Import Question JSON

$\text{Hint: Relations between Equilibrium Constants for a General Reaction and its Multiples.}$ $\text{Step 1:}$ $\text{Given,}$ $\text{HI(g)} \rightleftharpoons \frac{1}{2}\text{H}_2\text{(g)} + \frac{1}{2}\text{I}_2\text{(g)}$ $\text{K}_1 = \frac{[\text{H}_2]^{1/2}[\text{I}_2]^{1/2}}{[\text{HI}]} = 8.0$ $\text{Step 2:}$ $\text{We need to find the equilibrium constant for the following reaction:}$ $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons \text{2HI(g)}$ $\text{K}_2 = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ $\text{We notice that the second reaction is the reverse of the first reaction, but with all coefficients multiplied by 2.}$ $\text{For the reverse reaction, the equilibrium constant is the reciprocal of the original:}$ $\text{K}_{\text{reverse}} = \frac{1}{\text{K}_{\text{forward}}}$ $\text{When we multiply all coefficients in a reaction by a factor } n\text{, the equilibrium constant is raised to the power } n\text{:}$ $\text{K}_{n\times\text{coefficients}} = (\text{K}_{\text{original}})^n$ $\text{For our case, the second reaction is the reverse of the first with all coefficients doubled, so:}$ $\text{K}_2 = \frac{1}{(\text{K}_1)^2} = \frac{1}{(8.0)^2} = \frac{1}{64}$ $\text{Therefore, the equilibrium constant of the reaction H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons \text{2HI(g) is } \frac{1}{64}\text{.}$ $\text{The correct answer is option 2: }\frac{1}{64}$