Question

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 1

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 $1\;bar\;pressure$ .
$\Delta {H^ \circ }_{rxn} = \sum {\Delta H_f^ \circ } (products) - \sum {\Delta H_f^ \circ } (reactants)$
Where, $\Delta {H^ \circ }_{rxn}$ is the rate of change of enthalpy of reaction
$\Delta H_f^ \circ (products)$ is the rate of change enthalpy of the product
$\Delta H_f^ \circ (reactants)$ is the rate of change enthalpy of the reactants.

Complete Step By Step Answer:
As we know the enthalpy of combustion is $\Delta {H^ \circ } = - 890.7kJ\;mo{l^{ - 1}}$
We also know that the enthalpy is associated with the combustion reaction, i.e.
$C{H_4}(g) + {O_2}(g) \to C{O_2}(g) + {H_2}O(l) + {\Delta _{rxn}}$
We have to balance the given chemical reaction:
$C{H_4}(g) + {O_2}(g) \to C{O_2}(g) + 2{H_2}O(l) + {\Delta _{rxn}}$
With heat of formation values given by the following table:

Substance	$\begin{gathered} \Delta H_f^ \circ \\\ (kJmo{l^{ - 1}}) \\\ \end{gathered}$
$C{H_4}(g)$	$- 74.85$
$C{O_2}(g)$	$- 393.5$
${H_2}O(l)$	$- 286$

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:
$\Delta {H^ \circ }_{rxn} = \sum {\Delta H_f^ \circ } (products) - \sum {\Delta H_f^ \circ } (reactants)$
By keeping the given values we will get:
\Delta H_{rxn}^ \circ = \left\\{ {\left( { - 393.5 - 2 \times 286} \right) - \left( { - 74.85} \right)} \right\\}kJ\;mo{l^{ - 1}}
$\Rightarrow \Delta H = - 890.7kJ\;mo{l^{ - 1}}$
Hence, the enthalpy of combustion $\Delta {H^ \circ }$ is $- 890.7kJ\;mo{l^{ - 1}}$ .

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.