Q59 / 180 · NEET 2026 (Re-NEET) · Chemistry
For an elementary chemical reaction, the Arrhenius plot is given below.
If the energy of activation is $6.64$ kJ mol$^{-1}$ and R $=8.3$ J K$^{-1}$ mol$^{-1}$, the temperature at which the rate constant becomes $e^2$ min$^{-1}$, is
If the energy of activation is $6.64$ kJ mol$^{-1}$ and R $=8.3$ J K$^{-1}$ mol$^{-1}$, the temperature at which the rate constant becomes $e^2$ min$^{-1}$, is
- A$250$ K
- B$125$ K
- C$150$ K
- Correct answerD$200$ K
Solution
According to the Arrhenius equation:
$\ln k = \ln A - \dfrac{E_a}{RT}$
For a plot of $\ln k$ versus $1/T$, the y-intercept corresponds to $\ln A$. From the given graph, the line intersects the y-axis at $6$.
Therefore, $\ln A = 6$.
We are given:
$E_a = 6.64 \text{ kJ mol}^{-1} = 6640 \text{ J mol}^{-1}$
$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$
$k = e^2 \text{ min}^{-1}$, which gives $\ln k = \ln(e^2) = 2$
Substituting these values into the Arrhenius equation:
$2 = 6 - \dfrac{6640}{8.3 \times T}$
$\dfrac{6640}{8.3 \times T} = 6 - 2$
$\dfrac{6640}{8.3 \times T} = 4$
$T = \dfrac{6640}{4 \times 8.3}$
$T = \dfrac{6640}{33.2}$
$T = 200 \text{ K}$
Answer: $200$ K
$\ln k = \ln A - \dfrac{E_a}{RT}$
For a plot of $\ln k$ versus $1/T$, the y-intercept corresponds to $\ln A$. From the given graph, the line intersects the y-axis at $6$.
Therefore, $\ln A = 6$.
We are given:
$E_a = 6.64 \text{ kJ mol}^{-1} = 6640 \text{ J mol}^{-1}$
$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$
$k = e^2 \text{ min}^{-1}$, which gives $\ln k = \ln(e^2) = 2$
Substituting these values into the Arrhenius equation:
$2 = 6 - \dfrac{6640}{8.3 \times T}$
$\dfrac{6640}{8.3 \times T} = 6 - 2$
$\dfrac{6640}{8.3 \times T} = 4$
$T = \dfrac{6640}{4 \times 8.3}$
$T = \dfrac{6640}{33.2}$
$T = 200 \text{ K}$
Answer: $200$ K
Hence, the correct answer is option: 'D'.