Two reactions A$$_{1}$$ and A$$_{2}$$ have identical pre-exponential factors. The activation energy of A$$_{1}$$ is more than A$$_{2}$$ by 10 kJ mol$$^{-1}$$. If k$$_{1}$$ and k$$_{2}$$ are the rate constants for reactions A$$_{1}$$ and A$$_{2}$$, respectively at 300 K, then $$\ln\frac{k_{2}}{k_{1}}$$ is equal to
(R = 8.314 J mol$$^{-1}$$ K$$^{-1}$$)

We begin with the Arrhenius equation, which gives the temperature-dependence of a rate constant:

$$k = A\,e^{-\dfrac{E_a}{RT}}$$

Here $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

For the two reactions we are told that the pre-exponential factors are identical, so we can write

$$k_1 = A\,e^{-\dfrac{E_{a1}}{RT}} \qquad\text{and}\qquad k_2 = A\,e^{-\dfrac{E_{a2}}{RT}}$$

The activation energy of reaction $$A_1$$ is greater than that of reaction $$A_2$$ by $$10\ \text{kJ mol}^{-1}$$. Converting to joules, we have

$$E_{a1} = E_{a2} + 10\ \text{kJ mol}^{-1} = E_{a2} + 10000\ \text{J mol}^{-1}$$

We want the natural logarithm of the ratio of the two rate constants:

$$\ln\frac{k_2}{k_1}$$

Substituting the expressions for $$k_1$$ and $$k_2$$ gives

$$\ln\frac{k_2}{k_1} = \ln\!\left(\frac{A\,e^{-\dfrac{E_{a2}}{RT}}}{A\,e^{-\dfrac{E_{a1}}{RT}}}\right)$$

The factors $$A$$ cancel, and using the property $$\ln\left(\dfrac{e^x}{e^y}\right)=x-y$$ we obtain

$$\ln\frac{k_2}{k_1} = -\frac{E_{a2}}{RT} -\!\left(-\frac{E_{a1}}{RT}\right) = \frac{E_{a1}-E_{a2}}{RT}$$

But $$E_{a1}-E_{a2}=10000\ \text{J mol}^{-1}$$, so

$$\ln\frac{k_2}{k_1} = \frac{10000\ \text{J mol}^{-1}}{R\,T}$$

Now we substitute $$R = 8.314\ \text{J mol}^{-1}\ \text{K}^{-1}$$ and $$T = 300\ \text{K}$$:

$$\ln\frac{k_2}{k_1} = \frac{10000}{8.314 \times 300}$$

First calculate the denominator:

$$8.314 \times 300 = 2494.2$$

Then perform the division:

$$\ln\frac{k_2}{k_1} = \frac{10000}{2494.2} \approx 4.01$$

The value is extremely close to $$4$$, and among the given options $$4$$ is the only matching choice.

Hence, the correct answer is Option C.