# 5.3 Enthalpy  (Page 9/25)

 Page 9 / 25

We also can use Hess’s law    to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with ∑ representing “the sum of” and n standing for the stoichiometric coefficients:

$\text{Δ}{H}_{\text{reaction}}^{°}=\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{products}\right)-\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{reactants}\right)$

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:

$3{\text{NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)+\text{NO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=?$

## Solution: using the equation

Use the special form of Hess’s law given previously:

$\text{Δ}{H}_{\text{reaction}}^{°}=\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\text{(products)}-\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{reactants}\right)$
$\begin{array}{l}\\ \\ \\ =\left[2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{-207.4\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}\left(aq\right)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol NO}\left(g\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{+90.2 kJ}}{\overline{)\text{mol NO}\left(g\right)}}\right]\\ -\left[3\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{+33.2 kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{-285.8\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)}}\right]\\ =2\left(-207.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+1\left(\text{+90.2 kJ}\right)-3\left(\text{+33.2 kJ}\right)-1\left(-285.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)\\ =-138.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$

## 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 $\text{Δ}{H}_{\text{f}}^{°}$ values for these compounds (from Appendix G ), we have:

$3{\text{NO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3/2N}}_{2}\left(g\right)+{\text{3O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{1}^{°}=-99.6\phantom{\rule{0.2em}{0ex}}\text{kJ}$
${\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{2}^{°}=\text{+285.8 kJ}\phantom{\rule{0.2em}{0ex}}\left[-1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\right)\right]$
${\text{H}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{3}^{°}=-414.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\phantom{\rule{0.2em}{0ex}}\left[2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{HNO}}_{3}\right)\right]$
$\frac{1}{2}{\text{N}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{4}^{°}=\text{+90.2 kJ}\phantom{\rule{0.2em}{0ex}}\left[1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{NO}\right)\right]$

Summing these reaction equations gives the reaction we are interested in:

${\text{3NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)+\text{NO}\left(g\right)$

Summing their enthalpy changes gives the value we want to determine:

$\begin{array}{cc}\hfill \text{Δ}{H}_{\text{rxn}}^{°}& =\text{Δ}{H}_{1}^{°}+\text{Δ}{H}_{2}^{°}+\text{Δ}{H}_{3}^{°}+\text{Δ}{H}_{4}^{°}=\left(-99.6\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(\text{+285.8 kJ}\right)+\left(-414.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(\text{+90.2 kJ}\right)\hfill \\ & =-138.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \end{array}$

So the standard enthalpy change for this reaction is Δ H ° = −138.4 kJ.

Note that this result was obtained by (1) multiplying the $\text{Δ}{H}_{\text{f}}^{°}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the $\text{Δ}{H}_{\text{f}}^{°}$ 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.

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.

−1368 kJ/mol

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