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Q.

t1/4can be taken as the time taken for the concentration of a reactant to drop to 34of its initial         value. If the rate constant for a first order reaction is K, the t1/4 of its initial value. If the rate constant for a first order reaction is K, the t1/4 can be written as

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a

0.29/k

b

0.75/k

c

  0.10/k

d

0.69/k

answer is C.

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Detailed Solution

t1/4 = 2.303 k log 13/4 = 2.303k log 43

= 2.303k(log 4  log3) = 2.303k(2 log 2  log3)

= 2.303k(2 × 0.301  0.4771) = 0.29k

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