I missed a couple of classes when they went over this so I needsome help pleeez.

Question

I missed a couple of classes when they went over this so I needsome help pleeez. The ? is...
Given H2(g) +(1/2)O2(g)---->H2O(l), H=-286 kJ/mol, determine the standard enthalpy change for thereaction2H2O(l)---->2H2(g)+O2(g)
This is not in the book it's a handout. Any step by step is greatlyappreciated, my brain is old and tired right now!!!!!
Thank you
Lori

Explanation / Answer

Given H2(g) +(1/2)O2(g)---->H2O(l), H=-286 kJ/mol, Inverse it: H2O(l) ---> H2 (g)+1/2O2 (g) ==> H = 286kJ/mol But the reaction gives:          2H2O(l)---->2H2(g)+O2(g) means double the enthalpy             => H = 2 * 286kJ/mol = 572kJ/mol

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