A mixture consisting of 1 mol of H2O(g) and 1 mol CO(g) is placed in a 11 L reaction vessel at 800 K. At equilibrium, 0.422 mol CO2(g) is present as a result of the reaction
CO(g) + H2O(g)<->CO2(g) + H2(g).
What is Kc at 800 K?
(H2O) = 1/11 = 0.0909M
(CO) = 1/11 = 0.0909
(0.422/11) = 0.0384
......CO(g) + H2O(g)<->CO2(g) + H2(g)
I...0.0909...0.0909.....0
C....-x........-x.......x
E....0.0909-x.0.0909-x..x
The problems tells you that x = 0.0384.
Use that to determine (CO) and (H2O) ast equilibrium, substitute into Kc expression and solve for Kc.
To find the value of Kc, the equilibrium constant, we need to write down the balanced chemical equation and then use the given information to calculate the concentrations of the reactants and products at equilibrium.
The balanced chemical equation for the reaction is:
CO(g) + H2O(g) <-> CO2(g) + H2(g)
From the given information, we know that initially, the reaction mixture consists of 1 mol of CO(g) and 1 mol of H2O(g). We can assume that there are no products (CO2(g) and H2(g)) initially.
At equilibrium, we are told that there is 0.422 mol of CO2(g) present. However, we have to determine the concentration of H2(g) since it's not given.
To determine the equilibrium concentrations, we can use the stoichiometric coefficients of the balanced equation.
The stoichiometry tells us that for every 1 mol of CO(g) that reacts, 1 mol of CO2(g) is formed. Therefore, at equilibrium, the concentration of CO(g) will be 1 - 0.422 = 0.578 mol/L.
Similarly, since 1 mol of CO(g) reacts with 1 mol of H2O(g) to produce 1 mol of CO2(g) and 1 mol of H2(g), at equilibrium, the concentration of H2(g) will also be 0.422 mol/L.
Now that we have the equilibrium concentrations of CO(g), H2(g), and CO2(g), we can plug these values into the equilibrium expression (Kc) to calculate its value.
Kc = [CO2(g)]/[CO(g)] * [H2(g)]/[H2O(g)]
Kc = (0.422 mol/L) / (0.578 mol/L) * (0.422 mol/L) / (1 mol/L)
Simplifying the expression, we get:
Kc = 0.422^2 / (0.578 * 1)
Kc = 0.181
Therefore, Kc at 800 K for the given reaction is approximately 0.181.