Answer:
Option B,D
Explanation:
plan $RCOOR'+H_{2}0 $ $\underrightarrow{H^{+}}$ $ RCOOH+R'OH$
Acid hydrolysis of ester is follows first order kinetics
For same concentration of ester in each case rate in dependent on [H+] from acid
Rate = k[RCOOR']
Also for weak acid HA $ \rightleftharpoons H^{+}+A^{-}$
$K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$
[Rate]HA =k[H+]HA
[Rate]HX= k[H+]HX
(Rate)HX =100(Rate)HA
$\therefore$ Also in strong acid [H+]=[HX]=1M
$\frac{(Rate)_{HX}}{(Rate)_{HA}}=100=\frac{[H^{+}]_{HX}}{[H^{+}]_{HA}}$
$=\frac{1}{[H^{+}]_{HA}}$
$\therefore$ $[H^{+}]_{HA}=\frac{1}{100}$
$HA\rightleftarrows H^{+}+A^{-}$
1 0 0
(1-x) x x
x= 0.01
$\therefore$ $k_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$
$=\frac{0.01\times 0.01}{0.99}
$=1.01\times 10^{-4}$
