The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.
Using hess’s law
What is the standard enthalpy change for the reaction:
$3{\text{NO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)+\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=?$
Solution: using the equation
Use the special form of Hess’s law given previously:
$\text{\Delta}{H}_{\text{reaction}}^{\xb0}={\displaystyle \sum n\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}\text{(products)}}-{\displaystyle \sum n}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}(\text{reactants})$
$\begin{array}{l}\\ \\ \\ =\left[2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\mathrm{-207.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}(aq)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol NO}(g)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{+90.2 kJ}}{\overline{)\text{mol NO}(g)}}\right]\\ -\left[3\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{+33.2 kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\mathrm{-285.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)}}\right]\\ =2\left(\mathrm{-207.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+1(\text{+90.2 kJ})-3\left(\text{+33.2 kJ}\right)-1\left(\mathrm{-285.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)\\ =\mathrm{-138.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$
Solution: supporting why the general equation is valid
Alternatively, we can write this reaction as the sum of the decompositions of 3NO
_{2} (
g ) and 1H
_{2} O(
l ) into their constituent elements, and the formation of 2HNO
_{3} (
aq ) and 1NO(
g ) from their constituent elements. Writing out these reactions, and noting their relationships to the
$\text{\Delta}{H}_{\text{f}}^{\xb0}$ values for these compounds (from
Appendix G ), we have:
$3{\text{NO}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{3/2N}}_{2}(g)+{\text{3O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{1}^{\xb0}=\mathrm{-99.6}\phantom{\rule{0.2em}{0ex}}\text{kJ}$
${\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{2}^{\xb0}=\text{+285.8 kJ}\phantom{\rule{0.2em}{0ex}}\mathrm{[-1}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}({\text{H}}_{2}\text{O})]$
${\text{H}}_{2}(g)+{\text{N}}_{2}(g)+\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{3}^{\xb0}=\mathrm{-414.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}\phantom{\rule{0.2em}{0ex}}[2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}({\text{HNO}}_{3})]$
$\frac{1}{2}{\text{N}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{4}^{\xb0}=\text{+90.2 kJ}\phantom{\rule{0.2em}{0ex}}[1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}(\text{NO})]$
Summing these reaction equations gives the reaction we are interested in:
${\text{3NO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)+\text{NO}(g)$
Summing their enthalpy changes gives the value we want to determine:
$\begin{array}{cc}\hfill \text{\Delta}{H}_{\text{rxn}}^{\xb0}& =\text{\Delta}{H}_{1}^{\xb0}+\text{\Delta}{H}_{2}^{\xb0}+\text{\Delta}{H}_{3}^{\xb0}+\text{\Delta}{H}_{4}^{\xb0}=(\mathrm{-99.6}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\text{+285.8 kJ})+(\mathrm{-414.8}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\text{+90.2 kJ})\hfill \\ & =\mathrm{-138.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \end{array}$
So the standard enthalpy change for this reaction is Δ
H ° = −138.4 kJ.
Note that this result was obtained by (1) multiplying the
$\text{\Delta}{H}_{\text{f}}^{\xb0}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the
$\text{\Delta}{H}_{\text{f}}^{\xb0}$ of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.
Check your learning
Calculate the heat of combustion of 1 mole of ethanol, C
_{2} H
_{5} OH(
l ), when H
_{2} O(
l ) and CO
_{2} (
g ) are formed. Use the following enthalpies of formation: C
_{2} H
_{5} OH(
l ), −278 kJ/mol; H
_{2} O(
l ), −286 kJ/mol; and CO
_{2} (
g ), −394 kJ/mol.
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