Explanation

(a)
$k = \frac{2.303}{t}\log\frac{a}{a - x}$
Given, reaction is75% completed is 32 min
A=100,x=75
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ k = \frac{2.303}{32}\log\frac{100}{100 - 75}$ …(1)
For 50% completion of reaction
A=100, x=50
$\therefore\ \ \ \ \ \ \ \ \ \ \ k = \frac{2.303}{t}\log\frac{100}{100 - 50}$ …(2)
$\because\ \ \ \ \ \ \ LHSof\ Eq.(1) = Eq.(2)$
$\therefore\ \ \ \ RHSof\ Eq.(1) = Eq.(2)$
$\therefore\ \ \ \frac{2.303}{32}\log\frac{100}{100 - 75} = \frac{2.303}{t}\log\frac{100}{100 - 50}$
$or\frac{2.303}{32}\log{4 = \frac{2.303}{t}\log 2}$
Or $\frac{t}{32} = \frac{\log 2}{\log 4}$ or $t = \frac{32 \times \log 2}{2\log 2}$
$\therefore\ \ \ \ t = 16\ min$
$\therefore\ \ reaction\ will\ be\ 50\%\ completed\ in\ 16\ min$