calculate the enthalpy of formation of methane

C +O2 --CO2 deltaH = -393.7 KJ/MOL

H2+1/2 O2 ---H2O H= -285.8 KJ/MOL

CH4 +2O2 ---CO2+2H2O H = -890.4 KJ/MOL

C(s) + 2 H

_{2}(g) → CH

_{4}(g)

Given equations:

1. C(s) + O

_{2}(g) $\to $ CO

_{2}(g)..................................... ΔH

_{1}= –393.7 kJmol-1

2. H

_{2}(g) + 1/2 O

_{2}(g) $\to $ H

_{2}O(g)............................ ΔH2 = –285.8 kJmol-1

3. CH

_{4}(g) + 2 O

_{2}(g) $\to $ CO

_{2}(g) + 2 H

_{2}O(g)........ ...ΔH3 = –890.4 kJmol-1

Equation 1 already has carbon on the left side.

Equation 2 already has hydrogen on the left side, but we need two hydrogen molecules, so we multiply second equation by 2.

Equation 3 has methane on the left side instead of on the right side, so we multiply it by -1

1. C(s) + O

_{2}(g) $\to $ CO

_{2}(g)..................................... ΔH

_{1}= –393.7 kJmol

^{-1}

^{ }

2. 2H

_{2}(g) + O

_{2}(g) $\to $ 2 H

_{2}O(g)............................. .ΔH

_{2}

_{ }= –571.6 kJmol

^{-1}

^{ }

3. CO

_{2}(g) + 2 H

_{2}O(g) $\to $ CH

_{4}(g) + 2 O

_{2}(g).............ΔH

_{3}= +890.4 kJmol

^{-1}

^{ }

Eliminating oxygen, carbon dioxide, and water on both sides in all the three equations, we get:

Total: C(s) + 2 H

_{2}(g) $\to $ CH

_{4}(g).............................ΔH = –74.9 kJmol

^{-1}

^{ }

The standard heat of formation for methane is –74.9 kJmol

^{-1}.

**
**