In an acid catalysed hydrolysis of an organic compound A, the half life of A changes from 100 min (when carried out in a buffer solution of pH •• 3) to 1 min (when carried out in a buffer solution of pH = 4). Both the half lives are independent of the concentration of organic compound. If the rate law is represented as $R = K {[H^{+}]}^{α} [A]^{b}$ [At then the values of a and b will be :

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a= -1, b = 1

a= 1, b = 1

a= 2, b = 1

a= -2, b = 1

answer is D.

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Detailed Solution

$r_{1} = k_{1} [A]^{b}; b = 1; k_{1} = k \times {(10^{- 3})}^{a}; t_{1 / 2} = 100 \min$

$r_{2} = k_{2} [A]; k_{2} = k \times {(10^{- 4})}^{a}; t_{1 / 2} = 1 \min$

$∴ \frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = \frac{k_{2}}{k_{1}}$

$\begin{array}{llrl} \Rightarrow \frac{100}{1} = {[\frac{10^{- 4}}{10^{- 3}}]}^{a} \\ \Rightarrow a = - 2 \end{array}$

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