Question

4 moles of $HI$ is taken in a liter closed vessel and heated to equilibrium. At equilibrium, the concentration of $H_2$ is one $molli{t}^{-1}$.

Then the equilibrium constant is:

• A. $4$
• B. $0.5$
• C. $2$
• D. $0.25$

Answer 1

Answer: (D)

$0.25$

The equilibrium reaction is $H_2\left(g\right)$ + $I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$.

The number of moles of various species presents initially, and at equilibrium, are given in the following table.

	HI	$H_2$	$I_2$
Initial	4	0	0
change	-2	1	1
equilibrium	4-2=2	1	1

The expression for the equilibrium constant is $K_c=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}$.

Substitute values in the above expression.

$K_c=\frac{\left[H_2\right]\left[I_2\right]}{[HI{]}^2}=\frac{1\times 1}{2^2}=0.25$.

Therefore, the equilibrium constant is 0.25