It has been found that for a chemical reaction with rise in temperature by $$9$$ K the rate constant gets doubled. Assuming a reaction to be occurring at $$300$$ K, the value of activation energy is found to be ______ kJ mol$$^{-1}$$. [nearest integer] (Given $$\ln 10 = 2.3, R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$, $$\log 2 = 0.30$$)

We are given that for a rise of 9 K in temperature, the rate constant doubles. We need to find the activation energy at T = 300 K.

Using the Arrhenius equation in logarithmic form:

$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Substituting the given values $$T_1 = 300$$ K, $$T_2 = 309$$ K, and $$\frac{k_2}{k_1} = 2$$ yields

$$\log 2 = \frac{E_a}{2.303 \times 8.3}\left(\frac{1}{300} - \frac{1}{309}\right)$$

Noting that

$$\frac{1}{300} - \frac{1}{309} = \frac{309 - 300}{300 \times 309} = \frac{9}{92700}$$

This leads to

$$0.30 = \frac{E_a}{2.303 \times 8.3} \times \frac{9}{92700}$$

$$0.30 = \frac{E_a \times 9}{19.1149 \times 92700}$$

Next,

$$19.1149 \times 92700 = 1772,951.23$$

Approximating: $$2.303 \times 8.3 = 19.1149$$

$$19.1149 \times 92700 \approx 1,772,951$$

Solving for $$E_a$$:

$$E_a = \frac{0.30 \times 1,772,951}{9}$$

$$E_a = \frac{531,885.3}{9}$$

$$E_a = 59,098.4 \text{ J mol}^{-1}$$

Converting to kJ,

$$E_a \approx 59.1 \text{ kJ mol}^{-1}$$

Therefore, the activation energy is approximately 59 kJ mol$$^{-1}$$.