The equilibrium constant for the reaction HI(g)⇌12H2(g)+12I2(g) is 7.4. The value of equilibrium constant of the reaction 2HI(g)⇌H2(g)+I2(g) will be

The equilibrium constant for the reaction $HI (g) ⇌ \frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g)$ is 7.4. The value of equilibrium constant of the reaction $2 HI (g) ⇌ H_{2} (g) + I_{2} (g)$ will be

A
7.4
B
$1 / 7.4$
C
54.76
D
$\sqrt{7.4}$

Solution:

$K_{new} = {(K_{old})}^{2} = {7.42}^{2} = 54.76$

A number that represents the proportionality of the reactant and product concentrations at equilibrium in a reversible chemical reaction at a certain temperature.

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