5.3: Enthalpy

Learning Objectives

• Define the system and surroundings and explain heat flows in endothermic and exothermic processes.
• Define enthalpy as a state function related to internal energy, pressure, and volume.
• Interpret thermochemical equations and relate the magnitude and sign of ΔH to the reaction as written.
• Calculate the amount of heat transferred using a thermochemical equation and the quantity of reactant or product involved.

As discussed in the previous section, internal energy can be transferred as heat or work. Heat is the transfer of energy due to a temperature difference between the system and its surroundings. Work is the transfer of energy that does not require temperature differences. There are several types of work, but we will focus on pressure-volume work. Like heat, work describes energy in transit - it occurs only when energy is being transferred.

To further understand the relationship between heat (\(q\)) and the change in internal energy (\(ΔE\)), let's consider two sets of limiting conditions: reactions that occur at constant volume and reactions that occur at constant pressure. 

Assuming pressure-volume work is the only kind of work possible for the system, we can substitute \(w=-P\Delta V\) into the expression for change in internal energy, (\(\Delta E = q + w\)), giving:

\[ΔE = q − PΔV \label{5.4.1} \]

Constant Volume

In a closed, rigid container, the volume of the system does not change, so \(\Delta V = 0\). At constant volume, no pressure-volume work can be done, and \(w=-P\Delta V=0\). Under these conditions, the amount of heat transferred between the system to the surroundings (denoted \(q_{\textrm v}\) for constant volume) is equal to the change in internal energy:

\[\underset{\textrm{constant volume}}{q_{\textrm v}=\Delta E} \label{5.4.2} \]

Constant Pressure

Many chemical reactions are performed in open containers at approximately constant atmospheric pressure (about 1 atm). Heat under these conditions is given the symbol \(q_p\) to indicate constant pressure. Replacing \(q\) in Equation \(\ref{5.4.1}\) by \(q_p\) and rearranging the equation gives:

\[\underset{\textrm{constant pressure}}{q_{\textrm p}=\Delta E+P\Delta V} \label{5.4.3} \]

Thus, at constant pressure, heat is equal to the change in internal energy plus the pressure-volume work.

Because conditions of constant pressure are so common in chemistry, a new state function called enthalpy (\(H\)) is defined as the internal energy plus the product of pressure and volume.

\[H =E + PV \nonumber \]

At constant pressure, the change in the enthalpy of a system is:

\[\Delta H = \Delta E + \Delta (PV) = \Delta E + P \Delta V \label{5.3.5} \]

\(\Delta (PV)\) simplifies to \( P \Delta V\) because pressure is constant. This means that, at constant pressure, the change in enthalpy is equal to the heat transferred: \(\Delta H = q_p\). This matches our definition of enthalpy as the heat absorbed or released during a process at constant pressure. Importantly, under these conditions, \(\Delta H\) can be determined by measuring only heat; work does not need to be measured, which simplifies experimental procedures.

As with internal energy, it is the change in enthalpy that matters. Enthalpy, like internal energy, is a state function - its value depends only on the current state of the system, not the path taken to reach that state.

The Relationship between ΔH and ΔE

Enthalpy is often similar to internal energy, as many reactions involve little pressure-volume work. The following examples help illustrate when they are approximately equal and when they differ.

1. Reactions involving only solids, liquids, or solutions
   The volume does not change appreciably in these systems (\(\Delta V ≈ 0\)), and so we can simplify Equation \(\ref{5.3.5}\) to \(\Delta H ≈ \Delta E\). 
    
2. Reactions where the number of moles of gas does not change
   If the total number of gas molecules is the same on both sides of the equation, then the volume remains reasonably constant (\(\Delta V ≈ 0\)), so again \(\Delta H ≈ \Delta E\).
    
3. Reactions where the number of moles of gas changes
   In these cases \(\Delta V ≠ 0\), so \(\Delta H\) and \(\Delta E\) are not equal, however they still tend to be similar. To explore this idea, let's consider the conversion of oxygen gas to ozone at 298 K which involves the conversion of 3 moles of \(O_2(g)\) to 2 moles of \(O_3(g)\):
   \[3O_2(g) \rightarrow 2O_3(g) \nonumber \]
   For this reaction, \(\Delta H = 285.4 kJ\) and \(P\Delta V = -2.5 kJ\). This is an endothermic reaction (\(\Delta H > 0\)) with a decrease in the number of moles of gas and so the system will decrease in volume (\(\Delta V < 0\)). Using these values we can calculate the change in internal energy of the system:
   \[\Delta E = \Delta H - P\Delta V = 285.4 kJ - (- 2.4 kJ) = 287.9 kJ \nonumber \]
   We see that the change in internal energy (\(\Delta E = 287.9 kJ \)) is greater than the enthalpy change (\(\Delta H = 285.4 kJ \)), which is consistent with a decrease in volume as the surroundings did work on the system when its volume decreased. \(\Delta E \) and \(\Delta H \) are not identical, but they are very close, less than 1% different, and so even in cases where there is a change in volume \(\Delta H \) is often a good approximation of \(\Delta E \).

From these examples we can see that in the majority of reactions most of the energy change appears as heat, so \(\Delta H \) can be used as a good approximation of \(\Delta E \).

Example \(\PageIndex{1}\): The Relationship Between ΔH and ΔE

The combustion of graphite to produce carbon dioxide is described by the equation:

\[C_{(graphite, s)} + O_{2(g)} → CO_{2(g)} \nonumber \]

At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction. 

Would the change in internal energy (ΔE) be greater than, less than, or equal to the enthalpy change (ΔH)?

Strategy:

Use the balanced chemical equation to calculate the change in the number of moles of gas during the reaction.

Solution

This reaction produces 1 mol of gas (CO₂) and consumes 1 mol of gas (O₂), so there is no net change in the number of gas molecules:

Δn = 1 − 1 = 0.

Because ΔV is approximately zero, the pressure-volume work term is negligible, and ΔE ≈ ΔH.

Exothermic and Endothermic Processes

The sign of \(\Delta H \) indicates whether heat is absorbed or released in a process. If heat flows from a system to its surroundings, the enthalpy of the system decreases, so \(\Delta H \) is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so \(\Delta H \) is positive. Thus \(\Delta H < 0\) for an exothermic reaction, and \(\Delta H > 0\) 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.

Figure \(\PageIndex{1}\): The Enthalpy of Reaction. Energy changes in chemical reactions are usually measured as changes in enthalpy. (a) If heat flows from a system to its surroundings, the enthalpy of the system decreases, \(\Delta H \) is negative, and the reaction is exothermic. (b) Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, \(\Delta H \) is positive, and the reaction is endothermic.

Thermochemical Equations

Chemists use thermochemical equations to represent both the chemical change and the associated energy change of a reaction. In a thermochemical equation, the enthalpy change is written as a \(\Delta H \) value following the balanced chemical equation. This \(\Delta H \) value refers to the enthalpy change per mole of reactant and product in the reaction as written. How we manipulate \(\Delta H \) values associated with chemical equations relate to the fact that \(\Delta H \) is a state function and an extensive property. Extensive properties scale with the amount of a substance present. Mass, moles, and volume are all examples of extensive properties. The alternative is an intensive property, such as concentration or density, which do not scale with the amount of substance present.

Here are some key conventions to remember when dealing with \(\Delta H \) values associated with chemical equations:

1. The sign of \(\Delta H \) indicates the direction of heat flow
   A negative \(\Delta H \) indicates an exothermic reaction where heat is released by the system (the chemical reaction) and transferred to the surroundings (often most noticeably to the solvent and container in which the reaction is occurring). A positive \(\Delta H \) indicates an endothermic reaction where heat is absorbed by the system from the surroundings.
    
2. If the direction of a chemical equation is reversed, the sign of \(\Delta H \) also reverses
   This relates to \(\Delta H \) being a state function, the amount of heat released in one direction is exactly the same as the amount of heat absorbed in the reverse direction.
    
3. If all coefficients in the balanced equation are multiplied or divided, the \(\Delta H \) must be scaled by the same factor
   This relates to \(\Delta H \) being an extensive property. When the stoichiometry of the reaction is scaled by a factor, the enthalpy change must be scaled by the same factor. Let's use the reaction of hydrogen and oxygen to produce water as an example:
   \[\ce{H2(g) + 1/2 O2(g) -> H2O(l)} \quad \Delta H=-286 kJ/mol \nonumber \]
   This equation means that if 1 mol of \(\ce{H2(g)}\) was to react with 0.5 mol of \(\ce{O2(g)}\) to form 1 mol of \(\ce{H2O(l)}\), 286 kJ of heat is released.  If we double the coefficients, we have doubled the stoichiometry of the reaction and twice as much of heat would be released per reaction as written:
   \[\ce{2H2(g) + O2(g) -> 2H2O(l)} \quad \Delta H=2 \times(-286 kJ/mol )=-572 kJ/mol \nonumber \]
   If we halve the stoichiometry we need to halve \(\Delta H \):
   \[\ce{ 1/2 H2(g) + 1/4 O2(g) -> 1/2 H2O(l)} \quad \Delta H=\frac{1}{2} \times(-286 kJ/mol )=-143 kJ/mol \nonumber \]
   The enthalpy change must always correspond the stoichiometric coefficients in the chemical equation as written.
    
4. Units of \(\Delta H \), J/mol or kJ/mol, refer to the moles of reactants and products in the balanced chemical reaction
   Using the example above we can see that the amount of energy generated per mole of hydrogen consumed is consistent as long as the per mole component of the \(\Delta H \) unit takes the balanced stoichiometry of the reaction into account:
   \[\dfrac{−286 kJ}{1 \: mol \: \ce{H2(g)}} = \dfrac{−572 kJ}{2 \: mol \: \ce{H2(g)}} = \dfrac{−143 kJ}{\frac{1}{2} \: mol \: \ce{H2(g)}} \nonumber \]
5. Stoichiometric coefficients in the balanced equation are unitless
   These coefficients refer to the ratio of molecules of reactants and products in the reaction, not the actual amounts present. They do not carry units and should not be assigned “mol.”  
    
6. The \(\Delta H \) of a reaction depends on the physical states of the substances involved.
   The \(\Delta H \) for the production of water is different if the reaction produces liquid water (\(\ce{H2O(l)}\)) or gas phase water (\(\ce{H2O(g)}\)):
   \[\ce{H2(g) + 1/2 O2(g) -> H2O(l)} \quad \Delta H=-286 kJ/mol \nonumber \]
   \[\ce{H2(g) + 1/2 O2(g) -> H2O (g)} \quad \Delta H=-242 kJ/mol \nonumber \]
   Condensing water from gas to liquid releases additional heat energy because of the formation of stronger intermolecular forces, so the production of liquid water has a more negative \(\Delta H \). Because the physical state affects the enthalpy change, the physical states of all reactants and products must be specified in a thermochemical equation.

Example \(\PageIndex{2}\): Writing Thermochemical Equations

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 is released. What is ΔH for the following reaction?

\[\ce{HCl(aq) + NaOH (aq) -> NaCl(aq) + H2O (l)} \nonumber \]

Solution

For the reaction of 0.0500 mol acid (HCl), q = −2.9 kJ. The reactants are present in stoichiometric amounts (matching the 1:1 molar ratio in the balanced equation), and so the amount of acid (or base) may be used to calculate the molar enthalpy change. Since \(ΔH\) is an extensive property, it is proportional to the amount of substance reacted:

\[\Delta H= \frac{-2.9 kJ }{0.0500\,\ce{mol\, HCl} }=-58 kJ/mol \nonumber \]

This gives the heat released per mole of HCl neutralized, consistent with the stoichiometry of the reaction.

The corresponding thermochemical equation is:

\[\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O (l)} \quad \Delta H=-58 kJ/mol \nonumber \]

The heat released or absorbed in a chemical reaction depends on how much of each substance reacts. When calculating ΔH, it's important to identify the limiting reactant, since it determines how much product forms—and how much energy is involved. 

Exercise \(\PageIndex{1}\)

When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat is produced. What is ΔH for the following reaction?

\[\ce{ Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g)} \nonumber \]

Answer

Step 1: Calculate the moles of each reactant:

\[\ce{ Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g)} \nonumber \]

\[1.34g\;Zn\left(\frac{mol}{\;65.38g}\right)=0.0205\;mol\;Zn \\ 0.750\;\frac{mol}{L}HCl(0.0600L)=0.0450\;mol\; HCl \nonumber\]

According to the balanced chemical equation, 2 mol of HCl are required for every 1 mol of Zn. The amount of HCl present is more than twice the amount of Zn, so Zn is the limiting reactant.

Step 2: Use the limiting reactant to determine ΔH:

\[\Delta H =\frac{-3.14kJ}{0.0205mol}=-153kJ/mol\nonumber\]

This is the enthalpy change for the reaction as written, per mole of Zn reacted.

Example \(\PageIndex{3}\): Writing Thermochemical Equations

A gummy bear contains 2.67 g sucrose, C₁₂H₂₂O₁₁. When it reacts with 7.19 g potassium chlorate, KClO₃, 43.7 kJ of heat is produced. What is ΔH for the following reaction?

\[\ce{ C12H22O11(aq) + 8KClO3(aq) -> 12 CO2(g) + 11 H2O (l) + 8 KCl(aq)} \nonumber \]

Solution

The limiting reactant must be identified: it determines how much heat is released.

Step 1: Calculate moles of each reactant:

\[\begin{aligned}
(2.67\, \text{g}) \left(\dfrac{1\, \text{mol}}{342.3\, \text{g} } \right) &= 0.00780\, \text{mol} \, \ce{C12H22O11} \\[4pt]
(7.19\, \text{g}) \left(\dfrac{1\, \text{mol}}{122.5\, \text{g} } \right) &= 0.0587\, \text{mol} \, \ce{KClO3}
\end{aligned} \nonumber \]

Step 2: Use stoichiometry to find the limiting reactant

From the balanced equation, 1 mol of sucrose reacts with 8 mol of KClO₃:

\[0.00780\, \text{mol} \, \ce{C12H22O11}\left(\dfrac{8\, \text{mol } \ce{KClO3}}{1\, \text{mol}\, \ce{C12H22O11}}\right) = 0.0624\,\text{mol } \ce{KClO3}\;\text{(amount needed to react with all of sucrose)} \nonumber \]

Since only 0.0587 mol of KClO₃ are available, KClO₃ is the limiting reactant and must be used to calculate the enthalpy change:

\[\Delta H = \dfrac{-43.7\, \text{kJ}} {0.0587\, \text{mol} \ce{KClO3}} = -744\, \text{kJ/mol}\nonumber \]

This is the enthalpy change per mole of KClO₃ consumed as written in the balanced equation. Since the equation involves 8 mol KClO₃, the total ΔH is:

\[-744\, \text{kJ/mol} \times 8 = -5960\, \text{kJ/mol}\nonumber\]

The thermochemical equation is:

\[\ce{C12H22O11 + 8 KClO3 -> 12 CO2 + 11 H2O + 8 KCl} \quad \Delta H=-5960\, \text{kJ/mol} \nonumber \]

Exercise \(\PageIndex{2}\)

When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of \(\ce{FeCl2(s)}\) and 8.60 kJ of heat is produced. What is ΔH for the following reaction?

\[\ce{Fe(s) + Cl2 (g) \rightarrow FeCl2(s)}\nonumber\]

Answer

Since the amount of product is given, we can find the enthalpy change without considering the limiting reactant.

Step 1: Convert mass of product to moles:

\[3.22\;g\;\ce{FeCl2(s)}\left(\frac{mol}{126.75\;g}\right)=0.0254\,mol \nonumber\]

Step 2: Calculate ΔH per mole of product formed:

\[\Delta H = \frac{-8.60\,\text{kJ}}{0.0254\,mol}=-339\,\text{kJ/mol} \nonumber\]

Summary

• At constant volume, no pressure-volume work is done, so the heat transferred equals the change in internal energy: q_v ​= ΔE
• At constant pressure, the heat transferred equals the change in enthalpy: q_p = ΔH
  • Enthalpy (H) is a state function defined as: H = E + PV
  • Therefore, ΔH = ΔE + PΔV at constant pressure
• In many reactions—especially those involving only solids, liquids, or solutions—the volume change is negligible, so: ΔH ≈ ΔE
  • Even when gas is involved, if pressure-volume work is small, this approximation often remains valid.
• The sign of ΔH indicates the direction of heat flow:
  • ΔH < 0: Exothermic (heat is released to the surroundings)
  • ΔH > 0: Endothermic (heat is absorbed from the surroundings)
• A thermochemical equation includes both the balanced chemical reaction and the enthalpy change (ΔH). Key points to remember:
  • ΔH is associated with the reaction as written.
  • If the reaction is reversed, the sign of ΔH is reversed.
  • If the reaction is multiplied or divided by a factor, ΔH is scaled by the same factor.
  • The physical states of reactants and products must be specified, as they affect the enthalpy change.
  • The magnitude of ΔH refers to the amount of heat transferred following one mole of the reaction as written.
• The heat transferred in a reaction depends on the amount of limiting reactant, since it determines how much product forms and, therefore, how much energy is released or absorbed.