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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

Respuesta :

Answer :

-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

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