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(a−x) k1=2.303tloga1a1−x1……(i) k2=2.303tloga2a2−x2……(ii) x1=60100a1,t1=60 x2=50100a2,t2=? From Eqs. (i) and (ii) 2.303tloga1a1−x1=2.303tloga2a2−x2 2.30360loga1(a1−60100a1)=2.303t2loga2(a2−50100a2) 2.30360log100a110a1=2.303t2log100a250a2 160log10040=1t2log10050 t2=60log10050log10040=60(log10−log5)(log10−log4) =60(1−0.69)(1−0.60)=60×0.310.40 =1.5×31=46.5≈45min