Question

Reaction between Nitrogen and Oxygen takes place as follows:-
$2N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N_{2}O\left(g\right)$
If a mixture of $0.482mol$ of $N_2$ and $0.933mol$ of $O_2$ is placed in a reaction vessel of volume $10L$ and allowed to form $N_{2}O$ at a temperature for which $Ke=2.0\times {10}^{-37}$, determine the composition of the equilibrium mixture.

Answer 1

$2N_2+O_2\rightleftharpoons 2N_{2}O\left(g\right)$

initial concentration of $N_2\left[N_2\right]=\frac{0.482}{10}$

$=0.0482$

$\left[O_2\right]=\frac{0.933}{2}=0.0933$

$2N_2+O_2\rightleftharpoons 2N_{2}O$

$0.0482-2x0.0933-x$ $2x$

$2\times {10}^{-37}=\frac{[2x{]}^2}{(0.0482-2x{)}^2(0.0933-x)}$

$2\times {10}^{-37}=\frac{4x^2}{\left(0.0482{)}^2\right(0.0933)}$

$x=0.0329\times {10}^{-19}$

$=3.29\times {10}^{-21}$

$\left[N_{2}O\right]=2\times 3.29\times {10}^{-21}=6.58\times {10}^{-21}$

$\left[N_2\right]\simeq 0.0482$

$\left[O_2\right]\simeq 0.0933$