Question

The half-life periods of a first order reaction at $300\, K$ and $400\, K$ are $50\, s$ and $10\, s$ respectively. The activation energy of the reaction in $kJ \; mol^{-1}$ is (log 5 = 0.70)

• A. 4
• B. 8
• C. 16.1
• D. 20.1

Answer 1

Answer: (C)

16.1

Answer 2

Given, For first order reaction, (i) half-life $\left(t_{1 / 2}\right)$ at temperature $300\, K =50\, s$ (ii) half-life $\left(t_{1 / 2}\right)$ at temperature $400\, K =10\, s$ $\therefore K_{1}($ at $300\, K )=\frac{0.693}{50}=0.014 \,s ^{-1}$ and $K_{2}($ at $400 \,K )=\frac{0.693}{10}=0.07\, s ^{-1}$ $\left[\because K=0.693 / t_{(1 / 2)}\right.$ for lst order reaction $]$ where, $K_{1}$ and $K_{2}$ are rate constant at $300 \,K$ and $400\, K$ respectively. Also, according to Arrhenius theory $\log \frac{K_{2}}{K_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} \cdot T_{2}}\right]$ where, $E_{a}=$ activation energy $T_{1} =$ temperature $(300\, K )$ $T_{2} =$ temperature $(400\, K )$ $R =$ gas constant $\left(8.314\, J \,mol ^{-1}\right)$ Thus, $\log \frac{0.07}{0.014}=\frac{E_{a}}{2.303 \times 8.314}\left[\frac{400-300}{400 \times 300}\right]$ or, $\log\, 5=\frac{E_{a}}{2.303 \times 8.314}\left[\frac{100}{400 \times 300}\right]$ or, $0.70=\frac{E_{a}}{19.15 \times 1200}$ or, $E_{a}=0.7 \times 19.15 \times 1200$ $=16086 \,J$ $=16.08 \,kJ \,mol ^{-1}$ $E_{a} \approx 16.10 \,kJ \,mol ^{-1}$