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The enthalpy change of formation of carbon dioxide is –394 kJ mol–1. The enthalpy change of formation of water is –286 kJ mol–1. The enthalpy change of formation of methane is –74 kJ mol–1. What is the enthalpy change of combustion of methane?

User LeoGalante
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2 Answers

7 votes

Answer:

-655

Step-by-step explanation:

ye

User Jakob Harteg
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The enthalpy change of combustion of methane : -892 kJ/mol

Further explanation

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

Reaction for combustion of Methane (CH₄) :

CH₄ + 2O₂ → CO₂ + 2H₂O.

The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero(∆Hf °O₂ =0)

∆Hf ° CO₂ = -394 kj/mol

∆Hf ° H₂O=-286 kj/mol

∆Hf ° CH₄=-74 kJ/mol

The enthalpy change of combustion of methane :


\tt \Delta H_(rxn)=\Delta H_f~CO_2+2.\Delta H_fH_2O-(\Delta H_fCH_4)\\\\\Delta H_(rxn)=(-394-2.-286)-(-74)\\\\\Delta H_(rxn)=-892~kJ/mol

User Ddmps
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