JEE MAIN - Chemistry (2009 - No. 7)
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 25oC is :
($$\Delta _fG^oH^+_{(aq)}$$ = 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 25oC is :
-22.88 kJ
-228.88 kJ
+228.88 kJ
-343.52 kJ
Explanation
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
$$57.32 = \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \left( { - 286.20} \right)$$
$$\Delta {H^ \circ }_f = - 286.20 + 57.32$$
$$ = - 228.88\,kJ$$
$$(i)$$ $$\,\,\,\,\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right);$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = 57.32\,kJ$$
$$(ii)$$ $$\,\,\,\,\,\,{H_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right);$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = - 286.20\,kJ$$
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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