To determine the equilibrium constant (Kc) at 800 K, you need to use the provided information about the initial and equilibrium states of the reaction. The equilibrium constant is defined as the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.
First, let's write the balanced chemical equation for the reaction:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Next, let's analyze the given information in terms of initial and equilibrium concentrations:
Initial:
[H2O(g)] = 1 mol
[CO(g)] = 1 mol
At equilibrium:
[CO2(g)] = 0.369 mol
Now, let's assume the equilibrium concentrations of H2(g) is "x" (in mol) since it is not given.
Using the equation for the equilibrium constant, Kc, we have:
Kc = ([CO2(g)] * [H2(g)]) / ([CO(g)] * [H2O(g)])
Plugging in the given values and "x" as the concentration of H2(g):
Kc = (0.369 * x) / (1 * 1)
Now, the volume of the reaction vessel is given as 7 L, which allows us to relate the concentrations to the molar amounts using the ideal gas law equation:
PV = nRT
Since the reaction occurs at 800 K, we can use the ideal gas constant R = 0.082 L·atm/(mol·K).
For CO2(g),
[P(CO2(g))] * V = n(CO2(g)) * R * T
(0.369 / 7) * V = 0.369 * 0.082 * 800
For CO(g),
[P(CO(g))] * V = n(CO(g)) * R * T
(1 / 7) * V = 1 * 0.082 * 800
For H2(g),
[P(H2(g))] * V = n(H2(g)) * R * T
(P(H2(g)) / 7) * V = x * 0.082 * 800
Now, we can rearrange the equations to solve for the partial pressures of CO2(g) and CO(g) in terms of V:
P(CO2(g)) = (0.369 / 7) * (0.369 * 0.082 * 800)
P(CO(g)) = (1 / 7) * (1 * 0.082 * 800)
Substituting these values and the initial concentration of H2O(g) into the Kc equation:
Kc = (0.369 * x) / (1 * 1)
Kc = (0.369 * x) / (1)
Kc = 0.369 * x
Thus, the equilibrium constant, Kc, at 800 K for the given reaction is 0.369.