Question

When $N_2{O}_5$ is heated at certain temperature, is dissociates as $N_2{O}_3\left(g\right)\rightleftharpoons N_2{O}_3\left(g\right)+O_2\left(g\right);K_c=2.5$. At the same time $N_2{O}_3$ also decomposes as:
$N_2{O}_3\rightleftharpoons N_{2}O\left(g\right)+O_3\left(g\right)$. If initially $4.0$ moles of $N_2{O}_3$ are taken in $1.0$ litre flask and allowed to dissociate. Concentration of $O_2$ at equilibrium is $2.5M$. Equilibrium concentration of $N_2{O}_5$ is?

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

Answer 1

The correct option is A $1.0M$

$N_2{O}_5\rightleftharpoons N_2{O}_5+O_2$
$4-x$ $x-y$ $x+y$
$N_2{O}_3\rightleftharpoons N_{2}O+O_2$
$x-y$ $y$ $y+x$
$\because \left[O_2\right]=x+y=2.5$
and $2.5=\frac{(x+y)(x-y)}{4-x}$
$\therefore x=2.166$
$\left[N_2{O}_5\right]4-x=1.846$