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Chemistry · EQUILIBRIUM
हिन्दीAt \( 731\text{ K} \), if the value of \( K_c \) for \( \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g}) \) is \( 46.4 \), what is the value of \( K_c \) for the reaction \( 3\text{HI}(\text{g}) \rightleftharpoons \frac{3}{2}\text{H}_2(\text{g}) + \frac{3}{2}\text{I}_2(\text{g}) \)?
Correct answer · Option 1
\( 3.16 \times 10^{-3} \)
Explanation
The reaction is inverted and multiplied by \( 3/2 \). Therefore, \( K_c' = (K_c)^{-3/2} = (46.4)^{-1.5} = \frac{1}{(46.4)^{1.5}} = \frac{1}{46.4 \times \sqrt{46.4}} \approx \frac{1}{46.4 \times 6.812} = \frac{1}{316.06} \approx 3.16 \times 10^{-3} \).
