Question

For decomposition of $N_2{O}_5$, rate constant were determined at two different temperature. At $27{}^{\circ }C,$ rate constant was $2\times {10}^{-4}{\text{sec}}^{-1}$, at $47{}^{\circ }C$ the rate constant is $6\times {10}^{-4}{\text{sec}}^{-1}$.
Then identify the correct statement(s).
(Given $ln3=1.1$)

• A. Activation energy is $10.56\text{cal/mol}$
• B. On increasing temperature, rate of above reaction will increase
• C. Reaction must be exothermic
• D. Rate of reaction is $1.2\times {10}^{-3}{\text{M sec}}^{-1}$ at $\left[N_2{O}_5\right]=2M$ at $47{}^{\circ }C$

Answer 1

Answer: (B), (D)

The correct options are
B On increasing temperature, rate of above reaction will increase
D Rate of reaction is $1.2\times {10}^{-3}{\text{M sec}}^{-1}$ at $\left[N_2{O}_5\right]=2M$ at $47{}^{\circ }C$

(a) $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2{k}_{T_1}=2\times {10}^{-4}{\text{sec}}^{-1}\left(at300K\right)k_{T_2}=6\times {10}^{-4}{\text{sec}}^{-1}\left(at320K\right)ln\frac{k_{T_2}}{k_{T_1}}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})ln\frac{6\times {10}^{-4}}{2\times {10}^{-4}}=\frac{E_a}{R}(\frac{1}{300}-\frac{1}{320})ln3=\frac{E_a}{2}\times \frac{20}{300\times 320}{E}_a=\frac{1.1\times 300\times 320}{10}=10560\text{cal}{E}_a=10.56\text{kcal/mol}$

(b) On increasing temperature rate constant increases so rate of reaction increase.

(c) Increasing the temperature, increases the reaction rate. Hence the reaction is endothermic.

(d) Reaction rate $=k_{reaction}[N_2{O}_5{]}^1\text{reaction at}47{}^{\circ }C=k_{{47}^{\circ }}\times 2=6\times {10}^{-4}\times 2=1.2\times {10}^{-3}{\text{M sec}}^{-1}$