Be sure to answer all parts. Consider the following equilibrium process at 686°C:CO2 (g) + H2 (g) ⇌ CO (g) + H2O (g) The equilibrium concentrations of the reacting species are [CO] = 0.0520 M, [H2] = 0.0400 M, [CO2] = 0.0810 M, and [H2O] = 0.0360 M. (a) Calculate Kc for the reaction at 686°C. (b) If we add CO2 to increase its concentration to 0.500 mol / L, what will the concentrations of all the gases be when equilibrium is reestablished?

a) Kc = 0.578

b) [CO2] = 0.4764 M

[H2] = 0.0164 M

[CO] = 0.0756 M

[H2O]=0.0596 M

Explanation:

Step 1: Data given

Temperature = 686 °C

The equilibrium concentrations of the reacting species are:

[CO] = 0.0520 M

[H2] = 0.0400 M

[CO2] = 0.0810 M

[H2O] = 0.0360 M

Step 2: The balanced equation

CO2 (g) + H2 (g) ⇌ CO (g) + H2O (g)

For 1 mol CO2 we need 1 mol H2 to produce 1 mol CO and 1 mol H2O

Step 3: Calculate Kc

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

Kc = (0.0520 * 0.0360) / (0.0810 * 0.0400)

Kc = 0.578

b) If we add CO2 to increase its concentration to 0.500 mol / L, what will the concentrations of all the gases be when equilibrium is reestablished?

adding CO2 the equilibrium will shift on the right:

Concentrations at the equilibrium will be:

[CO2] = 0.500 - x

[H2] = 0.0400 - x

[CO] = 0.0520+x

[H2O] = 0.0360+x

0.578 = (0.0520+x) (0.0360+x) / (0.500-x) (0.0400-x)

0.578 = (0.0520+x) (0.0360+x) / (0.02 - 0.500x - 0.0400x + x²)

0.01156 - 0.26588 + 0.578x² = 0.001872 + 0.0880x + x²

x = 0.0236

[CO2] = 0.500 - 0.0236 = 0.4764 M

[H2] = 0.0400 - 0.0236 = 0.0164 M

[CO] = 0.0520+0.0236 = 0.0756 M

[H2O] = 0.0360 + 0.0236 = 0.0596 M