Question

Consider \( A \xrightarrow{k_1} B \) and \( C \xrightarrow{k_2} D \) are two reactions. If the rate constant (\(k_1\)) of the \( A \rightarrow B \) reaction can be expressed by the following equation
\[ \log_{10} k = 14.34 - \frac{1.5 \times 10^{4}}{T/K} \] and activation energy of \( C \rightarrow D \) reaction (\(E_{a2}\)) is \( \dfrac{1}{5} \)th of the \( A \rightarrow B \) reaction (\(E_{a1}\)), then the value of (\(E_{a2}\)) is ______________ kJ mol\(^{-1}\) (Nearest Integer).

Hint:

When a rate equation is given in logarithmic Arrhenius form, directly compare coefficients with the standard Arrhenius equation to extract activation energy.

Answer 1

Correct Answer: 144

To determine the activation energy (\(E_{a2}\)) for the reaction \(C \rightarrow D\), we begin with the relationship for the rate constant (\(k_1\)) of the reaction \(A \rightarrow B\). The given equation is:
\[ \log_{10} k = 14.34 - \frac{1.5 \times 10^{4}}{T} \] This equation relates to the Arrhenius equation, which is given as:
\[ k = A \cdot e^{-\frac{E_a}{RT}} \] Taking logarithms on both sides, it can be expressed as:
\[ \log_{10} k = \log_{10} A - \frac{E_a}{2.303RT} \] Comparing the two logarithmic equations given, we find:
\[ -\frac{E_a}{2.303RT} = -\frac{1.5 \times 10^{4}}{T} \] Solving for \(E_a\), we get:
\[ E_{a1} = 1.5 \times 10^{4} \times 2.303 \times R \] Assuming \(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\), convert this to kJ/mol:
\[ E_{a1} = 1.5 \times 10^{4} \times 2.303 \times 0.008314 \approx 287.55 \, \text{kJ mol}^{-1} \] Since the activation energy of \(C \rightarrow D\) is \(\frac{1}{5}\) of \(E_{a1}\):
\[ E_{a2} = \frac{287.55}{5} \approx 57.51 \, \text{kJ mol}^{-1} \] Rounding to the nearest integer, \(\boxed{58}\). This value falls outside the provided range, indicating possible discrepancies. Re-evaluations confirm accurate calculations without rounding during computations.