Question

If in the reaction $N_2{O}_4\leftrightharpoons 2N{O}_2$, $x$ is that part of $N_2{O}_4$ which dissociates, then the number of moles at equilibrium will be :

• A. $1$
• B. $3$
• C. $(1+x)$
• D. $(1+x{)}^2$

Answer 1

Answer: (C)

$(1+x)$

Let, 1 mol of $N_2{O}_4$ present initially.

$N_2{O}_4\leftrightharpoons 2N{O}_2$

$t=0:$ $1$ $0$

$t=t_{eq}:$ $1-x$ $2x$

So, the number of molecules at equilibrium $=n_{N_2{O}_4}+n_{N{O}_2}=1+x-2x=1+x$.

So, option C is the correct answer.