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 ${NO}_2\left(g\right)$. 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

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

Moles of unreacted $N_2{O}_4=1(1-0.2)=0.8$

Moles of $N{O}_2=2\times 0.2$

$=0.4$

Total moles $=0.8+0,4$

$=1.2$

$\frac{P_1}{T_1{n}_1}=\frac{P_2}{T_2{n}_2}$

$\frac{1}{300\times 1}=\frac{P_2}{600\times n_2}$

$\frac{1}{300}=\frac{P_2}{600\times 1.2}$

$P_2=2.4atm$

So (B) option is correct.