Consider the following equilibrium.
A(g) + 3 B(g) ⇌ 2 C(g)
If the initial concentrations are [A] = 1.00 M,[B] = 3.00 M,and [C] = 0,at equilibrium it is found that [C] = 0.980 M.Calculate K_c for this reaction.

Question

Quizplus · Accepted Answer

The equilibrium constant $K_c$ for the reaction $A(g) + 3B(g) \rightleftharpoons 2C(g)$ can be calculated using the equilibrium concentrations of the reactants and products. Given that at equilibrium, $[C] = 0.980 \, M$, we can determine the changes in concentrations of $A$ and $B$ based on the stoichiometry of the reaction. For every 2 moles of $C$ produced, 1 mole of $A$ and 3 moles of $B$ are consumed.Since $[C]$ increases by $0.980 \, M$, the change in $[A]$ is $-0.490 \, M$ (half of the change in $[C]$) and the change in $[B]$ is $-3 \times 0.490 \, M = -1.470 \, M$. Therefore, the equilibrium concentrations are:- $[A]_{eq} = 1.00 \, M - 0.490 \, M = 0.510 \, M$- $[B]_{eq} = 3.00 \, M - 1.470 \, M = 1.530 \, M$- $[C]_{eq} = 0.980 \, M$The equilibrium constant $K_c$ is given by:$$K_c = \frac{[C]^2}{[A][B]^3}$$Plugging in the equilibrium concentrations:$$K_c = \frac{(0.980)^2}{(0.510)(1.530)^3} = \frac{0.9604}{0.510 \times 3.5809} = \frac{0.9604}{1.8276} \approx 0.526$$Therefore, the correct answer is $0.526$, which corresponds to choice A.

Consider the Following Equilibrium