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75% of first order reaction is completed in 30 min. Calculate its rate constant

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7.7 x 10^-4min^-1

7.7 x 10^-4M.sec^-1

7.7 x 10^-4sec^-1

7.7min^-1

answer is C.

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Detailed Solution

$\begin{array}{l} t_{75 %} = 2 \times t_{50 %} = 30 \min, t_{1 / 2} = 15 \min \\ K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{15} = 4.62 \times 10^{- 2} \min^{- 1} \\ or \frac{0.693}{15 \times 60} = 7.7 \times 10^{- 4} \sec^{- 1} \end{array}$

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