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 $\mathrm{HI}\left(\mathrm{g}\right)⇌\frac{1}{2}{\mathrm{H}}_{2}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{I}}_{2}\left(\mathrm{g}\right)$  is 7.4. The value of equilibrium constant of the reaction $2\mathrm{HI}\left(\mathrm{g}\right)⇌{\mathrm{H}}_{2}\left(\mathrm{g}\right)+{\mathrm{I}}_{2}\left(\mathrm{g}\right)$ will be

1. A

7.4

2. B

$1/7.4$

3. C

54.76

4. D

$\sqrt{7.4}$

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### Solution:

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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