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?

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

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 :

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.