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?

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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Annainn Annainn · Answer 2 · 2019-07-20T12:11:00+02:00

Answer:

ΔHrxn for the combustion of CH4 =802 kJ

Explanation:

Enthalpy of combustion is the amount of heat released when 1 mole of a substance buring completely in the presence of air or oxygen. The given reaction is:

[tex]CH4 + 2O2 \rightarrow CO2 + 2H2O[/tex]

The enthalpy of combustion of methane can be deduced using the following equation:

[tex]\Delta Hrxn = \sum n_{p}\Delta H_{f}^{0}(products)-\sum n_{r}\Delta H_{f}^{0}(reactants)[/tex]

where np and nr are the number of moles of products and reactants

ΔH⁰f are the standard enthalpies of formation of the respective reactants and products

[tex]\Delta Hrxn = [2\Delta H_{f}^{0}(H2O)+1\Delta H_{f}^{0}(CO2)]-[1\Delta H_{f}^{0}(CH4)+ 2\Delta H_{f}^{0}(O2)][/tex]

Substituting the given data:

ΔH = [2(928) + 1(1598)] - [1(1656) + 2(498)] = 802 kJ