Question

4.5 moles each of hydrogen and iodine are heated in a sealed 10-liter vessel. At equilibrium 3.0 moles of HI were found.

The equilibrium constant for $H_2\left(g\right)$ + $I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is:

• A. 1
• B. 10
• C. 3
• D. 0.33

Answer 1

The correct option is A 1

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

The following table lists the equilibrium concentrations of various species.

	$H_2$	$I_2$	$HI$
Initial	4.5	4.5	0
Change	-x	-x	2x
Equilibrium	4.5-x	4.5-x	2x

But for [HI], $2x=3$, So $x=1.5$

Hence, the equilibrium number of moles of hydrogen, iodine, and HI are 3,3 and 3 respectively.

The total volume is 10 L.

Hence, the equilibrium concentrations of hydrogen, iodine, and HI are 0.3M,0.3M and 0.3M respectively.

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

Substitute values in the above expression.

$K_c=\frac{(0.3{)}^2}{\left(0.3\right)\left(0.3\right)}=1$.