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Question

On the basis of the following thermochemical data:(ΔfG0(H)+(aq)=0)

H2O(l)H+(aq)+OH(aq); ΔH=57.32kJ
H2(g)+12O2(g)H2O(l); ΔH=286.20kJ
The value of enthalpy of formation of OH ion at 25oC is:

A
228.51 kJ
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B
198.51 kJ
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C
+238.5 kJ
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D
238.5 kJ
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Solution

The correct option is B 228.51 kJ
Solution:- (A) 228.88kJ
Given:-

H2O(l)H+(aq.)+OH(aq.)ΔH=57.32kJ.....(1)

H2(g)+12O2(g)H2O(l)ΔH=286.20kJ.....(2)

Adding eqn(1)&(2), we have

H2O(l)+H2(g)+12O2(g)H+(aq.)+OH(aq.)+H2O(l);ΔH=(57.32)+(286.20)kJ

H2(g)+12O2(g)H+(aq.)+OH(aq.);ΔH=228.88kJ

Hence the value of enthalpy of formation of OH ion at 25 is 228.88kJ.

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