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Question

H2(g)+12O2(g)H2O(l);ΔH295 K=285.8 kJ.
The molar enthalpy of vapourisation of water at 1 atm and 25oC is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at 25oC is:

A
241.8 kJ
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B
-241.8 kJ
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C
-329.8 kJ
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D
329.8 kJ
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Solution

The correct option is B -241.8 kJ
H2(g)+12O2(g)H2O(l); ΔH1=285.8kJ

H2O(l)H2O(g); ΔH2=44kJ

Formation of water vapour:

H2(g)+12O2(g)H2O(g); ΔH3

Here, ΔH3=ΔH1+ΔH2

ΔH3=285.8+44

ΔH3=241.8 kJ

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