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Chemical Kinetics

Question

A first order reaction takes 69.3 minutes for 50% completion. Set up an equation determining the time needed for 80% completion of the reaction.

Solution

Half life reaction can be given by
t_1/2 = 0.693/k
thus using above equation we get

t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{69.3} = \frac{69.3 \times 10^{- 2}}{69.3 \min} = 1 \times 10^{- 2} \min u \sin g f i r s t o r d e r e q u a t i o n, w e g e t K = \frac{2.303}{t} \log \frac{[R_{0}]}{[R]} t = \frac{2.303}{1 \times 10^{- 2}} \log \frac{[R_{0}]}{\frac{20}{180} [R_{0}]} t = \frac{2.303}{1 \times 10^{- 2}} \log 5 \frac{2.303}{10^{- 2}} \times 0.6990 = 1.6097 \times 10^{2} \min = 160.97 \min .