# calculate the enthalpy of formation of methane C +O2 --CO2 deltaH = -393.7 KJ/MOLH2+1/2 O2 ---H2O H= -285.8 KJ/MOLCH4 +2O2 ---CO2+2H2O H = -890.4 KJ/MOL

The formation of methane from its elements is given as,

C(s) + 2 H2(g) → CH4(g)

Given equations:

1. C(s) + O2(g) $\to$ CO2(g)..................................... ΔH1 = –393.7 kJmol-1
2. H2(g) + 1/2 O2(g) $\to$ H2O(g)............................ ΔH2 = –285.8 kJmol-1
3. CH4(g) + 2 O2(g) $\to$ CO2(g) + 2 H2O(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) + O2(g) $\to$ CO2(g)..................................... ΔH1 = –393.7​ kJmol-1
2. 2H2(g) + O2(g) $\to$ 2 H2O(g)............................. .ΔH2 = –571.6 kJmol-1
3. CO2(g) + 2 H2O(g) $\to$ CH4(g) + 2 O2(g).............ΔH3 = +890.4​ kJmol-1

Eliminating oxygen, carbon dioxide, and water on both sides in all the three equations, we get:
Total: C(s) + 2 H2(g) $\to$ CH4(g).............................ΔH = –74.9 kJmol-1

The standard heat of formation for methane is –74.9 kJmol-1.

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