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The value of equilibrium constant of the reaction HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g) is 8.0. The equilibrium constant of the reaction H2(g) + I2(g) ⇌ 2HI(g) will be:

1. 1/16
2. 1/64
3. 16
4. 1/8

Correct answer: 16

Solution

The equilibrium constant for the reverse reaction is the reciprocal of the given constant, and since the stoichiometry is doubled, the new constant is the square of the reciprocal: (1/8)^2 = 16.

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