The enthalpy of neutralization of NaOH by HCl | vedclass

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(C) The enthalpy of neutralization of a strong acid with a strong base is always $-57.1 \, kJ/mol$ (or $-55.84 \, kJ/mol$ as given). For a weak base l

Vedclass · Accepted Answer

$4.5$

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(C) The enthalpy of neutralization of a strong acid with a strong base is always $-57.1 \, kJ/mol$ (or $-55.84 \, kJ/mol$ as given). For a weak base like $NH_4OH$,some energy is consumed in the ionization of the weak base. The enthalpy of neutralization of a weak base is given by: $\Delta H_{	ext{neutralization}}(NH_4OH) = \Delta H_{	ext{neutralization}}(	ext{strong acid-strong base}) + \Delta H_{	ext{ionization}}(NH_4OH)$ Given: $\Delta H_{	ext{neutralization}}(NaOH) = -55.84 \, kJ/mol$ $\Delta H_{	ext{neutralization}}(NH_4OH) = -51.34 \, kJ/mol$ Therefore,$\Delta H_{	ext{ionization}}(NH_4OH) = \Delta H_{	ext{neutralization}}(NH_4OH) - \Delta H_{	ext{neutralization}}(NaOH)$ $\Delta H_{	ext{ionization}}(NH_4OH) = -51.34 - (-55.84) = 4.5 \, kJ/mol$ Thus,the enthalpy of ionization of $NH_4OH$ is $4.5 \, kJ/mol$.

The enthalpy of neutralization of $NaOH$ by $HCl$ is $-55.84 \, kJ/mol$ and that of $NH_4OH$ by $HCl$ is $-51.34 \, kJ/mol$. The enthalpy of ionization of $NH_4OH$ is ..... $kJ/mol$.

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