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Question

If 60% of a first order reaction was completed in 60 min, 50% of the same reaction would be completed in approximately [log 4 = 0.60, log 5 = 0.69

A
50 min
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B
45 min
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C
60 min
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D
40 min
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Solution

The correct option is B 45 min
From first order reaction,
Rate constant, k=2.303tlog10a(ax)
k1=2.303tloga1a1x1(i)
k2=2.303tloga2a2x2(ii)
x1=60100a1, t1=60
x2=50100a2, t2=?
From Eqs. (i) and (ii)
2.303tloga1a1x1=2.303tloga2a2x2
2.30360loga1(a160100a1)=2.303t2loga2(a250100a2)
2.30360log100a110a1=2.303t2log100a250a2
160log10040=1t2log10050
t2=60log10050log10040=60(log10log5)(log10log4)
=60(10.69)(10.60)=60×0.310.40
=1.5×31=46.545 min

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