Question

One mole of $N_2{O}_4\left(g\right)$ at $300\text{K}$ and $1.0atm$ is kept in a closed container. It is heated to $600\text{K}$ 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. 1.2 atm
• B. 2.4 atm
• C. 2.0 atm
• D. 1.0 atm

Answer 1

The correct option is B 2.4 atm

Let's assume initial mass of $N_2{O}_4$ is $100g$. Therefore, after $20\%$ decomposition, the mass of $N_2{O}_4$ left will be $80g$.

$\begin{array}{cccc}& N_2{O}_4\left(g\right)& \rightleftharpoons & 2N{O}_2\left(g\right)\\ \text{Initial}& \frac{100}{92}mol& & 0\\ & =1.08mol& & \\ \text{At equilibrium}& \frac{80}{92}\text{mol}& & \frac{20}{46}\text{mol}\\ & =0.86\text{mol}& & =0.43\text{mol}\end{array}$

From ideal gas equation,
At $300\text{K},P_{0}V=n_{0}R{T}_0$
$1\times V=1.08\times R\times 300...\left(1\right)$
At $600\text{K},P_{1}V=n_{1}R{T}_1$
$P_1\times V=(0.86+0.43)\times R\times 600...\left(2\right)$
Dividing equation (2) by (1),
$\frac{P_1}{1}=\frac{1.29\times 600}{1.08\times 300}$
$P_1=\frac{1.29\times 2}{1.08}=2.38\text{atm}\simeq 2.4\text{atm}$