Question

One mole of $N_2{O}_4\left(g\right)$ at $300K$ is kept in a closed container under one atmosphere. It is heated to $600K$ when $20$% by mass of $N_2{O}_4\left(g\right)$ decomposes to ${NO}_2\left(g\right)$. The resultant pressure is:

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

Answer 1

Answer: (B)

$2.4$ atm

Inital total pressure at 300 K is 1 atm. The system is heated to 600 K
$P\propto T$
$P_2=P_1\times \frac{T_2}{T_1}=1atm\times \frac{600K}{300K}=2atm$
Initial total pressure at 600 K is 2 atm.
Mass percent is proportional to mole percent which in turn is proportional to percent of pressure of $N_2{O}_4$ decomposed.
20% of $N_2{O}_4$ decomposed.
20 % of 2 atm is decomposed.
$2atm\times \frac{20}{100}=0.4atm$ is decomposed.
$2atm-0.4atm=1.6atm\text{of}{N}_2{O}_4$ is remaining.
$N_2{O}_4\rightleftharpoons 2N{O}_2$
$2\times 0.4atm=0.8atm$ of $N{O}_2$ is formed.
Total pressure $=1.6atm+0.8atm=2.4atm$