H2(g)+12O2(g)→H2O(l);ΔH295K=−285.8kJ. 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