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

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

2b2t

explanation:

2b2t