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Chemistry · EQUILIBRIUM
हिन्दीIn a \( 1.0\text{ L} \) vessel at \( 731\text{ K} \), pure \( \text{HI}(\text{g}) \) is introduced such that its initial concentration is \( 3.04 \times 10^{-2}\text{ mol L}^{-1} \). At equilibrium, the concentration of \( \text{H}_2(\text{g}) \) is \( 0.345 \times 10^{-2}\text{ mol L}^{-1} \). What is the equilibrium constant \( K_c' \) for the dissociation reaction: \( 2\text{HI}(\text{g}) \rightleftharpoons \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \)?
Correct answer · Option 2
0.0215
Explanation
From stoichiometry, \( [\text{I}_2] = [\text{H}_2] = 0.345 \times 10^{-2}\text{ M} \). \( [\text{HI}]_{\text{eq}} = 3.04 \times 10^{-2} - 2(0.345 \times 10^{-2}) = 2.35 \times 10^{-2}\text{ M} \). \( K_c' = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} = \frac{(0.345 \times 10^{-2})^2}{(2.35 \times 10^{-2})^2} = (\frac{0.345}{2.35})^2 \approx (0.1468)^2 \approx 0.0215 \).
