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21 October, 13:26

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

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  1. 21 October, 15:53
    0
    -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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