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 0.647 mol CO2 (g) is present as a result of the reaction:
CO(g) + H2O (g) --> CO2(g) + H2
What is K(equilibrium constant) at 800 K?
DrBob222 the correct answer is Kc = 69 moles per solution of pk
To determine the equilibrium constant (K) at 800 K for the given reaction:
Step 1: Write down the balanced equation for the reaction:
CO(g) + H2O(g) --> CO2(g) + H2(g)
Step 2: Write down the expression for the equilibrium constant:
K = [CO2(g)][H2(g)] / [CO(g)][H2O(g)]
Step 3: Determine the concentrations of the substances at equilibrium:
Given that 0.647 mol of CO2(g) is present at equilibrium, and since 1 mol of CO(g) and H2O(g) were initially mixed, we can assume that the number of moles of CO(g) and H2O(g) present at equilibrium is (1 - 0.647) mol = 0.353 mol.
Since H2(g) is a product, we do not know its concentration at equilibrium. However, we can determine it using the stoichiometry of the reaction.
Step 4: Calculate the concentration of H2(g):
Since 1 mol of CO(g) reacts with 1 mol of H2O(g) to form 1 mol of H2(g), and the initial concentration of H2(g) is 0 mol, at equilibrium the concentration of H2(g) will be 0.353 mol.
Step 5: Calculate the equilibrium constant:
Substituting the concentrations into the equilibrium constant expression, we get:
K = [0.647][0.353] / [1][1] = 0.2276
Therefore, the equilibrium constant (K) at 800 K is approximately 0.2276.
To find the equilibrium constant, K, we need to know the balanced equation for the reaction equation and the concentrations (or pressures) of the reactants and products at equilibrium.
The balanced equation for the given reaction is:
CO(g) + H2O(g) → CO2(g) + H2(g)
We are given that at equilibrium, 0.647 mol CO2(g) is present. However, we still need to determine the concentrations of CO(g), H2O(g), and H2(g) at equilibrium.
Since the reaction vessel has a volume of 11 L, we can assume this is the total volume at equilibrium.
Given that initially, 1 mol of H2O(g) and 1 mol of CO(g) were present and assuming no reaction has occurred before equilibrium, the total moles of the gases are 1 + 1 + 0.647 = 2.647 moles.
To find the equilibrium concentrations, we divide the moles of each gas by the total volume:
[CO] = (0.647 mol) / (11 L)
[H2O] = (1- 0.647 mol) / (11 L)
[CO2] = (0.647 mol) / (11 L)
[H2] = (0.647 mol) / (11 L)
Now we can plug the concentrations into the equilibrium expression:
K = ([CO2] x [H2]) / ([CO] x [H2O])
Substituting the values we found earlier:
K = [(0.647 mol/L) x (0.647 mol/L)] / [(0.647 mol/L) x (1 - 0.647 mol/L)]
Simplifying further:
K = (0.4187 mol^2/L^2) / (0.2359 mol^2/L^2)
Finally, calculating the value:
K ≈ 1.77 (rounded to two decimal places)
Therefore, the equilibrium constant, K, at 800 K is approximately 1.77.
...........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.