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 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:

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

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

Substituting the given data:

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

answer is a

explanation:

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