# 5.3 Enthalpy  (Page 3/25)

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The following conventions apply when we use Δ H :

1. Chemists use a thermochemical equation to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a Δ H value following the equation for the reaction. This Δ H value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation . For example, consider this equation:

${\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{2em}{0ex}}\text{Δ}\text{H}=-286\phantom{\rule{0.2em}{0ex}}\text{kJ}$

This equation indicates that when 1 mole of hydrogen gas and $\frac{1}{2}$ mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (Δ H is an extensive property):

$\begin{array}{l}\text{(two-fold increase in amounts)}\\ 2{\text{H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(-286\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)=-572\phantom{\rule{0.2em}{0ex}}\text{kJ}\\ \left(\text{two-fold decrease in amounts}\right)\\ \frac{1}{2}{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{4}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(-286\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)=-143\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$
2. The enthalpy change of a reaction depends on the physical state of the reactants and products of the reaction (whether we have gases, liquids, solids, or aqueous solutions), so these must be shown. For example, when 1 mole of hydrogen gas and $\frac{1}{2}$ mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released.

${\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=-242\phantom{\rule{0.2em}{0ex}}\text{kJ}$
3. A negative value of an enthalpy change, Δ H , indicates an exothermic reaction; a positive value of Δ H indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its Δ H is changed (a process that is endothermic in one direction is exothermic in the opposite direction).

## Measurement of an enthalpy change

When 0.0500 mol of HCl( aq ) reacts with 0.0500 mol of NaOH( aq ) to form 0.0500 mol of NaCl( aq ), 2.9 kJ of heat are produced. What is Δ H , the enthalpy change, per mole of acid reacting, for the acid-base reaction run under the conditions described in [link] ?

$\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NaCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)$

## Solution

For the reaction of 0.0500 mol acid (HCl), q = −2.9 kJ. This ratio $\frac{-2.9\phantom{\rule{0.2em}{0ex}}\text{kJ}}{0.0500\phantom{\rule{0.2em}{0ex}}\text{mol HCl}}$ can be used as a conversion factor to find the heat produced when 1 mole of HCl reacts:

$\text{Δ}\text{H}=1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol HCl}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{-2.9\phantom{\rule{0.2em}{0ex}}\text{kJ}}{0.0500\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol HCl}}}\phantom{\rule{0.2em}{0ex}}=-58\phantom{\rule{0.2em}{0ex}}\text{kJ}$

The enthalpy change when 1 mole of HCl reacts is −58 kJ. Since that is the number of moles in the chemical equation, we write the thermochemical equation as:

$\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NaCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=-58\phantom{\rule{0.2em}{0ex}}\text{kJ}$

When 1.34 g Zn( s ) reacts with 60.0 mL of 0.750 M HCl( aq ), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:

$\text{Zn}\left(s\right)+2\text{HCl}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ZnCl}}_{2}\left(aq\right)+{\text{H}}_{2}\left(g\right)$

Δ H = −153 kJ

Be sure to take both stoichiometry and limiting reactants into account when determining the Δ H for a chemical reaction.

## Another example of the measurement of an enthalpy change

A gummy bear contains 2.67 g sucrose, C 12 H 22 O 11 . When it reacts with 7.19 g potassium chlorate, KClO 3 , 43.7 kJ of heat are produced. Determine the enthalpy change for the reaction

${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(aq\right)+8{\text{KClO}}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}12{\text{CO}}_{2}\left(g\right)+11{\text{H}}_{2}\text{O}\left(l\right)+8\text{KCl}\left(aq\right).$

## Solution

We have $2.67\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}}{342.3\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}}\phantom{\rule{0.2em}{0ex}}=0.00780\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ available, and $7.19\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}}{122.5\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}}\phantom{\rule{0.2em}{0ex}}=0.0587\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{KClO}}_{3}$ available. Since $0.0587\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{KClO}}_{3}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}{8\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{KClO}}_{3}}\phantom{\rule{0.2em}{0ex}}=0.00734\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ is needed, C 12 H 22 O 11 is the excess reactant and KClO 3 is the limiting reactant.

The reaction uses 8 mol KClO 3 , and the conversion factor is $\frac{-43.7\phantom{\rule{0.2em}{0ex}}\text{kJ}}{0.0587\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{KClO}}_{3}},$ so we have $\text{Δ}\text{H}=8\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{-43.7\phantom{\rule{0.2em}{0ex}}\text{kJ}}{0.0587\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{KClO}}_{3}}\phantom{\rule{0.2em}{0ex}}=-5960\phantom{\rule{0.2em}{0ex}}\text{kJ}.$ The enthalpy change for this reaction is −5960 kJ, and the thermochemical equation is:

${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}+8{\text{KClO}}_{3}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}12{\text{CO}}_{2}+11{\text{H}}_{2}\text{O}+8\text{KCl}\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=-5960\phantom{\rule{0.2em}{0ex}}\text{kJ}$

When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of FeCl 2 ( s ) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of FeCl 2 ( s ) is produced?

Δ H = −338 kJ

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