Use bond enthalpy values to calculate the enthalpy change for the following reaction. Then, label if the reaction is exothermic or endothermic. CH4(g)+2O2(g)-->CO2(g)+2H2O(g)

Answer 1

Breaking bonds:
- 4 C-H bonds @ 413 kJ/mol = 1652 kJ/mol
- 2 O=O bonds @ 498 kJ/mol = 996 kJ/mol

Forming bonds:
- 2 C=O bonds @ 799 kJ/mol = 1598 kJ/mol
- 4 O-H bonds @ 463 kJ/mol = 1852 kJ/mol

Enthalpy change (ΔH) = (Σ bonds broken) - (Σ bonds formed)
ΔH = (1652 kJ/mol + 996 kJ/mol) - (1598 kJ/mol + 1852 kJ/mol)
ΔH = -652 kJ/mol

Since the value of ΔH is negative, the reaction is exothermic.

Answer 2

is that correct

Answer 3

Yes, that is correct!