Consider this equilibrium process at 686°C.

CO2(g) + H2(g) CO(g) + H2O(g)
The equilibrium concentrations of the reacting species are [CO] = 0.050 M, [H2] = 0.045 M, [CO2] = 0.086 M, and [H2O] = 0.040 M.

(a) Calculate Kc for the reaction at 686°C.

(b) If the concentration of CO2 to raised to 0.40 mol/L by the addition of CO2, what would be the concentrations of all the gases when equilibrium is reestablished?

Can you do the Kc; i.e., the first part?

Yes, the answer Kc=0.517

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

E......0.086....0.045.....0.050...0.040
---------------------------------------
I.......0.400....0.045...0.050...0.040
C.........-x......-x.......+x.....+x
E.....0.400-x..0.045-x..0.050-x..0.040+x

Substitute the E line into Keq expression and solve. Post your work if get stuck.

To calculate the equilibrium constant (Kc) for the given reaction, we need to use the concentrations of the reactants and products at equilibrium. The general formula for Kc is:

Kc = [C]^c * [D]^d / [A]^a * [B]^b

Where A, B, C, and D represent the reactants and products in the balanced equation, and a, b, c, and d represent their respective stoichiometric coefficients.

(a) To calculate Kc, we need to determine the exponents for each species based on their stoichiometric coefficients. Looking at the balanced equation:

CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

We can see that the stoichiometric coefficients of CO2, H2, CO, and H2O are 1, 1, 1, and 1, respectively. Therefore, the exponents for these species will be 1, 1, 1, and 1, respectively.

Now, substitute the given equilibrium concentrations into the expression for Kc:

Kc = ([CO]^1 * [H2O]^1) / ([CO2]^1 * [H2]^1)

Kc = (0.050 * 0.040) / (0.086 * 0.045)

Kc = 0.002 / 0.00387

Kc ≈ 0.516

So, the value of Kc for the reaction at 686°C is approximately 0.516.

(b) If the concentration of CO2 is increased to 0.40 mol/L, the new equilibrium concentrations of all gases can be determined using the initial concentrations and changes based on the stoichiometry of the reaction.

First, let's calculate the change in concentration of CO and H2O:

Change in [CO] = (0.086 - 0.050) = 0.036 M
Change in [H2O] = (0.045 - 0.040) = 0.005 M

Since the stoichiometric coefficient of CO2 in the reaction is 1, the change in [CO2] will be equal to the change in [CO]:

Change in [CO2] = Change in [CO] = 0.036 M

To find the change in [H2], we use the stoichiometric coefficient of H2 in the reaction, which is also 1. Therefore:

Change in [H2] = Change in [H2O] = 0.005 M

To find the new equilibrium concentrations, add the changes to the initial concentrations:

[CO] = 0.050 M + 0.036 M = 0.086 M
[H2] = 0.045 M + 0.005 M = 0.050 M
[CO2] = 0.086 M + 0.036 M = 0.122 M
[H2O] = 0.040 M + 0.005 M = 0.045 M

So, when equilibrium is reestablished after adding CO2, the concentrations of the gases would be [CO] = 0.086 M, [H2] = 0.050 M, [CO2] = 0.122 M, and [H2O] = 0.045 M.