Question
At $T(K)$, the equilibrium constant of
$A_{2}(\text{ }g) + B_{2}(\text{ }g) \rightleftharpoons C(g)$ is $2.7 \times 10^{- 5}$.
What is the equilibrium constant for
$\frac{1}{3}{\text{ }A}_{2}(\text{ }g) + \frac{1}{3}{\text{ }B}_{2}(\text{ }g) \rightleftharpoons \frac{1}{3}C(g)$ at the same temperature?
Options
Solution
Sol. [Key Concept: Raising a reaction's coefficients to $\frac{1}{3}$ raises its equilibrium constant to the power $\frac{1}{3}$: $K' = K^{1/3}$.]
$$A_{2}\left( \text{ }g \right) + B_{2}\left( \text{ }g \right) \rightleftharpoons C(g)\ K_{eq} = 2.7 \times 10^{- 5}$$
$$\begin{matrix} & \frac{1}{3}{\text{ }A}_{2}(\text{ }g) + \frac{1}{3}{\text{ }B}_{e}(\text{ }g) \rightleftharpoons \frac{1}{3}C(\text{ }g){\ \ \ \ \ \ \ \ ;K}_{eq}^{1} = \left( K_{eq} \right)^{1/3} \\ & {\text{ }K}_{eq}^{1} = \left( 2.7 \times 10^{- 5} \right)^{1/3} \\ & \ = 3 \times 10^{- 2} \end{matrix}$$
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