Question

For the following reaction
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
If initially $25mL$ of $H_2$ and $20mL$ of $I_2$ are present in a container and at equilibrium $30mL$ of $HI$ is formed then calculate equilibrium constant.

Answer 1

The given reaction is
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
$V_0$ $25mL$ $20mL$ $0$
$V_{eq.}$ $25-x$ $20-x$ $2x$
Hete, $V_0$ is the initial volume and $V_{eq}$ is the volume present at equilibrium. Volume of a gas is directly proportional to the number of moles and its molar concentration.
Since, at equilibrium 30mL of HI is formed,
$2x=30mL$
$\therefore$ $x=15mL$
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
$V_{eq.}$ $25-15=10$ $20-15=5$ $30$
Equilibrium constant may be calculated as
$K_c=\frac{{\left[HI\right]}^2}{\left[I_2\right]{\left[H_2\right]}^{}}$
$K_c=\frac{(30{)}^2}{10\times 5}$
$K_c=\frac{900}{50}$
$K_c=18$