy loss in the surroundings. " +"(Remember, this loss of entropy in the surroundings is due to the particles there dropping to lower levels as they release heat energy to the system.)","vgraphs.gif") CI["9-10:entropy (S), vector graph"]=new Array("_evg9") addScript(9,"Temperature and Bonding","_tb9,_evg9") //ch10 CI["_law10"]=new Array("Modern thermodynamics is based on an extremely simple idea: Whenever energy can be exchanged, it will exchange randomly. " +"No one way of distributing the energy is particularly better than any other. " +"What we observe is actually just the \"most probable\" distribution--others being so very much less likely. " +"Entropy is related to the number of ways energy can be distributed in the most probable (i.e. Boltzmann) distribution. So an increase in entropy marks a \"favorable\" reaction. " +"But we have to remember that both the system and the surroundings have to be taken into account. This is what we call the \"Second Law of Thermodynamics.\"","deltasu.gif") CI["10-1:entropy (S), of the universe"]=new Array("_law10") CI["10-1:Second Law"]=new Array("_law10") CI["_vgraph10"]=new Array("","vgraphm1.gif") CI["10-1:spontaneity, and entropy (S)"]=new Array("_vgraph10") CI["10-1:melting, vector graph"]=new Array("_vgraph10") CI["_g10"]=new Array("Gibbs energy (G) is directly related to the change in entropy of the universe. " +"A negative value of DELTA_G indicates a spontaneous reaction, that is, one for which the entropy of the universe will increase " +"and the Second Law will be fulfilled.","gands.gif") CI["10-4:Gibbs energy (G), defined"]=new Array("_g10") CI["10-4:Gibbs energy (G), and spontaneity"]=new Array("_g10") CI["10-4:Gibbs energy (G), and Second Law"]=new Array("_g10") CI["_sl10"]=new Array('At a temperature below the melting point (on the left) melting is not spontaneous. ' +'The heat energy required from the surroundings to break the "bonds" in the solid (q) ' +'reduces the entropy of the surroundings too much. Above the melting point there ' +'is no problem. Notice that TDELTA_Ssur is relatively constant. This is because DELTA_Ssur is qsur/T = −q/T, so TDELTA_Ssur is simply −q.' ,"vgraphw.gif") CI["10-3:melting, vector graph"]=new Array("_sl10") addScript(10,"Gibbs Energy and the Entropy of the Universe","_law10,_vgraph10,_g10,_sl10") CI["_gt10"]=new Array("A graph of G vs. T is a downward-sloping curve. " +"To a first approximation, we can consider H relatively constant. But entropy (the slope at any given temperature) increases significantly with temperature. ","gtgraph.gif") CI["10-6:G-T graph"]=new Array("_gt10") CI["_gtc10"]=new Array("When comparing two substances...|...at all temperatures, the compound with the lower Gibbs energy is favored. The arrow says, \"A will be converted to B.\"" ,"gtab1.gif" ,"When comparing two substances...|...if the two curves cross, then at that temperature both A and B can coexist because their free energies are equal." ,"gtab2.gif" ,"When comparing two substances...|...above the crossing temperature, the situation is reversed. Still, though, the observed reaction will be the one leading to the compound with the lower Gibbs energy. At this temperature in this case B will be converted to A." ,"gtab3.gif") CI["10-7:G-T graph, comparing substances"]=new Array("_gtc10") CI["_gtw10"]=new Array("Below the melting point, the solid-to-liquid reaction cannot happen, because if it did, the entropy of the universe would decrease. (The Gibbs energy would increase.) But above the melting point, there is no problem, because the Gibbs energy of the liquid is below the Gibbs energy of the solid. Any solid present will melt.","gtwater.gif") CI["10-7:G-T graph, melting"]=new Array("_gtw10") CI["10-7:Gibbs energy (G), and melting"]=new Array("_gtw10") CI["10-7:Gibbs energy (G), and freezing"]=new Array("_gtw10") CI["_gteq10"]=new Array("Two states can coexist in equilibrium with each other only at the temperature at which their curves cross. This is the equilibrium temperature.","gteq.gif") CI["10-8:Gibbs energy (G), and equilibrium"]=new Array("_gteq10") CI["10-8:equilibrium, G-T graph"]=new Array("_gteq10") addScript(10,"Introduction to G-T Graphs","_gt10,_gtc10,_gtw10,_gteq10") CI["_hs10"]=new Array("" ,"hsrule.gif" ,"" ,"gths1.gif" ,"" ,"gths2.gif" ) CI["10-8:low enthalpy/high entropy rule"]=new Array("_hs10") CI["10-8:enthalpy (H), low enthalpy/high entropy rule"]=new Array("_hs10") CI["10-8:entropy (S), low enthalpy/high entropy rule"]=new Array("_hs10") addScript(10,"The Low Enthalpy/High Entropy Rule","_hs10") CI["_mpc10"]=new Array( "The melting point can be determined if we know the actual values of DELTA_H and DELTA_S" +" for the melting reaction. Since there are no gases here, we will be able to use all standard" +" values. Thus, we want to know at what temperature the lines for liquid and solid cross, i.e." +" where DELTA_G = 0:" ,"gmelt.gif") CI["10-9:Gibbs energy (G), and melting point"]=new Array("_mpc10") CI["_tf10"]=new Array("\"normal\" melting points can be calculated if the standard enthalpy change and the standard entropy change for the melting \"reaction\" are known. Be careful to convert kJ to J! For example, for water:","tablefus.gif") CI["10-10:enthalpy (H), standard molar, of fusion"]=new Array("_tf10") CI["10-10:entropy (S) standard molar, of fusion"]=new Array("_tf10") CI["10-10:melting, table of data"]=new Array("_tf10") addScript(10,"Calculations Involving DELTA_G = 0","_mpc10,_tf10") CI["_vpg10"]=new Array("The entropy of a gas depends upon its vapor pressure. Higher vapor pressure (smaller volume per mole) leads to wider-spaced translational energy levels, lower entropy and a less steep curvature on a graph of G vs. T.","gtgas2.gif") CI["10-10:Gibbs energy (G), and vapor pressure"]=new Array("_vpg10") CI["_vap10"]=new Array( "|At room temperature the vapor pressure of water is quite low..." ,"gtvap3.gif" ,"|...but as the temperature is raised, the vapor pressure rises as well..." ,"gtvap2.gif" ,"|...until at the \"normal boiling point\" the vapor pressure is one bar." ,"gtvap1.gif" ) CI["10-11:boiling point, effect of pressure"]=new Array("_vap10") CI["10-11:water, vapor pressure"]=new Array("_vap10") CI["10-11:water, boiling point"]=new Array("_vap10") CI["_rtvap10"]=new Array( "|If we focus on 70 oC (343 Kelvin), we see that if the vapor pressure of water is low, liquid water will evaporate...|animate_loop" ,"gtrtvap3.gif" ,"|...but as the vapor pressure rises, the molar entropy of the gas will decrease, the curve for the gas will become less steep, and equilibrium will be reached..." ,"gtrtvap2.gif" ,"|...likewise, if the vapor pressure is too high, water vapor will condense, lowering the vapor pressure and making the gas curve steeper, until..." ,"gtrtvap1.gif" ,"|...the two curves cross again at 70 oC so that equilibrium is reached again." ,"gtrtvap2.gif" ) CI["10-12:G-T graph, evaporation"]=new Array("_rtvap10") CI["10-12:spontaneity, water evaporation"]=new Array("_rtvap10") CI["10-12:Gibbs energy (G), and evaporation"]=new Array("_rtvap10") CI["10-12:evaporation, water"]=new Array("_rtvap10") CI["10-12:water, vapor pressure"]=new Array("_rtvap10") CI["_vapd10"]=new Array("When you open a bottle containing a volatile liquid such as gasoline, some of the vapor in the bottle escapes and is replaced by air. " +"That creates a nonequilibrium situation. When the bottle is recapped, the pressure inside the bottle invariably increases as equilibrium is re-established. " +"The next time the bottle is opened, a hissing sound will be heard. If the vapor is flammable and there is an ignition source, look out!","gtvapdanger.gif") CI["10-12:evaporation, danger"]=new Array("_vapd10") CI["_calc10"]=new Array("Estimating the vapor pressure at a specific temperature is easy with an equation-solving calculator. We need to know DELTA_Ho, DELTA_So, and the temperature involved. Using a standard calculator we could solve the equation for either the equilibrium temperature or the natural log of the vapor pressure, and then put values in as needed. ","gteqn.gif") CI["10-14:equation-solving calculator, vapor pressure"]=new Array("_calc10") addScript(10,"Gibbs Energy and Vapor Pressure","_vpg10,_vap10,_rtvap10,_vapd10,_calc10") CI["_bar10"]=new Array("Barometric pressure is the pressure of the atmosphere, which changes with altitude and weather. " +"(The pressure reported on the evening news is NOT barometric pressure, but instead the result of a calculation designed to give the same number in Denver as in New York for the same sort of weather condition.) " +"The actual boiling point of a liquid depends upon the actual barometric pressure above that liquid.","pbar.gif") CI["10-12:barometric pressure"]=new Array("_bar10") CI["10-12:pressure, barometric"]=new Array("_bar10") CI["10-12:boiling, and Gibbs energy (G)"]=new Array("_bar10") CI["_boil10"]=new Array("Shown here are two nonequilibrium situations in which the water is boiling. At 100 o C, the vapor pressure of water is 760 torr. If the barometric pressure is ever so slightly less than 760 torr, bubbles will form, and boiling will occur. " +"At 25 oC, the pressure above the liquid can be reduced using a vacuum pump. Once the pressure drops below 24 torr, the water will start to boil. This is the principle by which a \"rotovap\" works.","gtboil.gif") CI["10-14:boiling, water"]=new Array("_boil10") addScript(10,"Boiling and Vapor Pressure","_bar10,_boil10") //ch11 CI["_s11.0"]=new Array("Reaction quotients involve pressures of gases in bars and concentrations of solutes in moles/liter. Liquids and solids are not included.","qexample.gif") CI["11-1:reaction quotient (Q), examples"]=new Array("_s11.0") CI["_s11.1"]=new Array("We focus on equilibrium simply because that is the end of the road for all chemical " +"reactions. Equilibrium is represented by the most probable distribution. " +"It is the point where no reaction can decrease the Gibbs energy of the system any further, when DELTA_G = 0.","gteq.gif") CI["11-2:Gibbs energy (G), and equilibrium"]=new Array("_s11.1") CI["_geq11"]=new Array("The value of the equilibrium constant for a reaction at a given temperature comes from setting DELTA_G = 0, using the standard enthalpy change for DELTA_H, and adjusting the entropy change for the appropriate concentrations of solutes and pressures of gases.","gequil.gif") CI["11-3:equilibrium constant (K), and Gibbs energy (G)"]=new Array("_geq11") CI["11-3:reaction quotient (Q), and equilibrium constant (K)"]=new Array("_geq11") addScript(11,"Gibbs Energy and Equilibrium","_s11.0,_s11.1,_geq11") CI["11-4:pH, and equilibrium"]=new Array("The idea that \"neutral\" water has a pH of 7.0 derives from setting [H+] = [OH−] and solving for −log[H+]. Notice that neutral water has a pH of 7.00 only at 25 oC. " +"Water at 100 oC, for example, has a pH = 6.0 (as well as a pOH of 6.0) because Kw = 1x10−12. (That is, in hot water there are higher concentrations of both H+ and OH− in solution.","gkw.gif") CI["_lnk11"]=new Array("If values of the equilibrium constant can be determined for more than one temperature, a graph of ln(Keq) vs. 1/Teq can be used to determine DELTA_Ho and DELTA_So for the reaction experimentally.","lnkvs1_t.gif") CI["11-6:enthalpy (H) standard, of reaction, experimental determination"]=new Array("_lnk11") CI["11-6:entropy (S) standard, of reaction, experimental determination"]=new Array("_lnk11") CI["11-6:experimental determination, of standard enthalpy of reaction"]=new Array("_lnk11") CI["11-6:experimental determination, of standard entropy of reaction"]=new Array("_lnk11") CI["_kt11"]=new Array("An endothermic reaction (one with a positive DELTA_Ho) will show an increase in K with increasing temperature." +" The opposite will be true of an exothermic reaction. For an exothermic reaction, an increase in temperature results in a shift in equilibrium toward reactants." ,"kandt.gif") CI["11-6:temperature (T), effect on equilibrium constant (K)"]=new Array("_kt11") CI["11-6:equilibrium constant (K), effect of temperature"]=new Array("_kt11") CI["_cc11"]=new Array("The Clausius-Clapeyron equation relates to the effect of temperature on equilibrium constants. Notice that only enthalpy is involved. " +"A reaction with a positive DELTA_Ho will show an increase in K with increasing temperature." +"This equation is easily solved for any missing variable using an equation-solving calculator.","clausc.gif") CI["11-7:temperature (T), effect on equilibrium constant (K)"]=new Array("_lnk11") CI["11-7:equilibrium constant (K), effect of temperature"]=new Array("_lnk11") CI["11-7:Clausius-Clapeyron equation"]=new Array("_cc11") CI["11-7:equilibrium, equation-solving calculator"]=new Array("_cc11") CI["_kth11"]=new Array("","khight.gif") CI["11-7:equilibrium constant (K), high-temperature limit"]=new Array("_kth11") addScript(11,"Temperature and Equilibrium","_kt11,_lnk11,_kth11,_cc11") CI["_no11"]=new Array("|Consider a mixture of NO2 and N2O4 that is not at equilibrium. " +" As shown here, the partial pressure of NO2 is too high, and a reaction will occur that reduces the partial pressure of NO2 and increases the partial pressure of N2O4." +" Both are gases, so as the reaction occurs, both of their partial pressures will change. Click the next radio button to see the result of reaction." ,"gtno2a.gif" ,"|The curves now cross. Equilibrium has been reached." ,"gtno2b.gif" ) CI["11-8:nitrogen dioxide, equilibrium with dinitrogen tetroxide"]=new Array("_no11") CI["11-8:equilibrium, approach to"]=new Array("_no11") CI["11-8:equilibration"]=new Array("_no11") CI["_s11a3"]=new Array( "|Consider a system that is already at equilibrium. We focus specifically on the equilibrium temperature. Le Châtelier's principle states that if the system is somehow changed" +" so that it is no longer at equilibrium, it will automatically readjust so as to re-establish a new equilibrium state." +" What will this picture look like if the partial pressure of NO2 is suddenly increased?|animate" ,"gtlechat1.gif" ,"|The partial pressure of NO2 has been increased. Its curve flattens due to the decrease in its molar entropy. There is no longer an equilibrium. The system is no longer in its most probable state. What reaction will happen now?" ,"gtlechat2.gif" ,"|2 NO2 will react to give N2O4." ,"gtlechat2b.gif" ,"|Since both NO2 and N2O4 are gases, as their partial pressures change, their molar entropies will change as well. Both curves will adjust until a equilibrium is re-established at a new position." ,"gtlechat3.gif" ,"|Now consider an increase in temperature. How will the system adjust?" ,"gtlechat4.gif" ,"|N2O4 will react to give 2 NO2. Again, both curves will adjust to re-establish a new equilibrium position." ,"gtlechat5.gif" ,"|Equilibrium has been re-established. Le Châtelier's principle derives from the observation that no matter how a system is perturbed, it will automatically adjust to the most probable distribution of energy. Gibbs energy accounts for the changes in both the system and the surroundings when these adjustments occur." ,"gtlechat6.gif" ) CI["11-10:equilibrium, and Le Châtelier's principle"]=new Array("_s11a3") addScript(11,"An Example: 2NO2/N2O4","_no11,_s11a3") CI["_eqc11"]=new Array("By setting DELTA_G = 0, Q becomes the equilibrium constant. With an equation-solving calculator, solving this equation for Keq or Teq is easy. Alternatively, it can be rearranged to solve for Keq.","eqnk.gif") CI["11-6:equation-solving calculator, equilibrium"]=new Array("_eqc11") /* 11-13 equilibrium calculations. Could have something here? steps: 1) write the overall equilibrium equation 2) determine the value of K 3) write the equilibrium expression 4) determine the value of Q 5) tabulate the changes expected in terms of x 6) substitute the final values into the equilibrium expression 7) solve for x 8) determine the final values based on x 9) check that Q = K */ //CI["11-15:equilibration, at constant pressure"]=new Array("TEMP","p170.gif") /* again, coule have something here? steps: 1) write the overall equilibrium equation 2) determine the value of K 3) write the equilibrium expression 4) determine the value of Q 5) tabulate the changes expected in terms of x 6) include an adjustment to the final pressures 7) express alpha in terms of x 8) substitute the final values into the equilibrium expression 9) solve for x 10) determine the final values based on x 11) check that Q = K */ CI["_eqcc11"]=new Array("Predictions of actual concentrations or pressures of reactants and products at equilibrium should be treated " +"with some skepticism in mind for several reasons: " +"
  • Values of DELTA_Ho, DELTA_So, and especially DELTA_Go, are temperature-dependent. (This is because different substances have different heat capacities.)" +"Thus, if the temperature is not 298 Kelvin (the temperature usually given for tables of standard values), then the values we get from the tables will not be correct." +"
  • Most chemical reactions release or absorb heat, and so temperature is not always constant. The value of K will change as the reaction progresses." +"
  • Our use of DELTA_Go in relation to K hinges on the reaction being carried out at constant pressure. If this is not the case, then the value of K calculated from DELTA_Ho, DELTA_So, and T is suspect." +"
  • Many additional factors (deviations from ideality, solvent-solute interactions, etc.) can play subtle roles in chemical reactions. These factors may have significant effects on the value of K." +"
" +"Thus, any calculations involving equilibrium constants should be seen as estimates only and should not be expected to lead to exactly correct predictions." ,"") CI["11-16:equilibrium, calculations, complications"]=new Array("_eqcc11") addScript(11,"Equilibrium Calculations","_eqc11,_eqcc11") CI["11-16:Gibbs energy (G), of formation"]=new Array("If you take a close look at the expression for Keq in terms of standard molar enthalpy and" +" standard molar entropy, you will see that they can be rewritten in terms of the standard free" +" energy change for a given temperature. However, be careful! DELTA_Go is a function of temperature.","kg.gif") CI["_peq11"]=new Array("","kp.gif") CI["11-17:equilibrium, and probability"]=new Array("_peq11") CI["_seq11"]=new Array("","ksu.gif") CI["11-17:entropy (S), and equilibrium"]=new Array("_seq11") CI["_sku11"]=new Array("There is an amazingly simple relationship between DELTA_Suniverse, K, and Q:","skq.gif") CI["11-17:universe, entropy change, K and Q"]=new Array("_sku11") CI["11-17:entropy (S), of the universe"]=new Array("_sku11") CI["11-17:reaction quotient (Q), and equilibrium constant (K)"]=new Array("_sku11") addScript(11,"Equilibrium and Probability","_sku11,_peq11,_seq11") //ch12 CI["_"]=new Array("",".gif") //CI["12-1:phase change, G-T graph"]=new Array("TEMP","p408.gif") //CI["12-2:Gibbs energy (G), and melting"]=new Array("TEMP","p241.gif") //CI["12-2:Gibbs energy (G), and freezing"]=new Array("TEMP","p237.gif") //CI["12-2:G-T graph, melting"]=new Array("TEMP","p280.gif") //CI["12-2:G-T graph, freezing"]=new Array("TEMP","p275.gif") //CI["12-3:evaporation, and Gibbs energy (G)"]=new Array("TEMP","p207.gif") //CI["12-3:boiling, and Gibbs energy (G)"]=new Array("TEMP","p10.gif") //CI["12-3:G-T graph, evaporation"]=new Array("TEMP","p274.gif") //CI["12-3:G-T graph, boiling"]=new Array("TEMP","p268.gif") CI["_demo1-12"]=new Array( "||animate" ,"demo-balloon1.jpg" ,"" ,"demo-balloon2.jpg" ,"" ,"demo-balloon3.jpg" ,"" ,"demo-balloon4.jpg" ,"" ,"demo-balloon5.jpg" ) CI["vapor pressure, and temperature"]=new Array("_demo1-12") CI["balloon in an Erlenmeyer"]=new Array("_demo1-12") addScript(12,"Balloon in an Erlenmeyer","_demo1-12") CI["_demo2-12"]=new Array( "||animate" ,"demo-beans1.jpg" ,"" ,"demo-beans2.jpg" ,"" ,"demo-beans3.jpg" ) CI["vapor pressure, and temperature"]=new Array("_demo2-12") CI["pressure cooker"]=new Array("_demo2-12") addScript(12,"Can of Beans","_demo2-12") CI["_demo3-12"]=new Array( "||animate" ,"demo-rotovap1.jpg" ,"" ,"demo-rotovap2.jpg" ) CI["vapor pressure, and temperature"]=new Array("_demo3-12") addScript(12,"Rotary Evaporator","_demo3-12") CI["_demo4-12"]=new Array( "||animate" ,"demo-humidity1.jpg" ,"" ,"demo-humidity2.jpg" ,"" ,"demo-humidity3.jpg" ) CI["relative humidity"]=new Array("_demo4-12") CI["dew point"]=new Array("_demo4-12") addScript(12,"Relative Humidity and Dew Point","_demo4-12") CI["_demo5-12"]=new Array( "||animate" ,"demo-triplept1.jpg" ,"" ,"demo-triplept2.jpg" ) CI["triple point, of CO2"]=new Array("_demo5-12") addScript(12,"Triple Point of CO2","_demo5-12") CI["_demo6-12"]=new Array( "||animate" ,"demo-solubility1.jpg" ,"" ,"demo-solubility2.jpg" ) CI["solubility, and temperature"]=new Array("_demo6-12") addScript(12,"Solubility","_demo6-12") CI["_demo7-12"]=new Array( "||animate" ,"demo-supersat1.jpg" ,"" ,"demo-supersat2.jpg" ,"" ,"demo-supersat3.jpg" ) CI["supersaturation"]=new Array("_demo7-12") addScript(12,"Supersaturation","_demo7-12") CI["_demo8-12"]=new Array( "||animate" ,"demo-degas1.jpg" ,"" ,"demo-degas2.jpg" ) CI["degassing of solutions"]=new Array("_demo8-12") CI["solubility, of gases"]=new Array("_demo8-12") addScript(12,"Degassing of Solutions","_demo8-12") CI["_demo9-12"]=new Array( "||animate" ,"demo-fpdepress1.jpg" ,"" ,"demo-fpdepress2.jpg" ) CI["melting point depression"]=new Array("_demo9-12") CI["freezing point depression"]=new Array("_demo9-12") addScript(12,"Freezing Point Depression","_demo9-12") CI["_demo10-12"]=new Array( "" ,"demo-bpelevation.jpg" ) CI["boiling point elevation"]=new Array("_demo10-12") addScript(12,"Boiling Point Elevation","_demo10-12") //CI["12-4:Gibbs energy (G), and humidity"]=new Array("TEMP","p240.gif") //CI["12-4:Gibbs energy (G), and dew point"]=new Array("TEMP","p230.gif") //CI["12-4:dew point"]=new Array("TEMP","p81.gif") //CI["12-4:humidity"]=new Array("TEMP","p306.gif") //CI["12-4:G-T graph, humidity"]=new Array("TEMP","p278.gif") //CI["12-4:G-T graph, dew point"]=new Array("TEMP","p272.gif") //CI["12-4:vapor pressure, and humidity"]=new Array("TEMP","p517.gif") //CI["12-4:vapor pressure, and dew point"]=new Array("TEMP","p514.gif") //CI["12-5:Gibbs energy (G), and vapor deposition"]=new Array("TEMP","p250.gif") //CI["12-5:Gibbs energy (G), and sublimation"]=new Array("TEMP","p248.gif") //CI["12-5:Gibbs energy (G), and frost"]=new Array("TEMP","p239.gif") //CI["12-5:sublimation"]=new Array("TEMP","p477.gif") //CI["12-5:reaction quotient (Q), equation-solving calculator"]=new Array("TEMP","p443.gif") //CI["12-5:equation-solving calculator, reaction quotient (Q)"]=new Array("TEMP","p166.gif") //CI["12-5:G-T graph, frost"]=new Array("TEMP","p277.gif") //CI["12-5:frost formation"]=new Array("TEMP","p260.gif") //CI["12-5:vapor pressure, and frost"]=new Array("TEMP","p516.gif") //CI["12-5:vapor deposition"]=new Array("TEMP","p512.gif") //CI["12-6:triple point"]=new Array("TEMP","p506.gif") //CI["12-6:refrigeration"]=new Array("TEMP","p449.gif") //CI["12-6:G-T graph, vapor deposition"]=new Array("TEMP","p287.gif") //CI["12-6:G-T graph, sublimation"]=new Array("TEMP","p285.gif") //CI["12-7:Gibbs energy (G), and triple point"]=new Array("TEMP","p249.gif") //CI["12-7:carbon dioxide, triple point"]=new Array("TEMP","p44.gif") //CI["12-7:carbon dioxide, and Becker, Bob"]=new Array("TEMP","p42.gif") //CI["12-7:G-T graph, triple point"]=new Array("TEMP","p286.gif") //CI["12-7:Becker, Bob, and carbon dioxide"]=new Array("TEMP","p7.gif") //CI["12-7:vapor pressure, and triple point"]=new Array("TEMP","p518.gif") //CI["12-8:Gibbs energy (G), and critical point"]=new Array("TEMP","p229.gif") //CI["12-8:critical point"]=new Array("TEMP","p76.gif") //CI["12-8:G-T graph, critical point"]=new Array("TEMP","p271.gif") //CI["12-8:phase diagram"]=new Array("TEMP","p409.gif") //CI["12-9:Gibbs energy (G), and phase diagram"]=new Array("TEMP","p243.gif") //CI["12-9:G-T graph, phase diagram"]=new Array("TEMP","p243.gif") //CI["12-9:ice skating"]=new Array("TEMP","p328.gif") //CI["12-9:ice IX"]=new Array("TEMP","p327.gif") //CI["12-9:G-T graph, phase diagram"]=new Array("TEMP","p282.gif") //CI["12-10:Vonnegut, Kurt"]=new Array("TEMP","p541.gif") //CI["12-10:Gibbs energy (G), and solubility"]=new Array("TEMP","p246.gif") //CI["12-10:solubility"]=new Array("TEMP","p465.gif") //CI["12-10:G-T graph, solubility"]=new Array("TEMP","p284.gif") //CI["12-12:liquids, degassing"]=new Array("TEMP","p354.gif") //CI["12-13:equilibrium, in coral reefs"]=new Array("TEMP","p194.gif") //CI["12-13:carbon dioxide, and coral reefs"]=new Array("TEMP","p43.gif") //CI["12-13:Le Châtelier's principle, and coral reefs"]=new Array("TEMP","p348.gif") //CI["12-13:Kleypus, Joanie, and coral reefs"]=new Array("TEMP","p346.gif") //CI["12-15:freezing point depression"]=new Array("TEMP","p256.gif") //CI["12-15:liquids, impure"]=new Array("TEMP","p355.gif") //CI["12-15:Gibbs energy (G), and freezing point depression"]=new Array("TEMP","p238.gif") //CI["12-16:G-T graph, freezing point depression"]=new Array("TEMP","p276.gif") //CI["12-18:boiling point, elevation"]=new Array("TEMP","p14.gif") //CI["12-18:Gibbs energy (G), and boiling point elevation"]=new Array("TEMP","p228.gif") //CI["12-18:G-T graph, boiling point elevation"]=new Array("TEMP","p269.gif") //CI["12-19:Raoult's law"]=new Array("TEMP","p436.gif") //CI["12-19:Gibbs energy (G), and Raoult's law"]=new Array("TEMP","p244.gif") //CI["12-19:G-T graph, Raoult's law"]=new Array("TEMP","p283.gif") //ch13 CI["_uw13"]=new Array("Passing electrons trough a wire and returning them to the system serves as a \"non-PV\" way to change the energy of the system. We call it electrical work because, technically, heat is not involved. Ultimately, this energy is either used to physically move an object in the surroundings (using electromagnets) or transformed into heat in a secondary process (as with a light bulb or speaker).","workelec.gif") CI["13-4:internal energy (U), and electrical work"]=new Array("_uw13") CI["13-4:work (w), electrical"]=new Array("_uw13") CI["_s13.1"]=new Array("The presence of electrical work changes the relationship between " +"DELTA_H and q, and that results in a slightly different expression for Gibbs energy. In fact, \"minus DELTA_G\" will turn out to be " +"the maximum amount of electrical work that can be obtained from a chemical reaction.","hupv.gif") CI["13-1:Gibbs energy (G), and electrochemistry"]=new Array("_s13.1") //CI["13-1:equilibrium, in batteries"]=new Array("TEMP","p193.gif") //CI["13-1:battery, and equilibrium"]=new Array("TEMP","p6.gif") addScript(13,"Enthalpy and Electrical Work","_uw13,_s13.1") CI["_hf13"]=new Array("The H2/O2 fuel cell produces water as its only product.","h2o2fuel.gif") CI["13-10:hydrogen/oxygen, fuel cell"]=new Array("_hf13") CI["_fc13"]=new Array("","gtfuel.gif") CI["13-2:oxygen, reaction with hydrogen"]=new Array("_fc13") CI["13-2:hydrogen/oxygen, fuel cell"]=new Array("_fc13") CI["13-2:G-T graph, hydrogen/oxygen reaction"]=new Array("_fc13") CI["_gs13"]=new Array("For the hydrogen/oxygen fuel cell, both DELTA_H and DELTA_S are negative. At room temperature, the overall outcome is that DELTA_G is also negative, at least under standard conditions.","gandselec.gif") CI["13-2:entropy (S), and Gibbs energy (G)"]=new Array("_gs13") CI["13-3:hydrogen/oxygen reaction, vector graph"]=new Array("_gs13") CI["13-3:vector graph, Gibbs energy (G)"]=new Array("_gs13") addScript(13,"The Hydrogen-Oxygen Fuel Cell","_hf13,_fc13,_gs13") CI["_s13a2"]=new Array("|(a) The relationship between Gibbs energy and electrical work begins with the relationship of Gibbs energy to enthalpy and entropy under conditions of constant temperature and pressure." ,"vectorg2.gif" ,"|(b) But DELTA_H is the sum of q and welec. Both are shown as being negative here, indicating that this is an exothermic reaction that is producing electrical work." ,"vectorg3.gif" ,"|(c) Only q is related to the entropy increase in the surroundings. The production of electrical work reduces the entropy change of the surroundings." ,"vectorg4.gif" ,"|(d) For the Second Law to be satisfied--for our process to be heading toward the most probable distribution of energy in the universe--the TDELTA_S arrow (which is negative) must be shorter than the TDELTA_Ssur arrow. This implies that the two red lines must not cross!" ,"vectorg5.gif" ,"|(e) But to say the two red lines must not cross is equivalent to saying that the DELTA_G arrow must be longer than the arrow representing the electrical work performed. The fact that both of these arrows are negative leads to difficulties. Is it appropriate to say that \"DELTA_G is the maximum amount of electrical work\"? There should be no question that the DELTA_G arrow must be longer. It's just a problem of semantics. Perhaps the simplest way to say it is that both DELTA_G and welec are negative, and DELTA_G must be more negative than welec. Mathematically, we have simply that DELTA_G < welec." ,"vectorg6.gif" ) CI["13-6:Gibbs energy (G), and electrical work"]=new Array("_s13a2") CI["_s13a3"]=new Array("|For the reverse reaction, which is unfavorable, DELTA_G is positive. However, the application of electrical work to the system can drive the reaction. In this case, DELTA_G is the minimum amount of electical work that will be needed in order to get the job done. This process of applying electrical work to a system is what \"charges\" a battery. Note that in this case, as well as when the battery is discharging, we require DELTA_G < welec in order to satisfy the Second Law.","vectorgrev.gif") CI["13-7:Gibbs energy (G), positive change"]=new Array("_s13a3") addScript(13,"Gibbs Energy and Electrical Work","_s13a2,_s13a3") CI["_ti13"]=new Array("|The \"voltage\" of a battery measures its potential to do electrical work (when there is no current). " ,"circuit2.gif" ,"|Electrical \"current,\" which for historical reasons is seen as going the opposite direction as the electrons in a wire, is measured in units of Coulombs per second. Multiplying by the number of seconds gives us the amount of \"charge\" passing through the wire, in Coulombs." ,"circuit1.gif" ,"|The amount of actual work (−welec -- remember, welec itself is negative) done by the circuit depends upon what is connected to it. Multiplying the amount of charge passing through the circuit, Qelec, by the cell potential, Ecell, gives the maximum amount of work done by the cell in Joules." ,"circuit3.gif" ,"|We can think of Qelec x Ecell as the amount of energy \"freed\" from the cell by connecting the circuit. It is, in fact, −DELTA_G. We could calculate DELTA_G, then, by multiplying the current by the time and the cell potential. " +"Remember, though, DELTA_G must be negative if the cell is doing work! If the current is 0.2 Amperes (0.2 Coulombs/second), the time involved is 4 seconds, and the cell potential is 10 V (10 Joules/Coulomb), then the Gibbs energy change is −(0.20 x 4 x 10) = −8 Joules. " ,"circuit4.gif" ) CI["13-8:time (t), and electrical current (I)"]=new Array("_ti13.0") CI["13-8:electrical current (I), and electrical charge (Qelec)"]=new Array("_ti13.0") CI["13-8:electrical current (I)"]=new Array("_ti13.0") CI["13-8:electrical charge (Qelec)"]=new Array("_ti13.0") CI["13-8:cell potential"]=new Array("_ti13.1") CI["13-8:work (w), electrical"]=new Array("_ti13.2") CI["13-8:cell potential, and Gibbs energy (G)"]=new Array("_ti13.3") addScript(13,"The Electrical Connection","_ti13") CI["_bat13"]=new Array("|A battery consists of one or more chemical cells, each of which involves two separate chemical reactions. In the oxidation reaction, electrons are released to the anode and are passed into the circuit." +" Returning to the battery at the cathode, these electrons are \"consumed\" in the reduction reaction. In the course of the reaction, n moles of electrons are involved." ,"battery0.gif" ,"|The key to a chemical connection is that the charge passed through the wire," +" Qelec, comes from electrons released by chemical reactions" +" in the battery. Faraday's constant relates \"charge\" to number of moles of electrons." ,"battery1.gif" ,"|Since DELTA_G is related to Qelec, it is also related to the number of moles of electrons involved in the oxidation/reduction reactions. We can express the Gibbs energy change for a reaction to the cell potential without resort to measuring currents and times." ,"battery2.gif" ) CI["13-9:Faraday's constant"]=new Array("_bat13.1") CI["13-9:electrical charge (Qelec), and Faraday's constant"]=new Array("_bat13.1") CI["13-10:reduction"]=new Array("_bat13.0") CI["13-10:oxidation"]=new Array("_bat13.0") CI["13-10:cathode, and reduction"]=new Array("_bat13.0") CI["13-10:half-equation"]=new Array("_bat13.0") CI["13-10:anode, and oxidation"]=new Array("_bat13.0") CI["13-11:cell potential, and Gibbs energy"]=new Array("_bat13.2") addScript(13,"The Chemical Connection","_bat13") CI["_red13"]=new Array("Standard reduction potentials range from about −2 to +2 Volts. Notice that the most reactive species are on the bottom right and the upper left. " +"A general rule of thumb is that anything on the right will react with anything above it on the left. " +"For example, solution of Cu2+ will react with metallic zinc.","eredo.gif") CI["13-13:reduction potential, standard"]=new Array("_red13") CI["13-15:reduction potential standard, and reactivity"]=new Array("_13") CI["_hored13"]=new Array("The standard reduction potentials relating to the hydrogen/oxygen fuel cell are shown below.","h2o2stand.gif") CI["13-12:hydrogen/oxygen, fuel cell"]=new Array("_hored13") CI["_hox13"]=new Array("In order to use information from a table of standard reduction potentials, we must reverse one of the equations " +"along with its potential. In doing so, we call the voltage a standard oxidation potential","h2o2ox.gif") CI["13-13:oxidation potential, standard"]=new Array("_hox13") CI["13-12:cell potential, standard"]=new Array("_hox13") CI["_te13"]=new Array("Temperature can have a significant effect on cell potential, just as it can on DELTA_G. Notice that the a negative DELTA_G corresponds to a positive cell potential. When DELTA_G = 0, the cell potential is also 0.","h2o2temp.gif") CI["13-12:temperature (T), effect on cell potential"]=new Array("_te13") CI["13-12:cell potential, effect of temperature"]=new Array("_te13") addScript(13,"Standard Potentials","_red13,_hored13,_hox13,_te13") CI["_n13"]=new Array("|The Nernst equation relates an actual cell potential to a standard one. n here is the number of electrons produced in the oxidation or consumed in the reduction; Q is the reaction quotient." ,"nernst.gif" ,"|It's important to see where the Nernst equation comes from. Really all it is is a restatement of the definition of DELTA_G, now including the relationship between Gibbs energy and cell potential." ,"nernst0.gif" ,"|First, we rewrite the change in Gibbs energy in terms of an adjustment to the standard change in Gibbs energy. " ,"nernst1.gif" ,"|Next, we use the fact that DELTA_G = −nFEcell. A little rearranging does the job." ,"nernst2.gif" ,"|Thus, the Nernst equation is just a restatement of the definition of the change in Gibbs energy in terms of cell potentials." ,"nernst3.gif" ) CI["13-17:cell potential, effect of concentration"]=new Array("_n13") CI["13-17:Nernst equation"]=new Array("_n13") addScript(13,"The Effect of Concentration on Cell Potential (Nernst Equation)","_n13") CI["_kfe13"]=new Array("The equilibrium constant for a reaction is easily related to DELTA_Go." ,"kfrome0.gif" ,"The equilibrium constant for a chemical reaction can be determined if the standard cell potential is known. It is important to recognize that \"Volts\" are the same as \"Joules per Coulomb.\"" +" For example, if Eocell is 0.368 Volts, and one mole of electrons are produced in the oxidation half-reaction, then we can calculate K to be 1.7 x 106." ,"kfrome.gif" ,"|One of the most interesting aspects of electrochemistry is that it can be used to determine standard free energies and equilibrium constants for reactions that at first sight appear to have nothing to do with electrochemistry. " +"For example, consider the acid dissociation of water. What has this to do with electrochemistry, you say? " ,"kfrome1.gif" ,"|The fact is, we can write two half-reactions that add up to this overall equation! Thus, we can get Eo and, from that, calculate DELTA_Go and K." ,"kfrome2.gif" ,"|Can the argument be made that with a little creativity any reaction could be written as the sum of two half-reactions? Perhaps. If so, then who knows what fuel cell you might discover some day! It's possible!" ,"kfrome3.gif" ) CI["13-16:equilibrium constant (K), and cell potential"]=new Array("_kfe13") CI["13-16:cell potential, and equilibrium constant (K)"]=new Array("_kfe13") addScript(13,"Equilibrium Constants and Standard Cell Potentials","_kfe13") //CI["_esc13"]=new Array("--unfinished--","") // CI["13-19:electrochemistry, equation-solving calculator"]=new Array("_esc13") // CI["13-19:equation-solving calculator, electrochemistry"]=new Array("_esc13")