Question

From application of thermodynamics on chemical reaction,
we get $\Delta G=\Delta G^0+RT$ ln Q
Also $\Delta G=\Delta H-T\Delta S$.
If $\Delta G=0$, reaction is at equilibrium.
If $\Delta G>0$ reaction is non-spontaneous under given condition. $If\Delta G<0$,
reaction is spontaneous under given condition.
Consider a weak monobasic acid which is neutralised with KOH at 600 K. $\Delta G^0$ of the reaction is 16.581 K
cal/mol. Find dissociation constant of acid.
$(use\frac{16.581}{2.303\times 2\times 600}=6\times {10}^{-3}andR=2cal/mol-K)$

• A. ${10}^{-6}$
• B. ${10}^{-8}$
• C. ${10}^{-12}$
• D. ${10}^6$

Answer 1

The correct option is B ${10}^{-8}$

The equilibrium reaction for the neutralisation of monobasic acid is $HA+KOH\to KA+H_{2}O$.
The equilibrium constant for the above reaction is $\frac{Ka}{Kw}$.
The relationship between standard free energy change and equilibrium constant is $\Delta G^0=2.303RTlogK$.
Substitute values in the above expression.
$16.581\times 1000=2.303\times 2\times 600log\frac{K_a}{{10}^{-14}}$
But $(use\frac{16.581}{2.303\times 2\times 600}=6\times {10}^{-3}andR=2cal/mol-K)$
Hence, $6=log\frac{K_a}{{10}^{-14}}$
So, $K_a={10}^{-8}$.