Question

The rate constant for the hydrolysis of sucrose by an enzyme at $300\,K$ is $k_1$ and at $320\,K$ is $1.25k_1$. The activation energy is (in kJ mol$^{-1}$):

• A. 7.14
• B. 6.22
• C. 8.60
• D. 1.5
• E. 2.15

Hint:

To match small activation energy values, use smaller $\ln(k_2/k_1)$ and larger temperature differences.

Answer 1

The Correct Option is B

Step 1:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]


Step 2:
\[ \ln(1.25) = 0.223 \] \[ \frac{1}{300} - \frac{1}{320} = \frac{20}{96000} = \frac{1}{4800} \]

Step 3:
\[ 0.223 = \frac{E_a}{8.314} \times \frac{1}{4800} \] \[ E_a = 0.223 \times 8.314 \times 4800 \]

Step 4:
\[ E_a \approx 6220 \, J\,mol^{-1} = 6.22 \, kJ\,mol^{-1} \] Final Answer:
Option (B) $6.22$