Question

The enthalpy of neutralisation of a strong acid by a strong base is $-57.32$ kJ $mo{l}^{-1}$. The enthalpy of formation of water is $-285.84$ kJ $mo{l}^{-1}$. The enthalpy of formation of hydroxyl ion is:

• A. $+228.52kJ.mo{l}^{-1}$
• B. $-114.26kJ.mo{l}^{-1}$
• C. $-228.52kJ.mo{l}^{-1}$
• D. $+114.2kJ.mo{l}^{-1}$

Answer 1

The correct option is C $-228.52kJ.mo{l}^{-1}$

Solution:- (C) $-228.52KJ/mol$

Neutralisation of strong acid and strng base-

${H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)}⟶{H_{2}O}_{\left(l\right)};\Delta H=-57.32KJ/mol$

${H_{2}O}_{\left(l\right)}⟶{H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)};\Delta H=57.32KJ/mol.....\left(1\right)$

Formation of water-

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H_{2}O}_{\left(l\right)};\Delta H=-285.84KJ/mol.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we will get the enthalpy of formation of ${OH}^{-}$.

Therefore,

${H_{2}O}_{\left(l\right)}+{H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)}+{H_{2}O}_{\left(l\right)};\Delta H=-285.84+57.32$

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H^{+}}_{(aq.)}+{{OH}^{-}}_{(aq.)};\Delta H=-228.52KJ/mol$

Hence the enthalpy of formation of ${OH}^{-}$ ion is $-228.52KJ/mol$.