On the basis of the following thermochemical data : \; (Δ_f G^ {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 [AIEEE 2009]

Correct Answer is : −228.88 kJ-228.88\ {kJ}−228.88 kJ

Correct Answer is : −228.88 kJ-228.88\ {kJ}−228.88 kJ

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AIEEE 2009 Solved Paper

Section: Chemistry

Question : 5 of 89

Marks: +1, -0

\; (\Delta_f G^\circ \mathrm{H}_{(aq)}^{+}=0)

\; \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq}) ; \Delta H = 57.32\ \mathrm{kJ}

\; \mathrm{H}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l}) ; \Delta H = -286.20\ \mathrm{kJ}

\mathrm{OH}^{-}

25^{\circ}\mathrm{C}

Solution:

Given, for reaction

(i)

\;\; \mathrm{H}_2\mathrm{O}(\ell) \rightarrow \mathrm{H}^{+}(

(aq.

) + \mathrm{OH}^{-}(

(aq.

)

;

\Delta H_r = 57.32\ \mathrm{kJ}

(ii)

\mathrm{H}_2(\mathrm{g}) + \; \frac{1}{2} \ \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\ell)

;

\Delta H_r = -286.20\ \mathrm{kJ}

For reaction

(i)

\Delta H_r = \Delta H^{\circ}_f ( \mathrm{H}^{+} \cdot \mathrm{aq} ) +

\Delta H^{\circ}_f ( \mathrm{OH}^{-} \cdot \mathrm{aq} ) - \Delta H^{\circ}_f ( \mathrm{H}_2\mathrm{O}, \ell )

57.32 = 0 + \Delta H_f^{\circ} ( \mathrm{OH}^{-1}, \mathrm{aq} ) -

\Delta H^{\circ}_f ( \mathrm{H}_2\mathrm{O}, \ell ) \;\; \dots (iii)

For reaction

(ii)

\Delta H_r = \Delta H^{\circ}_f ( \mathrm{H}_2\mathrm{O}, \ell ) - \Delta H^{\circ}_f ( \mathrm{H}_2, \mathrm{g} ) -

\; \frac{1}{2} \Delta H^{\circ}_f ( \mathrm{O}_2, \mathrm{g} )

-286.20 = \Delta H^{\circ}_f ( \mathrm{H}_2\mathrm{O}, \ell )

On replacing this value in equ. ( iii

)

we have

\; 57.32 = \Delta H^{\circ}_f ( \mathrm{OH}^{-} \cdot \mathrm{aq} ) - (-286.20)

\; \Delta H^{\circ}_f = -286.20 + 57.32

\; = -228.88\ \mathrm{kJ}

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