Q9 / 125 · BITSAT 2009 · Chemistry
If the equilibrium constant of the reaction $2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}$ is $0.25,$ then the equilibrium constant for the reaction $\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}$ would be
- A1
- B2
- C3
- Correct answerD4
Solution
When the reaction is reversed,
$$
\mathrm{K}^{\prime}=\frac{1}{\mathrm{~K}}=\frac{1}{0.25}=4
$$
Hence, the correct answer is option: 'D'.