Combustion of methane. Refer to the second question in the attached image. Apparently with one mole of methane combustion the pressure and volume are different for gasses in the reaction. Course Contents » .. » Homework set #5 >Enthalpy of Combustion -- AlkanesWrite a balanced equation for the combustion of CH (g) (methane)i.e. its reaction with O,(g) forming the products CO2(g) and H20().Given the following standard heats of formation:AH of CO2(g) is -393.5 kJ/molAH of H20(1) is -286 ki/molAH of CH,(g) is -74.8 kJ/molWhat is the standard heat of reaction (AH) for the combustion reaction of CH(g)?-890.7 kJYou are correct.Previous TriesYour receipt no. is 164-687Calculate the difference, AH-AE=A(PV) for the combustion reaction of 1 mole of methane.(Assume standard state conditions and 298 K for all reactants and products.)-1094KJThe ideal gas law must be used.The incremental change in volume due to liquids is neglected (cf the volume of 1 mole of H,0Be careful to account for the total change in number of moles of gas molecules (be sure the sNote that certain alkanes are gases in the standard state.

Given, Heat of combustion of methane ∆H°= -890.7 kJUsing the equation : ∆H-∆E=∆(PV),where,(∆H-∆E) = change in energyP= pressureV = volume From ideal gas law :∆(PV) =nRT where,n = number of moles = 1(given)R= gas constant = 8.314Jmol KT= temperature = 298 KPutting values :⇒∆H-∆E =1mol ×8.314 Jmol K×298 K=22477.572 J⇒(∆H-∆E)=2.47776 kJ So, difference in the energy = 2.4776 kJ