Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane:
CH4+2O2→2H2O+CO2

ΔHrxn from a given table:
CH4 = 1656 kJ/mol
O2 = 498 kJ/mol
H2O = 928 kJ/mol
CO2 = 1598 kJ/mol
ΔHrxn = [(1656) + (2498)] - [(2928) + (1598)] = -802 kJ/mol ?

The heat of combustion is obtained by getting the difference between the summation of enthalpies of the products and the enthalpies of the reactants. Hence the equation is enthalpy change= 1598 kJ+2*928 kJ-1656 kJ-498 kJ. The answer is 1300 kJ.

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Blitztiger Blitztiger · Answer 2 · 2018-07-02T10:10:09+02:00

Blitztiger Blitztiger

02-07-2018

We subtract the enthalpies of the products by the enthalpies of the reactants, taking into account their stoichiometric coefficients:
2ΔH(H2O) + ΔH(CO2) - ΔH(CH4) - ΔH(2O2)= 2(928) + 1598 - 1656 - 2(498)= 802 kJ / 1 mol methane

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Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane: CH4+2O2→2H2O+CO2 ΔHrxn from a given table: CH4 = 1656 kJ/mol O2 = 498 kJ/mol H2O = 928 kJ/mol CO2 = 1598 kJ/mol ΔHrxn = [(1656) + (2*498)] - [(2*928) + (1598)] = -802 kJ/mol ?

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Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane:
CH4+2O2→2H2O+CO2

ΔHrxn from a given table:
CH4 = 1656 kJ/mol
O2 = 498 kJ/mol
H2O = 928 kJ/mol
CO2 = 1598 kJ/mol
ΔHrxn = [(1656) + (2498)] - [(2928) + (1598)] = -802 kJ/mol ?