Question

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

• A. 40 minutes
• B. 50 minutes
• C. 45 minutes
• D. 60 minutes

Answer 1

The correct option is C

45 minutes

$k=\frac{2.303}{60}log\frac{1}{0.4}=\frac{2.303}{60}log\frac{10}{4}=\frac{2.303}{60}\frac{5}{2}=\frac{2.303}{60}(log5-log2)=\frac{2.303}{60}log(0.69-0.3)=\frac{2.303}{60}\times 0.39$

$t_{\frac{1}{2}}=\frac{2.303\times 60}{2.303\times 0.39}=46.115\approx 45$ minutes