Question

At certain temperature, $K_C=1.0$ for reaction: $N{O}_2\left(g\right)+NO\left(g\right)\rightleftharpoons N_{2}O\left(g\right)+O_2\left(g\right)$ equal moles of $NO$ and $N{O}_2$ are to be placed in $5$ litre container until $N_{2}O$ concentration equilibrium is $0.5$M. How many mole of $(NO+N{O}_2)$ must be placed in the container.

Answer 1

The given reaction is :-

${NO}_2\left(g\right)+NO\left(g\right)\rightleftharpoons N_{2}O\left(g\right)+O_2\left(g\right)$

Initial moles : $M$ $M$ $0$ $0$ (let)

At eqm : $M-m$ $M-m$ $m$ $m$ (let)

Now, given that, at equation concentration of $N_{2}O=0.5M$

No. of moles of $N_{2}O=0.5\times 5=2.5$

Now, $K_C=\frac{\left[N_{2}O\right]\left[O_2\right]}{\left[{NO}_2\right]\left[NO\right]}$

$\Rightarrow 1=\frac{m.m}{(M-m)(M-m)}$

$\Rightarrow \frac{m}{M-m}=1or\frac{m}{M-m}=-1$

$\Rightarrow m=M-morm=-M+m$

$\Rightarrow M=2mor\Rightarrow M=0$ (not possible)

Now, $M=2\times m=2\times 2.5=5moles$

So, Total moles of $(NO+{NO}_2)=5+5=10moles$