Question

$1$mole of $H_2$, $2$moles of $I_2$ and $3$ moles of $HI$ were taken in a 1 litre flask. The equilibrium constant of the reaction $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is $50$ at ${330}^{\circ }C$.The concentration of $HI$ at equilibrium is ($moles.li{t}^{-1}$):

• A. $0.3$
• B. $1.3$
• C. $4.4$
• D. $2.7$

Answer 1

Answer: (C)

$4.4$

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$.

Initial	1	2	3
Change	$-x$	$-x$	$2x$
Equilibrium	$1-x$	$2-x$	$3+2x$

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.

$50=\frac{(3+2x{)}^2}{(1-x)(2-x)}$ or $x=0.7$.

Thus, the concentration of $HI$ at equilibrium is $3+2x=3+1.4=4.4moles.li{t}^{-1}$.