Question

One mole of $N_2{O}_4\left(g\right)$ at $300K$ is left in a closed container under one atm. It is heated to $600K$ when $20\%$ by mass of $N_2{O}_4\left(g\right)$ decomposes to $N{O}_2\left(g\right)$. The resultant pressure is:

• A. $1.2atm$
• B. $2.4atm$
• C. $2.0atm$
• D. $1.0atm$

Answer 1

Answer: (B)

$2.4atm$

Given:

Decomposition of $N_2{O}_4$ is $20\%$ by mass.

1 mole $\Rightarrow$ 28g

mass decomposed $=28\times \frac{20}{100}=5.6g$

mass left after decomposition $=28-5.6=22.4$

No. of moles of $N_2{O}_4$ left$=\frac{22.4}{28}=0.8mole$

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

	$N_2{O}_4$	$N{O}_2$
Initial moles	$1$	$0$
Change moles	$-0.2$	$0.4$
Equilibrium pressure	$0.8$	$0.4$

Total number of moles $=0.8+0.4=1.2$

Pressure at $300K$ is $1$ atm.

Pressure at $600K$ (without considering decomposition) will be $P_2=T_2\times \frac{P_1}{T_1}=600\times \frac{1}{300}=2$ atm.

Total pressure (when decomposition is considered) $=2\times \frac{\text{Final total number of moles}}{\text{Initial total number of moles}}=2\times \frac{1.2}{1}=2.4$ atm.