Question

Given H2(g) + (1/2)O2(g) ---> H2O(l), dH = -286 kJ/mol, determine the standard enthalpy change for the reaction 2h2O(l) ---> 2H2(g) + O2(g)

2H2O(l) ---> 2H2(g) + O2(g)

H2(g) + (1/2) O29g) ---> H2O (l) : dH = -286 kJ/mols

2H2O(l) ---> 2H2(g) + 2O2(g) : would dH be -572 or +572?

Answer 1

You want twice the reverse reaction given in the problem. So that's -(-286)*2 = ?