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 ${NO}_2\left(g\right)$. The resultant pressure is:

• A. $1,2atm$
• B. $2.4atm$
• C. $2.0atm$
• D. $1.0atm$

Answer 1

Answer: (B)

$2.4atm$

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

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 $n_2=0.8+0.4=1.2$

$\frac{P_1}{T_1\times n_1}=\frac{P_2}{T_2\times n_2}$

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

$P_2=2.4$ atm

Hence, the correct option is $\text{B}$