Equilibrium Sample Exercises Using the TI-85, TI-86, or TI-89 Equation Solver

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Full solutions are not given here. Instead, I just show how the equation gets put into the TI-85 and solved.

Page 554, Sample Exercise 15.10, Chemistry: The Central Science, 7/e

For the Haber process, N2(g) + 3 H2(g) <==> 2 NH3(g) , Kp = 1.45E-5 at 500 oC. In an equilibrium mixture of the three gases at 500 oC the partial pressure of H 2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture? 

eqn:Kp=x^2/(0.432*0.928^3)
 Kp=1.4E-5                                Kp=1.4E-5
 x=                 Press [SOLVE]==>     .x=.0022374216639981
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: The equilibrium pressure of NH3 is 0.00224 atm.

Page 555, Sample Exercise 15.11, Chemistry: The Central Science, 7/e

A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448 oC. The value of the equilibrium constant, Kc, for the reaction as written

H2(g) + I2(g) <==> 2 HI(g)

at 448 oC is 50.5. What are the concentrations of H2, I2, and HI in the flask at equilibrium? 

eqn:Kc=(HI+2x)^2/((H2-x)(I2-x))
 Kc=50.5                                  Kc=50.5
 HI=0                                     HI=0
 x=                 Press [SOLVE]==>     .x=.93498392612928
 H2=1                                     H2=1
 I2=2                                     I2=2
 bound={-1E99,1E99}                       bound={-1E99,1E99}

This gives us x, but what we really want are those equilibrium values:
Notice we put the [x-VAR] key here to good use. 

eqn:Kc=(HI+2x)^2/((H2-x)(I2-x))
 Kc=50.5                                  Kc=50.5
 HI=0+2x                                  HI=1.8699678522586
 x=                 Press [SOLVE]==>     .x=-6.5557910050015E-15
 H2=1-x                                   H2=.06501607387072
 I2=2-x                                   I2=1.06501607387072
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Now x is essentially 0, which shows that we have the correct equilibrium values. (If the 'E-15' is not showing on your calculator, press [2nd][>] to jump to the far end of the number to see what the exponent is.) Note that there are really two roots to this equation, but by some miraculous algorithm the TI calculators will always give you the smallest positive root, which is exactly what we want here!

Answer: [HI] = 0+2x = 1.870 M, [H2] = 1-x = 0.06501 M, and [I2] = 2-x = 1.06501 M

Additional Problems That Absolutely Require a Graphing Calculator

Ah, now the cool thing here is that we can do calculations that others can only dream of!
Without an equation-solving calculator, this sort of analysis is basically impossible.
For example, take Sample Exercise 15.10 and give it a twist:

For the Haber process, N2(g) + 3 H2(g) <==> 2 NH3(g) , Kp = 1.45E-5 at 500 oC. Starting with a partial pressure of H2 of 1.0 atm and a partial pressure of N2 of 2.0 atm, what are the partial pressures at equilibrium? 
eqn:Kp=(NH3+2x)^2/((N2-x)(H2-3x)^3)
 Kp=1.4E-5                                Kp=1.4E-5
 NH3=0                                    NH3=0
 x=                 Press [SOLVE]==>     .x=.0026586638154781
 N2=2                                     N2=2
 H2=1                                     H2=1
 bound={-1E99,1E99}                       bound={-1E99,1E99}
check: 
eqn:Kp=(NH3+2x)^2/((N2-x)(H2-3x)^3)
 Kp=1.4E-5                                Kp=1.4E-5
 NH3=0+2x                                 NH3=.00531732763095
 x=                 Press [SOLVE]==>     .x=0
 N2=2-x                                   N2=1.9973413361845
 H2=1-3x                                  H2=.99202400855357
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: [NH3] = 0.00532 atm, [N2] = 1.997 atm, and [H2] = 0.9920 atm.

OK, now what happens if we increase the partial pressure of NH3 to 2.0 atm? 

eqn:Kp=(NH3+2x)^2/((N2-x)(H2-3x)^3)
 Kp=1.4E-5                                Kp=1.4E-5
 NH3=2                                    NH3=2
 x=                 Press [SOLVE]==>     .x=-.97456743083544
 N2=1.9973413361845                       N2=1.9973413361845
 H2=.99202400855357                       H2=.99202400855357
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: The reaction runs backward (x < 0)! Check: 
eqn:Kp=(NH3+2x)^2/((N2-x)(H2-3x)^3)
 Kp=1.4E-5                                Kp=1.4E-5
 NH3=2+2x                                 NH3=0.0508651383291
 x=                 Press [SOLVE]==>     .x=1.4876168166883E-14
 N2=1.9973413361845-x                     N2=2.9719087670199
 H2=.99202400855357-3x                    H2=3.9157263010599
 bound={-1E99,1E99}                       bound={-1E99,1E99}

Final values: [NH3] = 0.0509 atm, [N2] = 2.972 atm, and [H2] = 3.916 atm.

Fine, now DOUBLE ALL THE PRESSURES (by reducing the volume). Let's see what happens: 

eqn:Kp=(NH3+2x)^2/((N2-x)(H2-3x)^3)
 Kp=1.4E-5                                Kp=1.4E-5
 NH3=0.0508651383291*2                    NH3=.1017302766582
 x=                 Press [SOLVE]==>     .x=.047692266897154
 N2=2.9719087670199*2                     N2=5.9438175340398
 H2=3.9157263010599*2                     H2=7.8314526021198
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: Forward again, to reduce the total number of moles of gas. Check: 
eqn:Kp=(NH3+2x)^2/((N2-x)(H2-3x)^3)
 Kp=1.4E-5                                Kp=1.4E-5
 NH3=.1017302766582+2x                    NH3=.19711481045251
 x=                 Press [SOLVE]==>     .x=-2.549999999999E-15
 N2=5.9438175340398-x                     N2=5.8961252671426
 H2=7.8314526021198-3x                    H2=7.6883758014283
 bound={-1E99,1E99}                       bound={-1E99,1E99}

Final values: [NH3] = 0.197 atm, [N2] = 5.896 atm, and [H2] = 7.688 atm.