For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of this reaction is ________ $$\times 10^{-3}$$ s$$^{-1}$$. (Round off to the Nearest Integer).
[Given: $$\log_{10} 2 = 0.301$$, $$\ln 10 = 2.303$$]

For a first-order reaction, the integrated rate law is $$\ln\frac{[A]_0}{[A]} = kt$$, or equivalently $$k = \frac{2.303}{t}\log_{10}\frac{[A]_0}{[A]}$$.

Given that 32% of the reactant remains after $$t = 570$$ s, we have $$\frac{[A]}{[A]_0} = 0.32$$ and $$\frac{[A]_0}{[A]} = \frac{1}{0.32} = \frac{100}{32} = \frac{25}{8}$$.

Now $$\log_{10}\frac{25}{8} = \log_{10}25 - \log_{10}8 = 2\log_{10}5 - 3\log_{10}2$$. Since $$\log_{10}2 = 0.301$$, we get $$\log_{10}5 = \log_{10}\frac{10}{2} = 1 - 0.301 = 0.699$$. So $$\log_{10}\frac{25}{8} = 2(0.699) - 3(0.301) = 1.398 - 0.903 = 0.495$$.

Therefore $$k = \frac{2.303 \times 0.495}{570} = \frac{1.140}{570} = 2.0 \times 10^{-3}$$ s$$^{-1}$$. The answer is 2.