For the thermal decomposition of N₂O₅(g) at constant volume, the following table can be formed, for the reaction mentioned below. 2 N₂O₅(g) → 2 N₂O₄(g) + O₂(g) [table] Sr.No | Time/s | Total pressure/(atm) ; 1 | 0 | 0.6 ; 2 | 100 | 'x' [/table] x = ... × 10⁻³ atm [nearest integer] Given : Rate constant for the reaction is $4.606 imes10^{-2}\ \mathrm{s^{-1}}$.

Question

For the thermal decomposition of N₂O₅(g) at constant volume, the following table can be formed, for the reaction mentioned below. 2 N₂O₅(g) → 2 N₂O₄(g) + O₂(g) [table] Sr.No | Time/s | Total pressure/(atm) ; 1 | 0 | 0.6 ; 2 | 100 | 'x' [/table] x = ... × 10⁻³ atm [nearest integer] Given : Rate constant for the reaction is $4.606	imes10^{-2}\ \mathrm{s^{-1}}$.

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] First order: $\ln\frac{P_0}{P_A}=kt=4.606=2\ln10\Rightarrow P_A=\frac{0.6}{100}=0.006$ atm. If $2x'$ atm of N₂O₅ decomposes, the total pressure is $P_0-2x'+2x'+x'=P_0+x'$. Here $2x'=0.6-0.006=0.594$, so $x'=0.297$ and $P_t=0.6+0.297=0.897$ atm $=897	imes10^{-3}$ atm. Answer: 897.

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?