Question

The value of equilibrium constant of the reaction $HI(g) \rightleftharpoons \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g)$ is 8.0. The equilibrium constant of the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be:
(a) $\frac{1}{16}$
(b) $\frac{1}{64}$
(c) 16
(d) $\frac{1}{8}$

Answer 1

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Answer 2

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