Samples of A (1.0 mol) and B (3.5 mol) are placed in a 1 L container and the following reaction takes place:A(g) + B(g) ⇄ 2C(g) At equilibrium, the concentration of A is 0.6 M. What is the value of K to 2 d.p?
Question
Samples of A (1.0 mol) and B (3.5 mol) are placed in a 1 L container and the following reaction takes place:A(g) + B(g) ⇄ 2C(g) At equilibrium, the concentration of A is 0.6 M. What is the value of K to 2 d.p?
Solution
To solve this problem, we need to use the formula for the equilibrium constant (K) which is the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.
The balanced chemical equation is: A(g) + B(g) ⇄ 2C(g)
At equilibrium, the concentration of A is given as 0.6 M. Since the initial amount of A was 1.0 mol in a 1 L container, the initial concentration of A was 1.0 M. Therefore, the change in concentration of A is 1.0 M - 0.6 M = 0.4 M.
According to the stoichiometry of the reaction, for every 1 mol of A that reacts, 1 mol of B also reacts and 2 mol of C are produced. Therefore, the change in concentration of B is also 0.4 M and the change in concentration of C is 2*0.4 M = 0.8 M.
The equilibrium concentrations of A, B, and C are therefore:
[A] = 0.6 M [B] = 3.5 M - 0.4 M = 3.1 M [C] = 0.8 M
Now we can substitute these values into the expression for K:
K = [C]^2 / ([A][B]) = (0.8)^2 / (0.6*3.1)
Calculating this gives K = 0.35 to 2 decimal places.
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