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Physics Question on work done thermodynamics

The initial rate of hydrolysis of methyl acetate (1M)(1 M) by a weak acid (HA,1M)(HA, 1M) is 1/100th of that of a strong acid (HX,1M)(HX, 1M), at 25∘C. 25^{\circ} C. The Ka(HA)K_a ( H A ) is

A

1×10−41 \times 10^{ - 4 }

B

1 ×10−5\times 10^{ - 5 }

C

1 ×10−6\times 10^{ - 6 }

D

1 ×10−3\times 10^{ - 3 }

Answer

1×10−41 \times 10^{ - 4 }

Explanation

Solution

PLAN RCOOR′+H2O→H+RCOOH+R′OHRCOOR' + H2_O \xrightarrow{ H^+ } RCOOH + R'OH
Acid hydrolysis of ester is follows first order kinetics.
For same concentration of ester in each case, rate is dependent on [H+[H^+ ]from acid.
Rate = k [RCOOR' ]
Also for weak acid, HA ⇌H+A−\rightleftharpoons H^+ A^-
Ka=[H+] [A−][HA]K_a = \frac{ [ H^+ ] \ [ A ^- ] }{ [ HA ]}
(Rate)HA=k[H+]HA( Rate)_{ HA } = k [H^+ ]_{ H A }
(Rate)Hx=k[H+]HX( Rate)_{H x } = k [ H^+ ] _{ HX }
(Rate)HX=100Rate)HA( Rate)_{ HX } = 100 Rate)_{ HA }
∴\therefore Also in strong acid, [H+][ H^+ ] = [ HX ] = 1M
(Rate)HX(Rate)HA=100=[H+]HX[H+]HA=1[H+]HA\frac{ (Rate )_{ HX} }{ (Rate )_{ HA}} = 100 = \frac{ [ H^+ ]_{ HX }}{ [ H^+ ]_{ HA }} = \frac{1}{ [ H^+ ]_{ HA }}
∴[H+]HA=1100\therefore [ H^+ ]_{ HA } = \frac{1}{ 100 }
HA ⇌H++A−\rightleftharpoons H^+ + A^-
\begin{array} \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0 \\\ (1 - x) \ \ \ \ \ \ x \ \ \ \ \ \ \ \ \ \ \ \ x \\\ \end{array}
∴Ka=[H+] [A−][HA]=0.01×0.010.99\therefore K_a = \frac{ [ H^+ ] \ [A^- ] }{ [HA] } = \frac{ 0.01 \times 0.01 }{ 0.99 }
1×10−41 \times 10^{ - 4 }