Question

When $N_2{O}_5$ is heated at temperature T, it dissociates as $N_2{O}_5\rightleftharpoons N_2{O}_3+O_2,K_e=2.5$. At the time $N_2{O}_3$ also decomposes as: $N_2{O}_3\rightleftharpoons N_{2}O+O^2$. If initially $4$ moles of $N_2{O}_5$ are taken in $1$ litre flask and allowed to attain equilibrium, concentration of $O_2$ was formed to be $2.5M$. Equilibrium concentration of $N_{2}O$ is:

• A. $1.0$
• B. $1.5$
• C. $2.166$
• D. $0.334$

Answer 1

The correct option is D $0.334$

$N_2{O}_5\rightleftharpoons N_2{O}_3+O_2,K_e=2.5$
The equilibrium concentration of $N_2{O}_5,N_2{O}_3$ and $O_2$ are $4-xM$, $x-yM$ and $x+yM$ respectively.

Here, $x$ is the concentration of dissociated $N_2{O}_5$ and $y$ is the concentration of dissociated $N_2{O}_3$.

But $\left[O_2\right]=2.5M$.

Hence, $x+y=2.5$ and $x-y=2x-2.5$.

The expression for the equilibrium constant becomes $K_c=2.5=\frac{(2x-2.5)\times 2.5}{4-x}$
Hence, $x=2.167$ and $\left[N_{2}O\right]=y=2.5-2.167=0.334$