Question

Calculate the equilibrium composition for the reaction Cgraphite + H2O(g) _<->CO(g) + H2(g) at T = 298 K, 500 K, 1000 K, and 2000 K and a pressure of 1 bar. For the initial number of moles, use 1 mole of graphite, 3 moles of H2O(g), and no CO or H2.

Answer 1

We will calcualte K at different given temperature first

Delta G⁰ = Sum of Delta G of products - Sum of Delta G reactants

DeltaG⁰ = [ DeltaG⁰CO + DeltaG⁰H2] - [ DeltaG⁰C + DeltaG⁰H2O]

             =      [-137.16 + 0 ] -[     0 +-228.61] = 91.45 KJ / mole

              Cgraphite + H2O(g) <->CO(g) + H2(g)

Initial         1               3                  0        0

Change                  -x                 +x      +x

Equilibrium           3-x                  x        x

Kc = x^2 / (3-x)

Delta G0 does not change with temperature

1) at 298 K

Delta G⁰ = -RT ln Kc

91.45 = -8.314 X 10^-3 X T x ln Kc

ln Kc = -36.91

Kc = 9.33 X 10^-17

Kc = x^2 / (3-x)

Kc is very small we can ignore x in denominator

9.33 X 10^-17 = x^2 / 3

x =1.67 X 10^-8 = [CO] = [H2]

[H2O] almost equal to 3

2) 500 K

Delta G⁰ = -RT ln Kc

91.45 = -8.314 X 10^-3 X T x ln Kc

ln Kc = -21.99

Kc = 2.8 X 10^-10

Kc = x^2 / (3-x)

Kc is very small we can ignore x in denominator

2.8 X 10^-10 = x^2 / 3

x = 2.898 X 10^-5= [CO] = [H2]

[H2O] = 3 - 0.00002898 = almost 3

3) 1000 K

Delta G⁰ = -RT ln Kc

91.45 = -8.314 X 10^-3 X T x ln Kc

ln Kc = -10.99

Kc = 1.687 X 10^-5

Kc = x^2 / (3-x)

Kc is very small we can ignore x in denominator

1.687 X 10^-5 = x^2 / 3

x =7.11 X 10^-3= [CO] = [H2]

[H2O] = 3 - 0.00711 = 2.992

4) 2000 K

Delta G⁰ = -RT ln Kc

91.45 = -8.314 X 10^-3 X T x ln Kc

ln Kc = -5.499

Kc = 0.0041

Kc = x^2 / (3-x)

0.0041(3-x) = x^2

0.0123 - 0.0041x = x^2

Solving for x

x = 0.1088 = [CO] = [H2]

[H2O] = 3 - .1088 = 2.891