Question

The standard enthalpies of formation of CO2 (g), H2O (l) and CH4 (g) are - 393× 5, - 286× 2 and - 74× 8 kJ mol-1 respectively. Calculate the enthalpy of combustion of methane.

Answer 1

CH4 + 2O2 → CO2 +2 H2O this is the reaction of combustion of methane.

C + O2 → CO2 ∆H1 = -393 Kj/mol
H2 + 1/2O2→ H2O; ∆ H2 = 286.2 Kj/mol 
C + 2H2 → CH4 ; ∆H3 = -74.8 kj/mol 

CH4 → C + 2H2 , ∆H3' = 74.8 
C + O2 → CO2 ∆ H1 = -393 
2( H2 + 1/2O2 → H2O ; ∆H2 = 2× 286.2 

Answer 2

Answer: The enthalpy of combustion of methane is -2517 kJ

Explanation:

Combustion reactions are defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water. The enthalpy change of this reaction is known as enthalpy of combustion.

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_{(product)}]-\sum [n\times \Delta H^o_{(reactant)}]$

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_{(CO_2(g))})+(2\times \Delta H^o_{(H_2O(g))})]-[(1\times \Delta H^o_{(CH_4(g))})+(2\times \Delta H^o_{(O_2(g))})]$

We are given:

$\Delta H^o_{(CH_4(g))}=-74\times 8=-592kJ/mol\\\Delta H^o_{(O_2(g))}=0kJ/mol\\\Delta H^o_{(CO_2(g))}=-393\times 5=-1965kJ/mol\\\Delta H^o_{(H_2O(g))}=-286\times 2=-572kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1965))+(2\times(-572))]-[(1\times (-592))+(2\times 0)]=-2517kJ$

Hence, the enthalpy of combustion of methane is -2517 kJ