Let's consider a balloon filled with H
2 and O
2. If a lighter is put under the ballon, 2H
2 + O
2 2H
2O
First, we're breaking:
1) Two H-H bonds (2 * 436.4 kJ/mol = Investing 872.8 kJ/mol)
2) One O=O bond (double bond) (1 * 489.7 kJ/mol = Investing 498.7 kJ/mol)
in total, 872.8 + 498.7 = 1371.5 kJ/mol
Then, four O-H bonds are formed:
4 * 460 kJ/mol = releasing 1840 kJ/mol.
My question is:
These molecules will release 1840 kJ/mol, and the lighter "provided" them with only 1371 kJ/ mol. Where does the extra 469 kJ/mol come from?