Question

The rate constant of a reaction at 400 K and 500 K are \( 0.04\,\text{min}^{-1} \) and \( 0.08\,\text{min}^{-1} \) respectively. The \( E_a \) of the reaction is about ( \( 2.303R = 19.15\,\text{J mol}^{-1}\text{K}^{-1} \) )

Hint:

To speed up calculations in competitive exams, memorize $\log 2 \approx 0.3$ and $\log 3 \approx 0.48$. Always be mindful of units, converting final energy values from Joules to kiloJoules as options usually require it.

Answer 1

Step 1: Use Arrhenius Equation
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \]

Step 2: Substitute Given Values
\[ k_1 = 0.04,\quad k_2 = 0.08,\quad T_1 = 400\,\text{K},\quad T_2 = 500\,\text{K} \] \[ \log\left(\frac{0.08}{0.04}\right) = \frac{E_a}{19.15} \left(\frac{500 - 400}{400 \times 500}\right) \]

Step 3: Simplify the Expression
\[ \log(2) = \frac{E_a}{19.15} \times \frac{100}{200000} \] \[ \log(2) = \frac{E_a}{19.15} \times \frac{1}{2000} \]

Step 4: Use Approximation
\[ \log(2) \approx 0.3 \] \[ 0.3 = \frac{E_a}{19.15 \times 2000} \]

Step 5: Calculate \( E_a \)
\[ E_a = 0.3 \times 19.15 \times 2000 \] \[ E_a = 11490 \,\text{J mol}^{-1} \]

Step 6: Convert into kJ/mol
\[ E_a = 11.49 \,\text{kJ mol}^{-1} \approx 11.5 \,\text{kJ mol}^{-1} \]

Step 7: Final Answer
\[ \boxed{E_a = 11.5 \,\text{kJ mol}^{-1}} \]