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Q.

t1/4 can be taken as the time taken for concentration of reactant to drop to 3/4 of its initial value. If the rate constant for a first order reaction is K, then t1/4 can be written as:

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a

0.10/K

b

0.29/K

c

0.69/K

d

0.75/K

answer is B.

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Detailed Solution

K=2.303tlog⁡aa−x=2.303t14log⁡4a3a=2.303t1/4log⁡43K=2.303×0.125t1/4=0.29t1/4
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