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