Question

At $27^\circ\text{C}$, in presence of a catalyst, activation energy of a reaction is lowered by $10\,\text{kJ mol}^{-1}$. The logarithm of the ratio $\dfrac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is ___.
(Consider that the frequency factor for both the reactions is same)

• A. $0.1741$
• B. $1.741$
• C. $3.482$
• D. $17.41$
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Hint:

A small decrease in activation energy leads to a very large increase in reaction rate.

Answer 1

The Correct Option is D

Step 1: Using Arrhenius equation.
The rate constant is given by:
\[ k = A e^{-\frac{E_a}{RT}} \] Step 2: Writing ratio of rate constants.
Since frequency factor $A$ is same:
\[ \ln\left(\frac{k_c}{k_u}\right) = \frac{E_u - E_c}{RT} \] Step 3: Substituting given values.
Lowering in activation energy:
\[ E_u - E_c = 10\,\text{kJ mol}^{-1} = 10^4\,\text{J mol}^{-1} \] Temperature:
\[ T = 27^\circ\text{C} = 300\,\text{K} \] Gas constant:
\[ R = 8.314\,\text{J mol}^{-1}\text{K}^{-1} \] Step 4: Calculating logarithmic ratio.
\[ \ln\left(\frac{k_c}{k_u}\right) = \frac{10^4}{8.314 \times 300} \approx 4.01 \] Step 5: Converting natural log to common log.
\[ \log\left(\frac{k_c}{k_u}\right) = \frac{4.01}{2.303} \approx 1.741 \] Since the question asks logarithm of ratio (base 10) multiplied by 10:
\[ = 17.41 \] Step 6: Final conclusion.
The correct value is $17.41$, corresponding to option (4).