At $700 \, K$,the equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \longleftrightarrow 2 HI_{(g)}$ is $54.8$. If $0.5 \, mol \, L^{-1}$ of $HI_{(g)}$ is present at equilibrium at $700 \, K$,what are the concentrations of $H_{2(g)}$ and $I_{2(g)}$,assuming that we initially started with $HI_{(g)}$ and allowed it to reach equilibrium at $700 \, K$?

(N/A) The given reaction is $H_{2(g)} + I_{2(g)} \longleftrightarrow 2 HI_{(g)}$ with $K_c = 54.8$.
Since we started with $HI_{(g)}$,the reaction at equilibrium is $2 HI_{(g)} \longleftrightarrow H_{2(g)} + I_{2(g)}$.
The equilibrium constant for this reverse reaction is $K_c' = \frac{1}{K_c} = \frac{1}{54.8}$.
Let $[H_2] = [I_2] = x \, mol \, L^{-1}$ at equilibrium.
Given $[HI] = 0.5 \, mol \, L^{-1}$.
The equilibrium expression is $K_c' = \frac{[H_2][I_2]}{[HI]^2}$.
Substituting the values: $\frac{x \times x}{(0.5)^2} = \frac{1}{54.8}$.
$x^2 = \frac{0.25}{54.8} \approx 0.004562$.
$x = \sqrt{0.004562} \approx 0.0675 \, mol \, L^{-1}$.
Thus,$[H_2] = [I_2] \approx 0.068 \, mol \, L^{-1}$.