QJEEM09Chcc5e: JEE Main 2009 Chemistry - Class 11 JEEM09Chcc5e Solution: On the basis of the following thermochemical data : ( = 0) H 2…

Question

Verified Solution • 2025–26 Session moderate

On the basis of the following thermochemical data :
(

\Delta _fG^oH^+_{(aq)}

= 0)

H₂O(l)

\to

H⁺(aq) + OH^-(aq);

\Delta H

= 57.32 kJ
H₂(g) +

{1 \over 2} O_2(g) \to

H₂O(l);

\Delta H

= -286.20 kJ

The value of enthalpy of formation of OH⁻ ion at 25^oC is :

Accepted Answer

Given, for reaction

(i)

\,\,\,\,\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right);

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = 57.32\,kJ

(ii)

\,\,\,\,\,\,{H_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right);

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

For reaction

(i)

\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H^ + }.aq} \right) +

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)

57.32 = 0 + \Delta {H^ \circ }_f\left( {O{H^{ - 1}},aq} \right) -

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)\,\,\,\,\,\,...\left( {iii} \right)

For reaction

(ii)

\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) - \Delta {H^ \circ }_f\left( {{H_2},g} \right) -

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1 \over 2}\Delta {H^ \circ }_f\left( {{O_2},g} \right)

- 286.20 = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)

On replacing this value in equ.

(iii)

we have

57.32 = \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \left( { - 286.20} \right)

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

= - 228.88\,kJ

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