Question

The enthalpy formation of $H_{2}O\left(l\right)$ is $-285.7$ kJ $mo{l}^{-1}$ and enthalpy of neutralization of strong acid and base is $-57.6$ kJ $mo{l}^{-1}$. what is the enthalpy of formation of $O{H}^{-1}$ ions

Answer 1

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H_{2}O}_{\left(l\right)};\Delta H=-285.7kJ/mol.....\left(1\right)$

Neutralisation of strong acid and base-

${H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)}⟶{H_{2}O}_{\left(l\right)};\Delta H=-57.6kJ/mol$

${H_{2}O}_{\left(l\right)}⟶{H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)};\Delta H=57.6kJ/mol.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}+{H_{2}O}_{\left(l\right)}⟶{H_{2}O}_{\left(l\right)}+{H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)};\Delta H=[(-285.7)+57.6]kJ/mol$

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)};\Delta H=-228.1kJ/mol$

Hence the enthalpy of formation of ${OH}^{-}$ ion is $-228.1kJ/mol$.