The rate constant for a first order reaction is given by the following equation :
$$\ln k = 33.24 - \frac{2.0 \times 10^4 K}{T}$$
The Activation energy for the reaction is given by ______ kJ mol$$^{-1}$$. (In Nearest integer) (Given : $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$)

We are given the rate constant equation for a first order reaction:

$$\ln k = 33.24 - \frac{2.0 \times 10^4 \text{ K}}{T}$$

We need to compare this with the Arrhenius equation in its logarithmic form.

The Arrhenius equation is:

$$k = A e^{-E_a/RT}$$

Taking the natural logarithm:

$$\ln k = \ln A - \frac{E_a}{RT}$$

$$\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$$

Comparing with the given equation $$\ln k = 33.24 - \frac{2.0 \times 10^4}{T}$$, we identify:

$$\frac{E_a}{R} = 2.0 \times 10^4 \text{ K}$$

Therefore:

$$E_a = R \times 2.0 \times 10^4 \text{ K}$$

$$E_a = 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \times 2.0 \times 10^4 \text{ K}$$

$$E_a = 16.6 \times 10^4 \text{ J mol}^{-1}$$

$$E_a = 166 \times 10^3 \text{ J mol}^{-1}$$

$$E_a = 166 \text{ kJ mol}^{-1}$$

Therefore, the activation energy is 166 kJ mol$$^{-1}$$.