t1/4 can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its value. If the rate constant for a first order reaction is k, the t1/4 can be written as [ln2 = 0.695, ln = 1.1]
A
0.69 / k
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B
0.75 / k
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C
0.10 / k
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D
0.29 / k
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Solution
The correct option is D0.29 / k t14=2.303Klog43=2.303KOlog4−log3=0.29K