Question

When $N_2{O}_5$ is heated at temp. T, it dissociates as $N_2{O}_5\rightleftharpoons N_2{O}_3+O_2,K_c=2.5$. At the same time $N_2{O}_3\rightleftharpoons N_{2}O+O_2$. If initially 4.0 moles of $N_2{O}_5$ are taken in 1.0 litre flask and allowed to attain equilibrium, concentration of $O_2$ was formed to be 2.5 M. Equilibrium concentration of $N_{2}O$ is:

• A. 1
• B. 1.5
• C. 2.166
• D. 0.334

Answer 1

Answer: (D)

0.334

Let x be the degree of dissociation for $N_2{O}_5$ dissociation and y be the degree of dissociation for $N_2{O}_3$ dissociation.
We have the following situation on reaching equilibrium for both the reactions,
$N_2{O}_5\rightleftharpoons N_2{O}_3+O_2$
(4-x) (x) (x)

$N_2{O}_3\rightleftharpoons N_{2}O+O_2$
(x-y) (y) (y)

Now, $x+y=2.5$,

and by definition of $K_c$,

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

On solving for x and y:

we get y = 0.334 M

Therefore, option D is correct.