Question

A mixture of hydrogen and iodine in the mole ration $1.5:1$ is maintained at ${450}^{o}C$. After the attainment of equilibrium $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$, it is found on analysis that the mole ratio of $I_2$ to $HI$ is $1:18$. Calculate the equilibrium constant and the number of moles of each species present under equilibrium, if initially, $127grams$ of iodine were taken.

Answer 1

R.E.F image

At

$t=0\frac{3}{2}x$ $x$

$t=req$ $\frac{3}{2}x-y$ $x-y$ $2y$

given at $t=teg\frac{I_2}{HI}=\frac{1}{18}$

$\frac{x-y}{2y}=\frac{1}{18}$

$9x-9y=y$

$9x=10y$

$\therefore \frac{x}{y}=\frac{10}{9}$

At, $t=0$

$x=127gm$

i.e $0.5$ moles of Iodine

$\therefore y=0.45moles$

Keq $=\frac{(2y{)}^2}{(\frac{3}{2}x-y)(x-y)}$

$=\frac{4\times (\frac{9x}{10}{)}^2}{(\frac{3x}{2}-\frac{9x}{10})(x-\frac{9x}{10})}$

$=\frac{4\times 51x^2}{100\times \frac{6x}{10}\times \frac{x}{10}}$

$=54$

$Keq=54$

moles of $H_2$ at equilibrium $=\frac{3}{2}x-y$

$\Rightarrow \frac{3}{2}\times 0.5-0.45=0.30$

$r_1{H}_{2}atequilibrium=0.30$

moles of $I_2$ at equilibrium $=x-y$

$\Rightarrow 0.5-0.45=0.05$

$r_1{I}_{2}atequilibrium=0.05$

moles of HI at equilibrium $=2y$

$\Rightarrow 2\times 0.45=0.90$

$r_{1}HIatequilibrium=0.90$