Question

1.50 mol each of hydrogen and iodine were placed in a sealed 10 L container maintained at 717 K. At equilibrium 1.25 mol each of hydrogen and iodine were left behind. The equilibrium constant, $K_c$ for the reaction
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)at717K$ is

• A. 0.40
• B. 0.16
• C. 25
• D. 50

Answer 1

The correct option is B 0.16

$\begin{array}{cccccc}& H_2\left(g\right)& +& I_2\left(g\right)& \rightleftharpoons & 2HI\left(g\right)\\ \text{At initial}& 1.5mol& & 1.5mol& & 0\\ \text{At equilibrium}& 1.25mol& & 1.25mol& & 0.5mol\end{array}$
At equilibrium 1.25 mol each of hydrogen and iodine are left behind. So amount of hydrogen and iodine reacted = 1.5 - 1.25 = 0.25 mol.
As per stochiometry, one mol of hydrogen and 1 mol of iodine reacts to form 2 mol of HI.
So 0.25 mol of hydrogen and iodine each will form 0.5 mol of HI
$\left[H_2\right]=\frac{1.25}{10},\left[I_2\right]=\frac{1.25}{10},\left[HI\right]=\frac{0.5}{10}$
$K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{(0.5/10{)}^2}{\frac{1.25}{10}\times \frac{1.25}{10}}$
$K_c=\frac{(0.5{)}^2}{(1.25{)}^2}=\frac{0.25}{1.5625}=0.16$