• # question_answer The acid catalyzed hydrolysis of an organic, compound 'A' at $30{}^\circ C$ has a time for half change of 100 minute when carried out in a buffer solution at $(pH=5)$ and 10 minute when carried out at $(pH=4)$. Both times of half change are independent of the initial concentration of A. If rate constant K is defined by $\frac{-d[A]}{dt}=K{{[A]}^{a}}{{[{{H}^{+}}]}^{b}},$ the values of a and b respectively are A) 1, 1   B) 1, 2 C) 0, 1                   D) 1, 0

 During any experiment, pH is constant, hence $\frac{-d[A]}{dt}=K'{{[A]}^{a}}$ where $K'=K{{[{{H}^{+}}]}^{b}}$ As ${{t}_{1/2}}$ is independent of initial conc. So $a=1$ Also K' can be given by $=\frac{0.693}{{{t}_{1/2}}}$ $\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}=\frac{K_{2}^{'}}{K_{1}^{'}}=\frac{K[{{H}^{+}}]_{2}^{b}}{K[{{H}^{+}}]_{1}^{b}}=\frac{[{{H}^{+}}]_{2}^{b}}{[{{H}^{+}}]_{1}^{b}}$ Now $\frac{100}{10}=\left( \frac{{{10}^{-4}}}{{{10}^{-5}}} \right)\text{or}\,b=1$