Standard Enthalpy of Reaction

The standard enthalpy of reaction, $\Delta H^\ominus _{rxn}$, is the change in enthalpy for a given reaction calculated from the standard enthalpies of formation for all reactants and products. The change in enthalpy does not depend upon the particular pathway of a reaction, but only upon the overall energy level of the products and reactants; enthalpy is a state function, and as such, it is additive. In order to calculate the standard enthalpy of a reaction, we can sum up the standard enthalpies of formation of the reactants and subtract this from the sum of the standard enthalpies of formation of the products. Stated mathematically, this gives us:

$\Delta H^\ominus _{rxn}=\sum \Delta H^\ominus _f\{products\}-\sum \Delta H^\ominus _f\{reactants\}$

Calculating the Standard Enthalpy of Reaction

Calculate the standard enthalpy of reaction for the combustion of methane:

$CH_4(g)+2\;O_2(g)\rightarrow CO_2(g)+2\;H_2O(g)\quad\quad \Delta H^\ominus _{rxn}=?$

In order to calculate the standard enthalpy of reaction, we need to look up the standard enthalpies of formation for each of the reactants and products involved in the reaction. These are typically found in an appendix or in various tables online. For this reaction, the data we need is:

$\Delta H^\ominus _f\{CH_4(g)\}=-75\; kJ/mol$

$\Delta H^\ominus _f\{O_2(g)\}=0\; kJ/mol$

$\Delta H^\ominus _f\{CO_2(g)\}=-394\; kJ/mol$

$\Delta H^\ominus _f\{H_2O(g)\}=-284\; kJ/mol$

Note that because it exists in its standard state, the standard enthalpy of formation for oxygen gas is 0 kJ/mol. Next, we sum up our standard enthalpies of formation. Keep in mind that because the units are in kJ/mol, we need to multiply by the stoichiometric coefficients in the balanced reaction equation.

$\sum\Delta H^\ominus _f\{products\}=\Delta H^\ominus _f\{CO_2(g)\}+\Delta H^\ominus _f\{H_2O(g)\}$

$=(1)(-394)+(2)(-284)=-962\;kJ/mol$

$\sum\Delta H^\ominus _f\{reactants\}=\Delta H^\ominus _f\{CH_4(g)\}+\Delta H^\ominus _f\{O_2(g)\}$

$=(1)(-75)+(2)(0)=-75\;kJ/mol$

Now, we can find the standard enthalpy of reaction:

$\Delta H^\ominus _{rxn}=\sum \Delta H^\ominus _f\{products\}-\sum \Delta H^\ominus _f\{reactants\}$

$=(-962)-(-75)=-887\;kJ/mol$

As we would expect, the standard enthalpy for this combustion reaction is strongly exothermic.