On the basis of the following thermochemical data: (Δ fG^° H_(aq)^+=0 ) H_2 O(ℓ) arrow H^+(aq)+OH^-(aq) ; Δ H=57.32 ~kJ H_2(~g)+1/2 O_2(~g) arrow H_2 O(ℓ) ; Δ H=-286.20 ~kJ The value of enthalpy of formation of OH^- ion at 25^° C is:

Correct answer: (B) -228.88 ~kJ. By adding the two given equations, we have H_2(~g)+1/2 O_2(~g) arrow H_(aq)^++OH_(aq)^- ; Δ H=-228.88 Kj Here Δ H_f^° of H_(aq)^+=0 ∴ Δ H_f^° of OH^-=-228.88 ~kJ

Q14 / 90 · JEE Main 2009 · Chemistry

On the basis of the following thermochemical data: $\left(\Delta \mathrm{fG}^{\circ} \mathrm{H}_{\text {(aq) }}^{+}=0\right)$ $$ \begin{aligned} & \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ; \Delta \mathrm{H}=57.32 \mathrm{~kJ} \\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286.20 \mathrm{~kJ} \end{aligned} $$ The value of enthalpy of formation of $\mathrm{OH}^{-}$ion at $25^{\circ} \mathrm{C}$ is:

A
$-22.88 \mathrm{~kJ}$
Correct answer
B
$-228.88 \mathrm{~kJ}$
C
$+228.88 \mathrm{~kJ}$
D
$-343.52 \mathrm{~kJ}$

Solution

By adding the two given equations, we have $$ \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} ; \Delta \mathrm{H}=-228.88 \mathrm{Kj} $$ Here $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}$ of $\mathrm{H}_{(\mathrm{aq})}^{+}=0$ $\therefore \Delta \mathrm{H}_{\mathrm{f}}^{\circ}$ of $\mathrm{OH}^{-}=-228.88 \mathrm{~kJ}$

Hence, the correct answer is option: 'B'.

Solution

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