The enthalpies of neutralization of HCN and H | vedclass

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(D) The enthalpy of neutralization of a weak acid with a strong base is given by $\Delta H_{neut} = -57.3 + \Delta H_{ion}$,where $\Delta H_{ion}$ is

Vedclass · Accepted Answer

$pK_a$ of $HCN$ is greater than $pK_a$ of $H_2S$

Vedclass · Answer

(D) The enthalpy of neutralization of a weak acid with a strong base is given by $\Delta H_{neut} = -57.3 + \Delta H_{ion}$,where $\Delta H_{ion}$ is the enthalpy of ionization of the weak acid.Since the enthalpy of neutralization is less exothermic for $HCN$ $(12.13 \ kJ \ equi^{-1})$ compared to $H_2S$ $(15.9 \ kJ \ equi^{-1})$,it implies that more energy is required to ionize $HCN$ than $H_2S$.This indicates that $HCN$ is a weaker acid than $H_2S$.Since $pK_a = -\log(K_a)$,a weaker acid has a smaller $K_a$ value and consequently a larger $pK_a$ value.Therefore,the $pK_a$ of $HCN$ is greater than the $pK_a$ of $H_2S$.

The enthalpies of neutralization of $HCN$ and $H_2S$ with a strong base are $12.13 \ kJ \ equi^{-1}$ and $15.9 \ kJ \ equi^{-1}$ respectively. Then:

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