Question

When 1.90 mol CO2 and 1.90 mol H2 are placed in a 3.00-L container at 395 ∘C, the following reaction occurs: CO2(g)+H2(g)⇌CO(g)+H2O(g).?

Part A
If Kc = 0.802, what are the concentrations of CO2 in the equilibrium mixture?
Part B
If Kc = 0.802, what are the concentrations of H2 in the equilibrium mixture?
Part C
If Kc = 0.802, what are the concentrations of CO in the equilibrium mixture?
Part D
If Kc = 0.802, what are the concentrations of H2O in the equilibrium mixture?

Answer 1

molarity of CO2 = 1.90 / 3 = 0.633 M

molarity of H2 = 1.90 / 3 = 0.633 M

CO2(g)    +     H2(g)   -------------------->   CO(g)   + H2O(g).

0.633 M           0.633 M                              0                0

0.633 -x            0.633 - x                             x                 x

Kc = [CO][H2O] / [CO2][H2]

0.802 = x^2 / (0.633-x)^2

0.896 = x / 0.633 - x

x = 0.567 - 0.896 x

x = 0.299

Equilibrium concentrations :

[CO] = x = 0.299 M

[H2O] = x = 0.299 M

[CO2] = 0.633 - x = 0.334 M

[H2] = 0.633 - x = 0.334 M