Question

When $N _{2} O _{5}$ is heated at certain temperature, it dissociates as
$N _{2} O _{5}( g ) \rightleftharpoons N _{2} O _{3}( g )+ O _{2}( g ) ; Kc =2 \cdot 5$. At the same time $N _{2} O _{3}$ also decomposes as :
$N _{2} O _{3}( g ) \rightleftharpoons N _{2} O ( g )+ O _{2}( g )$. If initially $4.0$ moles of $N _{2} O _{5}$ are taken in $1.0$ litre flask and allowed to dissociate. Concentration of $O _{2}$ at equilibrium is $2.5 \,M$. Equilibrium concentration of $N _{2} O _{5}$ is:

• A. 1.0 M
• B. 1.5 M
• C. 2.166 M
• D. 1.846 M

Answer 1

Answer: (D)

$\underset{4 - x}{N_2O_5} \rightleftharpoons \underset{x - y}{N_2O_3} + \underset{x + y}{O_2}$
$\underset{x - y}{N_2O_3} \leftrightharpoons \underset{y}{N_2O} + \underset{y + x}{O_2}$
$\because [O_2]= x + y =2 \cdot 5$
for $N _{2} O _{5}, Kc =\left[ N _{2} O _{5}\right]\left[ O _{2}\right] /\left[ N _{2} O _{5}\right]$
and $2.5=\frac{(x+y)(x-y)}{4-x}$
$\therefore x=2.166$
$\left[ N _{2} O _{5}\right]=4- x =1 \cdot 846$