Question

75% of a first order reaction was completed in 32 min. When would 50% of the reaction completed?

• A. 24 min
• B. 16 min
• C. 8 min
• D. 64 min

Answer 1

Answer: (B)

16 min

Answer 2

The first order rate constant is given as
$k=\frac{2.303}{t}\,log \frac{a_{0}}{a_{0}-X}$ ______(1) also, half life $t1/2 =\frac{2.303 log^{2}}{k}$ ______(2)
Equating K from equation (1) and (2)
$\therefore\frac{2.303 log2}{t_{1/2}}=\frac{2.303}{t}log \frac{a_{0}}{a_{0}-X},$Now given, for 75% completion of the reaction is 32 minutes
$\therefore\frac{2.303 log2}{t_{1/2}}=\frac{2.303}{32}log \frac{100}{100-75}\,or,\frac{log2}{t_{1/2}}=\frac{1}{32}log4,\,or\frac{log2}{t_{1/2}}=\frac{1}{32}\times2log2$
$\therefore t_{1/2} =$16 minutes.