Question

When $N_2{O}_5$ is heated at a certain temperature, it dissociates as:
$N_2{O}_5\left(g\right)\rightleftharpoons N_2{O}_3\left(g\right)+O_2\left(g\right);K_c=2.5$. At the same time $N_2{O}_3$ also decomposes as:
$N_2{O}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+O_2\left(g\right);$. If 4.0 moles of $N_2{O}_5$ are taken initially in a 1.0 litre flask and allowed to dissociate, the concentration of $O_2$ at equilibrium is 2.5 M. The equilibrium concentration of $N_2{O}_5$ is :

• A. 1.00 M
• B. 1.50 M
• C. 2.17 M
• D. 1.87 M

Answer 1

The correct option is D 1.87 M

$N_2{O}_5\rightleftharpoons N_2{O}_3+O_2$
4-x x-y x+y
$N_2{O}_3\rightleftharpoons N_{2}O+O_2$
x-y y x+y
$x+y=\left[O_2\right]=2.5$
$2.5=\frac{(x+y)(x-y)}{4-x}x=2.166\left[N_2{O}_5\right]=4-x=1.87$