Question

The value of equilibrium constant of the reaction HI(g)⇌12H2(g)+12I2 is 8.0. The equilibrium constant of the reaction H2(g)+I2(g)⇌2HI(g) will be :

Answer 1

value of equilibrium constant will be 1/8.

That is recipocal of the equilibrium constant in the first reaction.

Answer 2

The equilibrium constant of the reaction $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$ will be 0.125.

Explanation:

Equilibrium constant for a reaction is the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

For the given reaction: $2HI\rightleftharpoons H_2+I_2$

Equilibrium constant is given as:

$K_{eq}=\frac{[H_2]^1[I_2]^1}{[HI]^2}$

$8.0=\frac{[H_2]^1[I_2]^1}{[HI]^2}$

When the above reaction is reversed, the reverse equilibrium constant will be the reciprocal of the forward equilibrium constant.

$H_2+I_2\rightleftharpoons 2HI$

$K_eq'=\frac{1}{K_{eq}}=\frac{1}{8.0}=0.125$

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