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

The correct option is B 45 min

From first order reaction,
Rate constant, $k=\frac{2.303}{t}lo{g}_{10}\frac{a}{(a-x)}$
$k_1=\frac{2.303}{t}log\frac{a_1}{a_1-x_1}\dots \dots \left(i\right)$
$k_2=\frac{2.303}{t}log\frac{a_2}{a_2-x_2}\dots \dots \left(ii\right)$
$x_1=\frac{60}{100}{a}_1,t_1=60$
$x_2=\frac{50}{100}{a}_2,t_2=?$
From Eqs. (i) and (ii)
$\frac{2.303}{t}log\frac{a_1}{a_1-x_1}=\frac{2.303}{t}log\frac{a_2}{a_2-x_2}$
$\frac{2.303}{60}log\frac{a_1}{(a_1-\frac{60}{100}{a}_1)}=\frac{2.303}{t_2}log\frac{a_2}{(a_2-\frac{50}{100}{a}_2)}$
$\frac{2.303}{60}log\frac{100a_1}{10a_1}=\frac{2.303}{t_2}log\frac{100a_2}{50a_2}$
$\frac{1}{60}log\frac{100}{40}=\frac{1}{t_2}log\frac{100}{50}$
$t_2=\frac{60log\frac{100}{50}}{log\frac{100}{40}}=\frac{60(log10-log5)}{(log10-log4)}$
$=\frac{60(1-0.69)}{(1-0.60)}=\frac{60\times 0.31}{0.40}$
$=1.5\times 31=46.5\approx 45min$