If ${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{OH}}^{-}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_2\mathrm{O}+q_1$ and ${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}⟶{\mathrm{H}}_2\mathrm{O}+q_2$ then the enthalpy change for the reaction ${\mathrm{CH}}_3\mathrm{COOH}⟶{\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}$ is equal to:

1) ${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{OH}}^{-}\to {\mathrm{CHCOO}}^{-}+{\mathrm{H}}_2\mathrm{O\Delta H}=-{\mathrm{q}}_1$

${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\to {\mathrm{H}}_2\mathrm{O\Delta H}=-{\mathrm{q}}_2$

2) ${\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\mathrm{\Delta H}=+{\mathrm{q}}_2$

Adding 1 and 2

${\mathrm{CH}}_3\mathrm{COOH}\to {\mathrm{CHCOO}}^{-}+{\mathrm{H}}^{+}\mathrm{\Delta H}=-{\mathrm{q}}_1+{\mathrm{q}}_2$

Enthalpy change for reaction is a result. $={\mathrm{q}}_2-{\mathrm{q}}_1$

Hence, option C is correct.