Calculate ΔHrxn for the following reaction: CH4 (g) + 2O2 (g) →CO2 (g) + 2H2O (l) Use the following reactions and given ΔH values. CH4 (g) + O2 (g) →CH2O (g) + H2O (g), ΔH=-284 kJ CH2O (g) + O2 (g) →CO2 (g) + H2O (g), ΔH=-527 kJ H2O (l) →H2O (g), ΔH = 44.0 kJ

-899Kj

Explanation

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn = {ΔHH2O (l) + ΔHCO2} - ΔHCH4

We do not put that of Molecular oxygen in the equation as it is zero

ΔHH2O (g) - ΔHH2O (l) = 44

-527 = {ΔHH2O (g) + ΔHCO2} - ΔH{CH2O}

-284 = {ΔHCH2O + ΔHH2O (g) } - ΔH{CH4}

Adding the last 2 equations together will yield the following:

-527-284 = 2ΔH{H2O} (g) + ΔHCO2 - ΔHCH4

From the second equation,

ΔHH2O (g) = 44 + ΔHH2O (l)

Substitute this in the added equation

-811 = 2 (44 + ΔHH2O (l)) + ΔHCO2 - ΔHCH4

-811 = 88 + 2ΔHH2O (l) + ΔHCO2 - ΔHCH4

-899 = 2ΔHH2O (l) + ΔHCO2 - ΔHCH4

The heat of reaction is thus - 899kJ