Q5 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
Two chemical reactions of the same order have equal Frequency factor value.
Their Activation energies differ by $26.8 \mathrm{~kJ} / \mathrm{mol}$. At 300 K if $k_2=x k_1$ find the value of $x$.
- A$1.143 \times 10^3$
- B$4.665$
- Correct answerC$4.639 \times 10^4$
- D$2.286 \times 10^3$
Solution
The Arrhenius equation is given by $k = A e^{-E_a / (RT)}$.
Given that the frequency factors are equal, $A_1 = A_2 = A$.
The rate constants for the two reactions are $k_1 = A e^{-E_{a1} / (RT)}$ and $k_2 = A e^{-E_{a2} / (RT)}$.
Taking the ratio $k_2 / k_1 = e^{(E_{a1} - E_{a2}) / (RT)}$.
Given $E_{a1} - E_{a2} = 26.8 \text{ kJ/mol} = 26800 \text{ J/mol}$, $T = 300 \text{ K}$, and $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.
Substituting the values: $x = k_2 / k_1 = e^{26800 / (8.314 \times 300)}$.
$x = e^{26800 / 2494.2} = e^{10.7449}$.
$x \approx 46397$.
Answer: $4.639 \times 10^4$
Given that the frequency factors are equal, $A_1 = A_2 = A$.
The rate constants for the two reactions are $k_1 = A e^{-E_{a1} / (RT)}$ and $k_2 = A e^{-E_{a2} / (RT)}$.
Taking the ratio $k_2 / k_1 = e^{(E_{a1} - E_{a2}) / (RT)}$.
Given $E_{a1} - E_{a2} = 26.8 \text{ kJ/mol} = 26800 \text{ J/mol}$, $T = 300 \text{ K}$, and $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.
Substituting the values: $x = k_2 / k_1 = e^{26800 / (8.314 \times 300)}$.
$x = e^{26800 / 2494.2} = e^{10.7449}$.
$x \approx 46397$.
Answer: $4.639 \times 10^4$
Hence, the correct answer is option: 'C'.