Q43 / 180 · KCET 2012 · Chemistry
2 moles of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})$ is kept in a closed container at $298 \mathrm{~K}$ and under 1 atm pressure. It is heated to $596 \mathrm{~K}$ when $20 \%$ by mass of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})$ decomposes to $\mathrm{NO}_{2}$. The resulting pressure is
- A$1.2 \mathrm{~atm}$
- B$4.8 \mathrm{~atm}$
- C$2.8 \mathrm{~atm}$
- Correct answerD$2.4 \mathrm{~atm}$
Solution
$$
\begin{array}{cc}
\mathrm{N}_{2} \mathrm{O}_{4} & 2 \mathrm{NO}_{2} \\
1 \mathrm{~mol} & 92 \mathrm{~g} \\
2 \mathrm{~mol} & 2 \times 92 \mathrm{~g} \quad \text { Initially } \\
2 \times 92-2 \times 92 \times 0.2 & 2 \times 2 \times 92 \times 0.2 \text { At cqulibrium } \\
=147.2 & =73.6 \mathrm{~g} \\
=1.6 \mathrm{~mol} & =0.8 \mathrm{~mol}
\end{array}
$$
Total moles $=1.6+0.8=2.4$
At constant volume,
$$
\begin{aligned}
\frac{\mathrm{P}_{1}}{\mathrm{n}_{1} \mathrm{~T}_{1}} &=\frac{\mathrm{P}_{2}}{\mathrm{n}_{2} \mathrm{~T}_{2}} \\
\frac{1}{2 \times 298} &=\frac{\mathrm{P}_{2}}{2.4 \times 596} \\
\mathrm{P}_{2} &=\frac{2.4 \times 596}{596} \\
&=2.4 \mathrm{~atm}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.