3.4: Enthalpy of Reaction

We have stated that the change in energy ([latex]ΔU[/latex]) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work, (or just [latex]PV[/latex] work). Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:

[latex]\ce{Cu(s) + 4HNO3(aq) \rightarrow Cu(NO3)2(aq) + 2H_2O(l) + 2NO2(g)} \nonumber[/latex]

If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure 3.4.1). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of [latex]PV[/latex] work done by multiplying the external pressure [latex]P[/latex] by the change in volume caused by movement of the piston ([latex]ΔV[/latex]). At a constant external pressure (here, atmospheric pressure),

[latex]w = −PΔV \label{5.4.2}[/latex]

The negative sign associated with [latex]PV[/latex] work done indicates that the system loses energy when the volume increases. If the volume increases at constant pressure ([latex]ΔV > 0[/latex]), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases ([latex]ΔV < 0[/latex]), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.

The internal energy [latex]U[/latex] of a system is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy ([latex]H[/latex]) (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy [latex]U[/latex] plus the product of its pressure [latex]P[/latex] and volume [latex]V[/latex]:

[latex]H =U + PV \label{5.4.3}[/latex]

Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. So we can define a change in enthalpy ([latex]\Delta H[/latex]) accordingly

[latex]ΔH = H_{final} − H_{initial} \nonumber[/latex]

If a chemical change occurs at constant pressure (i.e., for a given [latex]P[/latex], [latex]ΔP = 0[/latex]), the change in enthalpy ([latex]ΔH[/latex]) is

[latex]\begin{align} ΔH &= Δ(U + PV) \[latex]4pt] &= ΔU + ΔPV \[latex]4pt] &= ΔU + PΔV \label{5.4.4} \end{align}[/latex]

Substituting [latex]q + w[/latex] for [latex]ΔU[/latex] (First Law of Thermodynamics) and [latex]−w[/latex] for [latex]PΔV[/latex] (Equation [latex]\ref{5.4.2}[/latex]) into Equation [latex]\ref{5.4.4}[/latex], we obtain

[latex]\begin{align} ΔH &= ΔU + PΔV \[latex]4pt] &= q_p + \cancel{w} −\cancel{w} \[latex]4pt] &= q_p \label{5.4.5} \end{align}[/latex]

The subscript [latex]p[/latex] is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation [latex]\ref{5.4.5}[/latex] we see that at constant pressure the change in enthalpy, [latex]ΔH[/latex] of the system, is equal to the heat gained or lost.

[latex]\begin{align} ΔH &= H_{final} − H_{initial} \[latex]4pt] &= q_p \label{5.4.6} \end{align}[/latex]

Just as with [latex]ΔU[/latex], because enthalpy is a state function, the magnitude of [latex]ΔH[/latex] depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.

To find [latex]ΔH[/latex] for a reaction, measure [latex]q_p[/latex].

When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction ([latex]ΔH_{rxn}[/latex]), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so [latex]ΔH_{rxn}[/latex] is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so [latex]ΔH_{rxn}[/latex] is positive. Thus:

• [latex]ΔH_{rxn} < 0[/latex] for an exothermic reaction, and
• [latex]ΔH_{rxn} > 0[/latex] for an endothermic reaction.

In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table:

If ΔH_rxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (Figure [latex]\PageIndex{2}a[/latex]). Conversely, if ΔH_rxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (Figure [latex]\PageIndex{2b}[/latex]). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.

Bond breaking ALWAYS requires an input of energy; bond making ALWAYS releases energy.y.

• Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):

  [latex]\begin{matrix}
  heat+ H_{2}O(s) \rightarrow H_{2}O(l) & \Delta H > 0
  \end{matrix} \label{5.4.7}[/latex]

  [latex]\begin{matrix}
  H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0
  \end{matrix} \label{5.4.8}[/latex]

  In both cases, the magnitude of the enthalpy change is the same; only the sign is different.

• Enthalpy is an extensive property (like mass). The magnitude of [latex]ΔH[/latex] for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation [latex]\ref{5.4.9}[/latex], which describes the reaction of aluminum with iron(III) oxide (Fe₂O₃) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al₂O₃, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe₂O₃ consumed:

  [latex]\ce{2Al(s) + Fe2O3(s) -> 2Fe (s) + Al2O3 (s) } + 815.5 \; kJ \label{5.4.9}[/latex]

  Thus ΔH = −851.5 kJ/mol of Fe₂O₃. We can also describe ΔH for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for ΔH, in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation [latex]\ref{5.4.10}[/latex], it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:

  [latex]\ce{ 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3 (s)} \quad \Delta H_{rxn}= - 851.5 \; kJ \label{5.4.10}[/latex]

  If 4 mol of Al and 2 mol of [latex]\ce{Fe2O3}[/latex] react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:

  [latex]- \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{5.4.6a}[/latex]

The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example 3.4.1.