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Question

Given, that,

H2O(l)H+(aq)+OH(aq);ΔH=57.32kJ

H2(g)+12O2(g)H2O(l);ΔH=286.02kJ

Then, calculate the enthalpy of formation of OH at 25oC is:

A
-228.8 kJ
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B
-343.53 kJ
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C
-33.53 kJ
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D
228.8 kJ
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Solution

The correct option is A -228.8 kJ
H2(g)+12O2(g)H2O(l);ΔH=286.02KJ

Then ΔHreaction=ΔHf(products)ΔHf(reactants)

ΔHreaction=ΔHf(H2O)ΔHf(H2)12ΔHf(O2)

286.02=ΔHf(H2O)012(0)

ΔHf(H2O)=286.02KJ

Now for ionization of water (H2O)

H2O(l)H+(aq)+OH(aq);ΔH=57.32KJ

ΔHreaction=ΔHf(products)ΔHf(reactants)

ΔHreaction=ΔHf(H+)+ΔHf(OH)ΔHf(H2O)

57.32=0+ΔHf(OH)(286.02)

ΔHf(OH)=286.0257.32

ΔHf(OH)=228.87

ΔHf(OH)=228.8 kJ

Hence, the correct option is A

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