Question

(a) $- \frac{d\lbrack A\rbrack}{dt} = K.\lbrack A\rbrack^{n}f = \frac{\left\lbrack A_{1} \right\rbrack - \left\lbrack A_{2} \right\rbrack}{\left\lbrack A_{1} \right\rbrack} = \frac{- d\lbrack A\rbrack}{\lbrack A\rbrack}$

from question, $f = \frac{- d\lbrack A\rbrack}{\lbrack A\rbrack}$

$\therefore\frac{f\lbrack A\rbrack}{t} = K\lbrack A\rbrack^{n} \Rightarrow \frac{f}{t} = K.\lbrack A\rbrack^{n - 1} $$$\text{or, }\log\left( \frac{f}{t} \right) = \log K + (n - 1).\log\lbrack A\rbrack$$

(b) $\frac{\left\lbrack A_{0} \right\rbrack^{1 - n} - \lbrack A\rbrack^{1 - n}}{1 - n} = Kt$

$= \lbrack A\rbrack^{1 - n} = \left\lbrack A_{0} \right\rbrack^{1 - n} + (n - 1).Kt$

(c) $\frac{t_{3/4}}{t_{1/2}} = \frac{\left\lbrack A_{0} \right\rbrack^{1 - n} - \left( \frac{\left\lbrack A_{0} \right\rbrack}{4} \right)^{1 - n}}{\left\lbrack A_{0} \right\rbrack^{1 - n} - \left( \frac{\left\lbrack A_{0} \right\rbrack}{2} \right)^{1 - n}} = \frac{1 - \left( 2^{2} \right)^{n - 1}}{1 - 2^{n - 1}} = 1 + 2^{n - 1}$