For the given first order reaction $$A \to B$$ the half life of the reaction is 0.3010 min. The ratio of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to _____. (Nearest integer)

The reaction considered is a first order reaction: $$A \to B$$, with half-life $$t_{1/2} = 0.3010$$ min and time $$t = 2.0$$ min.

For a first order reaction, the half-life is given by $$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$$, so that $$k = \frac{0.693}{0.3010} = \frac{\ln 2}{0.3010}$$.

Since $$\ln 2 = 2.303 \times \log 2 = 2.303 \times 0.3010 = 0.6932$$, this gives $$k = \frac{2.303 \times 0.3010}{0.3010} = 2.303 \text{ min}^{-1}$$.

In order to find the concentration ratio, we use the first order rate expression $$\ln\frac{[A_0]}{[A]} = kt$$, which leads to $$\ln\frac{[A_0]}{[A]} = 2.303 \times 2.0 = 4.606$$.

Converting to a common logarithm, we write $$2.303 \times \log\frac{[A_0]}{[A]} = 4.606$$, hence $$\log\frac{[A_0]}{[A]} = \frac{4.606}{2.303} = 2$$.

Therefore, $$\frac{[A_0]}{[A]} = 10^2 = 100$$.

This shows that the ratio of the initial concentration to the concentration at time 2.0 min is 100.