For the first order reaction A $$\to$$ B the half life is 30 min. The time taken for 75% completion of the reaction is _____ min. (Nearest integer)
Given: log 2 = 0.3010
log 3 = 0.4771
log 5 = 0.6989

For a first-order reaction, the integrated rate law gives the time for a given fraction of completion as $$t = \dfrac{2.303}{k}\log\dfrac{1}{1-x}$$, where $$x$$ is the fraction completed. The half-life is $$t_{1/2} = \dfrac{0.693}{k}$$, so $$k = \dfrac{0.693}{30} = 0.0231$$ min$$^{-1}$$.

For 75% completion ($$x = 0.75$$), we have $$t_{75} = \dfrac{2.303}{k}\log\dfrac{1}{0.25} = \dfrac{2.303}{k}\log 4 = \dfrac{2.303}{k} \times 2\log 2 = \dfrac{2.303 \times 2 \times 0.3010}{0.0231}$$.

Since $$t_{1/2} = \dfrac{2.303 \times 0.3010}{k} = 30$$ min, we get $$t_{75} = 2 \times 30 = \boxed{60}$$ min. Equivalently, 75% completion requires exactly 2 half-lives: after one half-life 50% remains, after two half-lives 25% remains (i.e., 75% has reacted).