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Heat of Formation

The enthalpy ...

Question

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

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Solution

${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 285.7 k J / m o l . . . . . (1)$
Neutralisation of strong acid and base-
${H^{+}}_{(a q .)} + {O H}^{-}_{(a q .)} ⟶ {H_{2} O}_{(l)}; Δ H = - 57.6 k J / m o l$
${H_{2} O}_{(l)} ⟶ {H^{+}}_{(a q .)} + {O H}^{-}_{(a q .)}; Δ H = 57.6 k J / m o l . . . . . (2)$
Adding ${e q}^{n} (1) & (2)$ , we have
${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} + {H_{2} O}_{(l)} ⟶ {H_{2} O}_{(l)} + {H^{+}}_{(a q .)} + {O H}^{-}_{(a q .)}; Δ H = [(- 285.7) + 57.6] k J / m o l$
${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H^{+}}_{(a q .)} + {O H}^{-}_{(a q .)}; Δ H = - 228.1 k J / m o l$
Hence the enthalpy of formation of ${O H}^{-}$ ion is $- 228.1 k J / m o l$ .

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