What will be the equilibrium constant at 717K for the reaction
`2HI(g) hArr H_2(g) +I_2(g)`
if K for`H_2(g) +I_2(g) hArr 2HI` at 717 K is 50.

Explanation

To find the equilibrium constant (K) for the reaction \(2HI(g) \rightleftharpoons H_2(g) + I_2(g)\) at 717 K, given that the equilibrium constant for the reverse reaction \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\) is 50, we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The given reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]
The equilibrium constant for this reaction at 717 K is:
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