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Calculate the enthalpy of combustion of methane, if the standard enthalpies of formation of methane, carbon dioxide, water are 74.85 , 393.5 and 286 ?

Answer
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Hint: Enthalpy of combustion is the change in enthalpy when one mole of the compound is completely burn in the presence of excess of oxygen with all the reactants and products in their normal states under normal conditions, i.e. 298K and 1barpressure .
 ΔHrxn=ΔHf(products)ΔHf(reactants)
Where, ΔHrxn is the rate of change of enthalpy of reaction
 ΔHf(products) is the rate of change enthalpy of the product
 ΔHf(reactants) is the rate of change enthalpy of the reactants.

Complete Step By Step Answer:
As we know the enthalpy of combustion is ΔH=890.7kJmol1
We also know that the enthalpy is associated with the combustion reaction, i.e.
 CH4(g)+O2(g)CO2(g)+H2O(l)+Δrxn
We have to balance the given chemical reaction:
 CH4(g)+O2(g)CO2(g)+2H2O(l)+Δrxn
With heat of formation values given by the following table:
Substance ΔHf(kJmol1)
CH4(g) 74.85
CO2(g) 393.5
H2O(l) 286

Hence according to the fact, the rate of change of enthalpy of the reaction is equal to the difference of the summation of rate of change of enthalpy of the product and the summation of rate of change of enthalpy of the reactants. This mathematically can be representing as:
 ΔHrxn=ΔHf(products)ΔHf(reactants)
By keeping the given values we will get:
 ΔHrxn={(393.52×286)(74.85)}kJmol1
 ΔH=890.7kJmol1
Hence, the enthalpy of combustion ΔH is 890.7kJmol1 .

Note:
Always remember, since enthalpy is a state function, the sign of the enthalpy of a reaction changes when the process is reversed, which means it does not depend upon the path. Hence, whatever reaction is used , the enthalpy of the final reaction is not affected.