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 0.7 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 0.7 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 the start of the reaction, we have 1.0 mol of A and 3.5 mol of B in a 1 L container. Therefore, the initial concentrations of A and B are 1.0 M and 3.5 M respectively.
At equilibrium, the concentration of A is given as 0.7 M. The change in concentration of A from the start of the reaction to equilibrium is 1.0 M - 0.7 M = 0.3 M.
Since the stoichiometric coefficient of A in the balanced chemical equation is 1, this means that 0.3 M of B has also reacted and 0.6 M of C has been formed (because 2 moles of C are formed for every mole of A that reacts).
Therefore, at equilibrium, the concentrations of A, B and C are 0.7 M, 3.5 M - 0.3 M = 3.2 M, and 0.6 M respectively.
Substituting these values into the formula for K gives:
K = [C]^2 / ([A][B]) = (0.6)^2 / (0.7 * 3.2) = 0.36 / 2.24 = 0.16
Therefore, the value of K to 2 decimal places is 0.16.
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