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Given: Equilibrium constant (K1) for the reaction: HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g); K1 = 8. To find equilibrium constant for the following reaction: H2(g) + I2(g) ⇌ 2HI(g); K2 = ?

1. 1/64
2. 64
3. 8^2 = 64
4. K2 = 1/64

Correct answer: 8^2 = 64

Solution

The given reaction is the reverse of the original reaction and its stoichiometry is doubled. The equilibrium constant for the reverse reaction is the reciprocal of K1, and doubling the stoichiometry squares the constant. Thus, K2 = (1/K1)^2 = (1/8)^2 = 64.

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