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
• B. 45 min
• C. 60 min
• D. 40 min

Answer 1

Answer: (B)

45 min

Answer 2

From first order reaction,
Rate constant, k=$\frac{2.303}{t} log_{10} \frac{a}{(a-x)}$
\hspace15mm k_1=\frac{2.303}{t_1} log \frac{a_1}{(a_1-x_1)} \hspace10mm ...(i)
\hspace15mm k_2=\frac{2.303}{t_2} log \frac{a_2}{(a_2-x_2)} \hspace10mm ...(ii)
\hspace15mm x_1=\frac{60}{100} a_1, t_1=60
\hspace15mm x_2=\frac{50}{100} a_2, t_2=?
From Eqs. (i) and (ii)
$\, \, \, \, \, \frac{2.303}{t_1} log \frac{a_1}{a_1-x_1}=\frac{2.303}{t_2} log \frac{a_2}{a_2-x_2}$
$\frac{2.303}{60} log \frac{a_1}{\big(a_1-\frac{60}{100}a_1\big)}=\frac{2.303}{t_2} log \frac{a_2}{\big(a_2-\frac{50}{100}a_2 \big)}$
$\, \, \, \, \, \, \, \frac{2.303}{60} log \frac{100 a_1}{40 a_1}=\frac{2.303}{t_2} log \frac{100 a_2}{50 a_2}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{1}{60} log \frac{100}{40} =\frac{1}{t_2} log \frac{100}{50}$
$\, \, \, \, \, t_2=\frac{60 \, log \, \frac{100}{50}}{log \, \frac{100}{40}} =\frac{60(log \, 10-log \, 5)}{(log \, 10-log \, 4)}$
$\, \, \, \, \, \, \, =\frac{60(1-0.69)}{(1-0.60)}=\frac{60 \times 0.31}{0.40}$
$\, \, \, \, \, \, \, \, \, \, =1.5 \times 31=46.5 \approx 45$ min