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Step-by-Step Solution
Step 1: Write the Arrhenius Equation
The rate constant k for a chemical reaction can be expressed using the Arrhenius equation:
$k = A e^{-\frac{E_a}{RT}}$
Here,
$A$ is the pre-exponential factor (frequency factor),
$E_a$ is the activation energy for the reaction,
$R$ is the universal gas constant, and
$T$ is the temperature in kelvin.
Step 2: Express the Rate Constants for Both Reactions
For reaction R1 with activation energy $E_{a_1}$:
$k_1 = A e^{-\frac{E_{a_1}}{RT}}$
For reaction R2 with activation energy $E_{a_2}$:
$k_2 = A e^{-\frac{E_{a_2}}{RT}}$
Given that the pre-exponential factors for both reactions are the same, we see that both have the same $A$ value.
Step 3: Form the Ratio of the Rate Constants
We divide $k_2$ by $k_1$ to find an expression for how the rate constants compare:
$\frac{k_2}{k_1} = \frac{A e^{-\frac{E_{a_2}}{RT}}}{A e^{-\frac{E_{a_1}}{RT}}}
= e^{-\frac{E_{a_2}}{RT}} \cdot e^{\frac{E_{a_1}}{RT}}
= e^{\frac{E_{a_1} - E_{a_2}}{RT}}$
Step 4: Relate the Activation Energy Difference to the Temperature
We know that $E_{a_1} - E_{a_2} = 10\,\text{kJ mol}^{-1}$, which is $10{,}000\,\text{J mol}^{-1}$. Substituting this difference into the logarithmic form,
we get:
$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_{a_1} - E_{a_2}}{R \, T}
= \frac{10{,}000\,\text{J mol}^{-1}}{(8.314\,\text{J mol}^{-1}\text{K}^{-1})(300\,\text{K})}$
Step 5: Calculate the Value
$\ln\left(\frac{k_2}{k_1}\right)
= \frac{10{,}000}{8.314 \times 300}
= \frac{10{,}000}{2494.2}
\approx 4$
This shows that the correct value of $\ln\left(\frac{k_2}{k_1}\right)$ is 4.
Step 6: Conclusion
Therefore, $\ln\left(\frac{k_2}{k_1}\right) = 4$. Hence, the correct answer to the question is 4.
