The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is $$1.0 \times 10^{-3}$$ s$$^{-1}$$ and the activation energy $$E_a = 11.488$$ kJ mol$$^{-1}$$, the rate constant at 200 K is ________ $$\times 10^{-5}$$ s$$^{-1}$$. (Round off to the Nearest Integer).
[Given $$R = 8.314$$ J mol$$^{-1}$$ K$$^{-1}$$]

We are given the rate constant at 300 K is $$k_1 = 1.0 \times 10^{-3}$$ s$$^{-1}$$, activation energy $$E_a = 11.488$$ kJ/mol $$= 11488$$ J/mol, and we need to find the rate constant at 200 K.

Using the Arrhenius equation: $$\ln\frac{k_1}{k_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$.

Substituting the values: $$\ln\frac{k_1}{k_2} = \frac{11488}{8.314}\left(\frac{1}{200} - \frac{1}{300}\right)$$.

Computing the temperature term: $$\frac{1}{200} - \frac{1}{300} = \frac{3 - 2}{600} = \frac{1}{600}$$.

So $$\ln\frac{k_1}{k_2} = \frac{11488}{8.314} \times \frac{1}{600} = \frac{11488}{4988.4} = 2.3026$$.

Since $$\ln 10 = 2.303$$, we get $$\frac{k_1}{k_2} = 10$$.

Therefore $$k_2 = \frac{k_1}{10} = \frac{1.0 \times 10^{-3}}{10} = 1.0 \times 10^{-4}$$ s$$^{-1}$$ $$= 10 \times 10^{-5}$$ s$$^{-1}$$.

The answer is $$\mathbf{10}$$.