A mixture consisting of 1 mol of H2O(g) and 1 mol CO(g) is placed in a 7 L reaction vessel at 800 K. At equilibrium, 0.369 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?
...........CO + H2O ==> H2 + CO2
initial....1.....1.......0....0
change.....-x....-x.....x.....x
equil.....1-x....1-x.....x....0.647
You know x = 0.647 which allows you to calculate moles CO, H2O, H2, and CO2 at equil. Then concn = moles/L, and substitute into Keq expression to solve for Keq.
To find the equilibrium constant Kc at 800 K, we need to write the balanced chemical equation and use the equilibrium concentrations of the reactants and products.
The balanced chemical equation for the reaction is:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
According to the given information, the initial moles of CO and H2O are both 1 mol. At equilibrium, 0.369 mol of CO2 is present. Since the stoichiometric coefficient of CO2 in the balanced equation is 1, the equilibrium concentrations of CO and H2O will be (1 - 0.369) = 0.631 mol.
The equilibrium concentration of H2 can be determined using the stoichiometry of the balanced equation. Since the stoichiometric coefficient of H2O is 1, the equilibrium concentration of H2 will also be 0.631 mol.
Now, we can write the expression for Kc using the equilibrium concentrations of the reactants and products:
Kc = [CO2] / ([CO] * [H2O])
Substituting the given values:
Kc = (0.369) / ((0.631) * (0.631))
Calculating this expression will give the value of Kc at 800 K.
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.