Question

A first order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. The activation energy of the reaction ($E_a$) is: [R = 314 J K$^{-1}$ mol$^{-1}$, log 2 = 3010; log 3 = 4771]

• A. $75.2\text{ kJ mol}^{-1}$
• B. $43.8\text{ kJ mol}^{-1}$
• C. $23.7\text{ kJ mol}^{-1}$
• D. $52.5\text{ kJ mol}^{-1}$

Hint:

Since rate constant $k$ is inversely proportional to half-life, a decrease in half-life from 30 to 10 minutes means the reaction rate tripled ($k_2/k_1 = 3$). Plugging $\log 3 \approx 0.477$ straight into the simplified Arrhenius multiplier leads directly to $43.8\text{ kJ mol}^{-1}$.

Answer 1

The Correct Option is B

Concept: The half-life ($t_{1/2}$) of a first-order reaction is related to its rate constant ($k$) by $t_{1/2} = \frac{\ln 2}{k}$, which implies that $k$ is inversely proportional to $t_{1/2}$. The temperature dependence of the rate constant is given by the Arrhenius equation: \[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \]

Step 1: Determine the ratio of the rate constants ($\frac{k_2}{k_1}$).
Given the half-lives at $T_1 = 300\text{ K}$ and $T_2 = 320\text{ K}$: \[ t_{1/2}(1) = 30\text{ min} \implies k_1 = \frac{\ln 2}{30} \] \[ t_{1/2}(2) = 10\text{ min} \implies k_2 = \frac{\ln 2}{10} \] Taking the ratio: \[ \frac{k_2}{k_1} = \frac{t_{1/2}(1)}{t_{1/2}(2)} = \frac{30}{10} = 3 \]

Step 2: Substitute values into the Arrhenius equation to calculate $E_a$.
\[ \log(3) = \frac{E_a}{2.303 \times 8.314} \left(\frac{320 - 300}{300 \times 320}\right) \] Using $\log 3 = 0.4771$: \[ 0.4771 = \frac{E_a}{19.147} \left(\frac{20}{96000}\right) \] \[ 0.4771 = \frac{E_a}{19.147} \left(\frac{1}{4800}\right) \] \[ E_a = 0.4771 \times 19.147 \times 4800 \approx 43848\text{ J mol}^{-1} \approx 43.8\text{ kJ mol}^{-1} \]