Given H2(g) + (1/2)O2(g) --> H2O, ∆Hº = -286 kJ/mol, determine the standard enthalpy change for the reaction 2H2O(l) --> 2H2(g) + O2(g).

Just checking... should end up with ∆Hº = 572 kJ/mol right??

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  1. Yes and no. Yes, the correct number is 572 kJ/reaction (not /mol since it's for two mols) AND I don't think you should call it dHo. dHo is -285 kJ/mol is for the H2 + 1/2 O2 ==> H2O. When you turn it around it becomes for the reaction and multiplying by 2 makes is for two mols. I know I'm picky but I'm earning my money. :-). Since I'm a volunteer there is no money but you get the point.

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