OnlineCBT
Chemistry · EQUILIBRIUM
हिन्दीFor the reaction \( \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g}) \) at \( 731\text{ K} \), initial concentrations are \( [\text{H}_2] = 2.40 \times 10^{-2}\text{ mol L}^{-1} \) and \( [\text{I}_2] = 1.38 \times 10^{-2}\text{ mol L}^{-1} \). If at equilibrium the concentration of \( \text{I}_2 \) is found to be \( 0.12 \times 10^{-2}\text{ mol L}^{-1} \), what is the calculated value of \( K_c \)?
Correct answer · Option 4
46.4
Explanation
Change in \( [\text{I}_2] \) = \( (1.38 - 0.12) \times 10^{-2} = 1.26 \times 10^{-2}\text{ M} \). Thus at equilibrium, \( [\text{H}_2] = 2.40 \times 10^{-2} - 1.26 \times 10^{-2} = 1.14 \times 10^{-2}\text{ M} \), and \( [\text{HI}] = 2 \times 1.26 \times 10^{-2} = 2.52 \times 10^{-2}\text{ M} \). \( K_c = \frac{(2.52 \times 10^{-2})^2}{(1.14 \times 10^{-2})(0.12 \times 10^{-2})} = \frac{6.3504 \times 10^{-4}}{1.368 \times 10^{-5}} \approx 46.4 \).
