Question

When $N_2{O}_3$ is heated at temp. $T$ it dissociates as $N_2{O}_5\to N_2{O}_3+O_2,K_c=2.5$. At the same time $N_2{O}_3$ also decomposes as : $N_2{O}_3\to N_{2}O+O_2$. If initially $4.0$ moles of $N_2{O}_3$ are taken in $1.0litre$ 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 $0.334$

$N_2{O}_5\rightleftharpoons N_2{O}_3+O_2{N}_2{O}_3\rightleftharpoons N_{2}O+O_2$

Initial $4$ $0$ $0$ $x$ $0$ $0$

At eq'b $4-x$ $x-y$ $x+y$ $x-y$ $y$ $x+y$

$x,y$ concentrations of $N_2{O}_5,N_2{O}_3$ decomposed

Now, $K_2=2.5$ & $x+y=2.5=\left[O_2\right]\to \left(1\right)$

$\frac{(x-y)(x+y)}{4-x}=2.5$

$\frac{(x-y)\left(2.5\right)}{(4-x)}=2.5$

$x-y=4-x$

$\Rightarrow 2x-y=4\to \left(2\right)$

Solving (1) & (2)

$x=2.166$

$y=0.33$

Equilibrium concentration of $N_{2}O=y=0.33$