Enthalpy: Heat Transfer at Constant Pressure

Most chemical reactions in the laboratory occur under constant-pressure conditions (isobaric processes) where the system experiences the constant external pressure of the atmosphere. When we perform reactions in an open beaker or flask, the reaction is allowed to change its volume by pushing against or being compressed by this atmospheric pressure.

While the First Law gives us the complete energy accounting through ΔU = q + w, in many laboratory situations it is more convenient to work with a different thermodynamic quantity called enthalpy (H).

Definition and Derivation

Enthalpy is defined as the sum of internal energy and the product of pressure and volume:

\[ H = U + PV \]

The enthalpy change, ΔH, during a process is:

\[ \begin{align*} \Delta H &= H_{\mathrm{final}} - H_{\mathrm{initial}} \\[1.5ex] &= (U_{\mathrm{final}} + P_{\mathrm{final}}V_{\mathrm{final}}) - (U_{\mathrm{initial}} + P_{\mathrm{initial}}V_{\mathrm{initial}}) \\[1.5ex] &= \Delta U + \Delta(PV) \end{align*} \]

NoteTerminology: “Enthalpy” vs. “Enthalpy Change”

Strictly speaking, H is enthalpy (a state function) and ΔH is the change in enthalpy. However, in casual usage, chemists often say “enthalpy” when they mean “enthalpy change.”

For example, you might hear “the enthalpy of combustion of methane is −890 kJ/mol” when the precise statement would be “the enthalpy change of combustion.” The context (a process with a sign and magnitude) makes clear that a change is meant.

Be aware of this shorthand. When someone refers to “the enthalpy of a reaction,” they mean ΔrH, not the absolute enthalpy H of any substance.

Using the product rule for the pressure-volume term and applying the First Law:

\[ \begin{align*} \Delta H &= \Delta U + P\Delta V + V\Delta P \\[1.5ex] &= (q + w) + P\Delta V + V\Delta P \end{align*} \]

The product rule from differential calculus states that for two variables x and y:

\[ d(xy) = x\,dy + y\,dx \]

Applied to thermodynamics:

  • If x = P and y = V
  • Then d(PV) = PdV + VdP
  • For finite changes: Δ(PV) = PΔV + VΔP

This rule allows us to separate the effects of pressure and volume changes in thermodynamic calculations.

For pressure-volume work, w = −PextΔV. Under conditions of constant pressure, Pext = P, ΔP = 0, and q → qp:

\[ \begin{align*} \Delta H &= (q_{\mathrm{p}} - P_{\mathrm{ext}}\Delta V) + P\Delta V + V(0) \\[1.5ex] &= q_{\mathrm{p}} - P\Delta V + P\Delta V \\[1.5ex] &= q_{\mathrm{p}} \end{align*} \]

At constant pressure:

\[ \Delta H = q_{\mathrm{p}} \]

where qp is the heat transferred under constant-pressure conditions. Like q, ΔH is an energy quantity with units of joules (J) or kilojoules (kJ). It represents the total enthalpy change for a process.

Enthalpy is a state function (path-independent), while heat qp remains a path function. The equality holds only at constant pressure. Compare this to constant volume, where ΔU = qV (derived on the First Law page).

Since most laboratory reactions occur in open containers at atmospheric pressure, ΔH is experimentally convenient: we can determine a state function change by measuring heat transfer directly.

NoteNotation: q vs qp

Strictly speaking, enthalpy equals heat transfer at constant pressure: ΔH = qp. However, since enthalpy calculations inherently assume constant-pressure conditions, the subscript is typically dropped. Throughout this text (and most others), q in enthalpy contexts means qp.

Enthalpy of Reaction

The enthalpy of reaction (ΔrH) is the enthalpy change when reactants are converted to products according to a balanced chemical equation. Unlike ΔH (total energy in kJ), ΔrH is a molar quantity with units of kJ mol−1, representing the enthalpy change per mole of reaction as written.

Consider the combustion of hydrogen:

\[ 2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{H_2O(l)} \qquad \Delta_{\mathrm{r}}H = -572~\mathrm{kJ~mol^{-1}} \]

The value −572 kJ mol−1 means 572 kJ of heat is released per mole of reaction as written. One mole of reaction consumes 2 mol H2 and 1 mol O2 while producing 2 mol H2O. The stoichiometric coefficients define what “one mole of reaction” means:

  • If 1 mol O2 reacts (one mole of reaction): 572 kJ released
  • If 1 mol H2 reacts (half a mole of reaction): 286 kJ released
  • If 2 mol O2 reacts (two moles of reaction): 1144 kJ released

The subscript in ΔrH indicates the type of process. Other common subscripts include:

This same principle applies to all thermodynamic state functions. The process subscript (r, f, c, etc.) indicates a molar quantity: ΔrU for internal energy, ΔrS for entropy, ΔrG for Gibbs energy. You’ll encounter entropy and Gibbs energy in General Chemistry II.

Strictly speaking, the molar enthalpy of reaction is defined as the partial derivative of enthalpy with respect to the extent of reaction (ξ, the Greek letter xi):

\[ \Delta_{\mathrm{r}} H = \left(\frac{\partial H}{\partial \xi}\right)_{T,P} \]

The extent of reaction ξ (in moles) measures how far a reaction has progressed. When ξ increases by 1 mol, the reaction proceeds by one “mole of reaction as written” - consuming reactants and forming products in their stoichiometric ratios.

This formal definition explains why ΔrH has units of kJ mol−1: it’s the rate of change of enthalpy (kJ) per mole of reaction progress. For practical purposes in general chemistry, thinking of ΔrH as “enthalpy change per mole of reaction as written” is equivalent and more intuitive.

The sign convention follows the same system-centric approach used throughout thermodynamics:

  • ΔrH < 0: Exothermic reaction - system releases heat to surroundings
  • ΔrH > 0: Endothermic reaction - system absorbs heat from surroundings

The enthalpy of reaction is calculated as:

\[ \Delta_{\mathrm{r}} H = \sum H_{\mathrm{products}} - \sum H_{\mathrm{reactants}} \]

NoteIUPAC Notation: Molar, Specific, and Total Quantities

IUPAC notation distinguishes thermodynamic quantities using case and subscripts:

Molar reaction quantities use uppercase letters with a process subscript (r, f, c, etc.):

  • ΔrH : Molar enthalpy of reaction (kJ mol−1)
  • ΔfH : Molar enthalpy of formation (kJ mol−1)
  • ΔcU : Molar internal energy of combustion (kJ mol−1)

The subscript indicates the process and implies “per mole of reaction as written.” These are intensive quantities.

Specific quantities (per unit mass) use lowercase letters:

  • Δcu : Specific internal energy of combustion (kJ g−1)
  • Δch : Specific enthalpy of combustion (kJ g−1)
  • cp : Specific heat capacity (J g−1 K−1)

Total (extensive) quantities use uppercase without a process subscript:

  • ΔH : Total enthalpy change for an experiment (kJ)
  • ΔU : Total internal energy change (kJ)
  • q : Heat transferred (kJ)

The relationship between specific and molar quantities:

\[ \text{Molar quantity} = \text{Specific quantity} \times \text{Molar mass} \]

Specific values are convenient when measuring mass directly in the lab. Molar values are standard for tabulating and comparing thermodynamic data.

Common usage: Many sources write ΔH where ΔrH would be more precise. When you see “ΔH = −572 kJ mol−1”, the units tell you it’s a molar quantity even without the subscript. Always check the units to know which quantity is meant.

Scaling Enthalpy: Heat for Any Amount

The molar enthalpy of reaction tells you the heat released or absorbed per mole of reaction as written. To find the total heat transfer for a specific amount of reactant or product, multiply the molar enthalpy by the number of moles involved, accounting for stoichiometry:

\[ q = n \Delta_{\mathrm{r}} H \]

where n is the moles of the substance and ΔrH is scaled by the appropriate stoichiometric ratio.

For example, the combustion of methane has ΔcH = −890 kJ mol−1:

\[ \mathrm{CH_4(g) + 2~O_2(g) \longrightarrow CO_2(g) + 2~H_2O(l)} \]

If 3 moles of CH4 combust, the total heat released is:

\[ q = (3~\mathrm{mol})\left(\frac{-890~\mathrm{kJ}}{1~\mathrm{mol~CH_4}}\right) = -2670~\mathrm{kJ} \]

If instead 3 moles of O2 react, the stoichiometric coefficient of 2 must be accounted for:

\[ q = (3~\mathrm{mol})\left(\frac{-890~\mathrm{kJ}}{2~\mathrm{mol~O_2}}\right) = -1335~\mathrm{kJ} \]

Tabulated molar values can then be applied to any amount of substance.

Practice

How much heat (in kJ) is required to melt 250. g of ice at exactly water’s melting point of 0 °C? The heat of fusion for water is ΔfusH = 6.01 kJ mol−1 at 273.15 K.

Solution

The phase transition is:

\[\mathrm{H_2O(s) \longrightarrow H_2O(l)}\]

Apply q = nΔH, converting mass to moles:

\[ \begin{align*} q &= n(\mathrm{H_2O})~\Delta_{\mathrm{fus}} H \\[1.5ex] &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} ~ \Delta_{\mathrm{fus}} H \\[1.5ex] &= 25\bar{0}.~\mathrm{g} \left ( \dfrac{1~\mathrm{mol}}{18.0\bar{2}~\mathrm{g}} \right ) \left ( \dfrac{6.0\bar{1}~\mathrm{kJ}}{1~\mathrm{mol}} \right ) \\[1.5ex] &= 83.\bar{3}7~\mathrm{kJ} \\[1.5ex] &= 83.4~\mathrm{kJ} \end{align*} \]

Practice

For the combustion of methane:

\[\mathrm{CH_4(g) + 2~O_2(g) \longrightarrow CO_2(g) + 2~H_2O(g)}\]

The enthalpy of combustion is ΔcH = −802.5 kJ mol−1. Calculate the heat released when:

  1. 2.00 moles of CH4(g) is combusted
  2. 5.00 moles of O2(g) is combusted
  3. 0.25 moles of H2O(g) is produced
Solution

a) 2.00 moles of CH4(g) combusted

The stoichiometric coefficient for CH4 is 1, so:

\[ \begin{align*} q &= n(\mathrm{CH_4})~\Delta_{\mathrm{c}} H \\[1.5ex] &= 2.0\bar{0}~\mathrm{mol~CH_4} \left ( \dfrac{-802.\bar{5}~\mathrm{kJ}}{1~\mathrm{mol~CH_4}} \right ) \\[1.5ex] &= -160\bar{5}~\mathrm{kJ} \\[1.5ex] &= -1.60 \times 10^{3}~\mathrm{kJ} \end{align*} \]

b) 5.00 moles of O2(g) combusted

The stoichiometric coefficient for O2 is 2, so the molar enthalpy must be scaled:

\[ \begin{align*} q &= n(\mathrm{O_2}) \left( \frac{\Delta_{\mathrm{c}} H}{2} \right) \\[1.5ex] &= 5.0\bar{0}~\mathrm{mol~O_2} \left ( \dfrac{-802.\bar{5}~\mathrm{kJ}}{2~\mathrm{mol~O_2}} \right ) \\[1.5ex] &= -200\bar{6}.25~\mathrm{kJ}\\[1.5ex] &= -2006~\mathrm{kJ} \end{align*} \]

c) 0.25 moles of H2O(g) produced

The stoichiometric coefficient for H2O is 2:

\[ \begin{align*} q &= n(\mathrm{H_2O}) \left( \frac{\Delta_{\mathrm{c}} H}{2} \right) \\[1.5ex] &= 0.2\bar{5}~\mathrm{mol~H_2O} \left ( \dfrac{-802.\bar{5}~\mathrm{kJ}}{2~\mathrm{mol~H_2O}} \right ) \\[1.5ex] &= -10\bar{0}.3~\mathrm{kJ} \\[1.5ex] &= -1.0 \times 10^{2}~\mathrm{kJ} \end{align*} \]

ImportantEnthalpy vs. Internal Energy in Reactions

While both ΔU and ΔH describe energy changes in reactions, they differ by the work done during the reaction:

\[ \begin{align*} \Delta H &= \Delta U + \Delta(PV) \\[1.5ex] &= \Delta U + \Delta n_{\mathrm{gas}}RT \end{align*} \]

For reactions involving gases, ΔH and ΔU differ by the PV work done. For reactions with only liquids and solids, where volume changes are minimal, ΔH ≈ ΔU.

Molar Quantities and Δνgas

In the Work chapter, we introduced Δngas as the change in moles of gas for a specific reaction, with units of mol. When working with molar reaction quantities (ΔrH, ΔrU), we need a dimensionless quantity instead.

Molar reaction quantities describe the energy change per mole of reaction as written. The appropriate quantity here is Δνgas, the change in stoichiometric coefficients of gaseous species. The symbol ν (Greek letter nu) represents stoichiometric coefficients, which are dimensionless by definition.

\[ \Delta \nu_{\mathrm{gas}} = \sum \nu_{\mathrm{products,~gas}} - \sum \nu_{\mathrm{reactants,~gas}} \]

This dimensionless value ensures the units work out correctly:

\[ \Delta_{\mathrm{r}} H = \Delta_{\mathrm{r}} U + \Delta \nu_{\mathrm{gas}}RT \]

where ΔrH and ΔrU are in kJ mol−1, R = 8.314 J mol−1 K−1, and T is in K.

WarningCommon Notation

Many textbooks use Δngas for both the change in moles (units of mol) and the change in stoichiometric coefficients (dimensionless). Context and units usually clarify the meaning, but this notation is imprecise. When you see Δngas in an equation with molar quantities like ΔrH, it is functioning as Δνgas.

Example: The combustion of hydrogen gas produces liquid water:

\[ \mathrm{2~H_2(g) + O_2(g) \longrightarrow 2~H_2O(l)} \]

At 298.15 K, ΔrH = −571.6 kJ mol−1. Calculate ΔrU for this reaction.

Solution:

Rearranging the relationship between molar enthalpy and internal energy:

\[ \Delta_{\mathrm{r}} U = \Delta_{\mathrm{r}} H - \Delta \nu_{\mathrm{gas}}RT \]

Step 1: Calculate Δνgas.

The product, H2O(l), is a liquid, contributing 0 to the gas coefficient sum. The reactants include 2 H2(g) and 1 O2(g).

\[ \begin{align*} \Delta \nu_{\mathrm{gas}} &= \Sigma\nu_{\mathrm{gas,~products}} - \Sigma\nu_{\mathrm{gas,~reactants}} \\[1.5ex] &= 0 - (2 + 1) \\[1.5ex] &= -3 \end{align*} \]

Step 2: Substitute values and calculate.

Using R = 8.314 J mol−1 K−1 and T = 298.15 K:

\[ \begin{align*} \Delta_{\mathrm{r}} U &= \Delta_{\mathrm{r}} H - \Delta \nu_{\mathrm{gas}}RT \\[1.5ex] &= -571.\bar{6}~\mathrm{kJ~mol^{-1}} - (-3) \left (\dfrac{8.31\bar{4}~\mathrm{J}}{\mathrm{mol~K}} \right )(298.1\bar{5}~\mathrm{K})\left( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= -571.\bar{6}~\mathrm{kJ~mol^{-1}} + 7.43\bar{6}45~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -564.\bar{1}63~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -564.2~\mathrm{kJ~mol^{-1}} \end{align*} \]

The difference between ΔrH and ΔrU is about 7.4 kJ mol−1, roughly 1.3 % of the total energy change. For reactions where Δνgas = 0 (equal stoichiometric coefficients of gas on both sides), ΔrH = ΔrU exactly.

The same problem can be solved using total quantities (ΔH, ΔU) instead of molar quantities. With total quantities, we use Δngas (in mol) rather than Δνgas (dimensionless). Suppose 2 moles of H2 react with 1 mole of O2 (one “mole of reaction”).

Given: ΔH = −571.6 kJ (total heat released for this specific reaction)

Find: ΔU

Here, Δngas is the actual change in moles of gas:

\[ \Delta n_{\mathrm{gas}} = 0~\mathrm{mol} - 3~\mathrm{mol} = -3~\mathrm{mol} \]

Using the total quantity relationship:

\[ \begin{align*} \Delta U &= \Delta H - \Delta n_{\mathrm{gas}}RT \\[1.5ex] &= -571.\bar{6}~\mathrm{kJ} - (-3~\mathrm{mol})\left (\dfrac{8.31\bar{4}~\mathrm{J}}{\mathrm{mol~K}} \right )(298.1\bar{5}~\mathrm{K})\left( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= -571.\bar{6}~\mathrm{kJ} + 7.43\bar{6}45~\mathrm{kJ} \\[1.5ex] &= -564.\bar{1}63~\mathrm{kJ} \\[1.5ex] &= -564.2~\mathrm{kJ} \end{align*} \]

Notice the dimensional analysis: (mol)(J mol−1 K−1)(K) = J. The mol units cancel, giving energy in joules.

Both approaches give the same numerical answer, but the molar approach (ΔrH, Δνgas) yields a value in kJ mol−1 that can be applied to any amount of reactants, while the total approach (ΔH, Δngas) gives the energy for that specific reaction.

Standard Enthalpy

Standard State

To help simplify thermodynamic evaluations, a reference point was adopted called the standard state, denoted with a Plimsoll (⦵) or degree symbol (°).

Standard state is defined to be for:

  • Gases: A hypothetical state a gas would have as a pure substance obeying the ideal gas equation at standard state pressure
  • Liquids: The state of a pure substance as a liquid under standard state pressure
  • Solids: The state of a pure, crystalline substance under standard state pressure
  • Solutes: A hypothetical state a solute would have in an ideal solution with a standard amount concentration, c°, of exactly 1 mol L−1

Standard state pressure, p°, is defined to be a pressure of 1 bar or 105 Pa.

Note that temperature is not included in the definition of standard state; however, most thermodynamic quantities are compiled at specific temperatures, usually at 25 °C (298.15 K).

For a comprehensive treatment of standard states, temperature corrections, and when approximations break down, see Standard States: Conventions, Corrections, and Real Chemistry.

NoteUnderstanding Standard States as Reference Conditions

Standard states serve as reference conditions for thermodynamic consistency, not necessarily the stable state of a substance at those conditions.

For example, water vapor at 25 °C and 1 bar is defined as a hypothetical ideal gas state, even though real water at these conditions would be liquid (the equilibrium vapor pressure is only 0.0313 bar). Similarly, diamond has a tabulated standard enthalpy of formation even though graphite is the more stable allotrope at standard conditions.

This convention works because standard values are used to calculate relative reaction properties (ΔrH°, ΔrG°), which are then converted to real conditions using equilibrium expressions and activities. Consistency is what matters: if everyone uses the same reference states, thermodynamic calculations produce correct predictions regardless of whether the reference states physically exist.

For a detailed exploration of how thermodynamic data are determined for “impossible” states and why this system works, see The Mystery of Hypothetical Standard States.

Standard Enthalpy of Formation

A standard enthalpy of formation (ΔfH° in kJ mol−1), also called the standard heat of formation, of a compound is the change in enthalpy during the formation of exactly 1 mole of a substance from its constituent elements in their reference state, with all substances in the standard state.

NoteWhy Do Elements Have ΔfH° = 0?

The standard enthalpy of formation for elements in their most stable form at standard conditions is defined to be zero. This is not a measured value. It’s a reference point convention.

Why this convention exists:

By definition, elements in their most stable form at standard conditions (298.15 K, 1 bar) are assigned ΔfH° = 0. This serves as the baseline against which all other enthalpies of formation are measured. It’s analogous to setting sea level as the zero point for measuring altitude (arbitrary but necessary for consistency).

Implications:

  • We can never know “absolute” enthalpies, only enthalpy differences
  • All tabulated ΔfH° values are relative to the elements
  • The choice of zero doesn’t affect calculated ΔrH values (the zeros cancel out in reaction calculations)

Important: Only the most stable form gets zero. For example:

  • Carbon: Graphite has ΔfH° = 0, but diamond has ΔfH° = +1.89 kJ mol−1 because graphite is slightly lower in energy
  • Oxygen: O2(g) has ΔfH° = 0, but O3(g) (ozone) has ΔfH° = 142.7 kJ mol−1 because it’s formed from O2(g)

This reference state convention is what makes thermochemical calculations possible using tabulated data.

Quick Reference

A table of Standard Thermodynamic Data provides standard values for ΔfH°, ΔfG°, S°, and Cp for hundreds of substances.

Consider the formation of carbon dioxide from its constituent elements in their reference states. This reaction is exothermic with a standard heat of reaction −393.5 kJ mol−1, and can be written as:

\[ \mathrm{C(s,~graphite) + O_2(g) \longrightarrow CO_2(g)} \qquad \Delta_{\mathrm{r}}H^{\circ} = -393.5\,\mathrm{kJ\,mol^{-1}} \]

Notice the reactants. Elemental carbon is given in its standard state (a crystalline solid in the form of graphite, an allotrope of solid carbon; solid carbon also exists in the form as diamond, another allotrope, but graphite is slightly lower in energy). Elemental oxygen exists naturally as oxygen gas at its standard state. Carbon dioxide exists as a gas at its standard state.

The standard heat of formation for carbon dioxide, ΔfH°(CO2, g), can be determined from the heats of formation of the reactants (where ν is the stoichiometric coefficient from the balanced chemical equation) using values tabulated in thermodynamic reference tables compiled at specific temperatures, usually at 25 °C (298.15 K).

TipA Deeper Look: Temperature and Pressure Dependence of Standard Values

Standard thermodynamic quantities are typically tabulated at 25 °C (298.15 K) and 1 bar pressure. While these values depend on both temperature and pressure, the practical importance differs dramatically for typical laboratory chemistry.

Pressure effects are usually negligible because:

  1. Most laboratory work occurs near 1 bar (0.95 to 1.05 bar depending on weather/altitude)
  2. Liquids and solids are nearly incompressible
  3. For gases, doubling pressure (1 bar → 2 bar) changes ΔG by only ~1.7 kJ mol−1, often within experimental uncertainty

Temperature effects are much more significant: A modest temperature change of 75 K (e.g., 25 °C to 100 °C) can change ΔH by 5-10%. Students routinely encounter reactions at non-standard temperatures (ice baths, body temperature, hot plates, combustion), making temperature corrections far more important than pressure corrections in general chemistry.

For detailed discussion of when corrections are needed, see Practical Decision Rules: When to Apply Temperature Corrections.

\[ \begin{align*} \Delta_{\mathrm{r}}H^{\circ} &= \bigg[1~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{CO_2,g})\bigg] - \bigg[1~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{C,~graphite}) + 1~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{O_2,g})\bigg] \\[1.5ex] -393.\bar{5}~\mathrm{kJ\,mol^{-1}} &= \bigg[1~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{CO_2,g})\bigg] - \bigg[1~(0) + 1~(0)\bigg] \\[1.5ex] -393.\bar{5}~\mathrm{kJ\,mol^{-1}} &= \Delta_{\mathrm{f}} H^{\circ}(\mathrm{CO_2, g}) \end{align*} \]

Therefore, to create CO2 from pure elements in the standard state is an exothermic process.

Practice

Write out a balanced chemical reaction that is associated with the standard enthalpy of formation of H2O(l).

Solution

\[\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2}(g) \longrightarrow \mathrm{H_2O(l)}\]

Note that we are asked to find the standard enthalpy of formation which is defined as being the change in heat for the formation of exactly 1 mole of a substance. Here, exactly 1 mole of product is represented which requires a fractional coefficient of 1/2 to appear in front of O2.

Practice

Determine the standard molar enthalpy of reaction (in kJ mol−1) for a reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas. Is the reaction endothermic or exothermic? What is the total enthalpy change (in kJ) if 10 moles of HCl were produced?

Thermodynamic data: ΔfH°(HCl, g) = −92.30 kJ mol−1

Solution

Write a balanced chemical equation:

\[\mathrm{H_2(g)} + \mathrm{Cl_2(g)} \longrightarrow 2~\mathrm{HCl(g)}\]

Calculate the standard molar enthalpy of reaction:

\[ \begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= \sum \nu_i \Delta_{\mathrm{f}} H^{\circ}(\text{products}) - \sum \nu_i \Delta_{\mathrm{f}} H^{\circ}(\text{reactants}) \\[1.5ex] &= \bigg [ 2 \times \Delta_{\mathrm{f}}H^{\circ}(\mathrm{HCl,g)} \bigg ] - \bigg [ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{H_2, g)} + \Delta_{\mathrm{f}}H^{\circ}(\mathrm{Cl_2, g)} \bigg ] \\[1.5ex] &= \bigg [ 2 \times (-92.3\bar{0}) \bigg ]~\mathrm{kJ~mol^{-1}} - \bigg [ 0 + 0 \bigg ]~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= \bigg [ -184.\bar{6}00 \bigg ]~\mathrm{kJ~mol^{-1}} - \bigg [ 0 \bigg ]~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -184.\bar{6}00~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -184.6~\mathrm{kJ~mol^{-1}} \end{align*} \]

Since ΔrH° is negative, the reaction is exothermic.

Calculate the total enthalpy change if exactly 10 moles of hydrogen chloride gas was produced:

\[ \begin{align*} \Delta H &= n(\mathrm{HCl}) \left( \frac{\Delta_{\mathrm{r}} H^{\circ}}{\nu_{\mathrm{HCl}}} \right) \\[1.5ex] &= 10~\mathrm{mol~HCl} \left ( \dfrac{-184.\bar{6}~\mathrm{kJ}}{2~\mathrm{mol~HCl}} \right ) \\[1.5ex] &= -923.\bar{0}00~\mathrm{kJ} \\[1.5ex] &= -923.0~\mathrm{kJ} \end{align*} \]

Standard Enthalpy of Reaction

The standard enthalpy of reaction (ΔrH°, in kJ mol−1) can be determined as the difference in the total standard molar product and total standard molar reactant enthalpies of formation.

WarningSpecies in Standard State

Standard enthalpy of reaction represents reactants and products in the standard state where the reactants are unmixed (i.e. in pure, isolated form) and the products are unmixed.

Therefore, the standard enthalpy of a reaction compares the state of separated/isolated reactants to the state of separated/isolated products.

For a general reaction:

\[ \nu_{\mathrm{A}}~\mathrm{A} + \nu_{\mathrm{B}}~\mathrm{B} \longrightarrow \nu_{\mathrm{C}}~\mathrm{C} + \nu_{\mathrm{D}}~\mathrm{D} \]

The standard enthalpy of reaction would be:

\[\Delta_{\mathrm{r}} H^{\circ} = \bigg[ \nu_{\mathrm{C}}~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{C}) + \nu_{\mathrm{D}}~ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{D}) \bigg] - \bigg [ \nu_{\mathrm{A}}~\Delta_{\mathrm{f}}H^{\circ}(\mathrm{A}) + \nu_{\mathrm{B}}~ \Delta_{\mathrm{f}}H^{\circ}(\mathrm{B}) \bigg ] \]

With tabulated formation enthalpies, you can calculate the enthalpy change for any reaction.

Example: Calculate the standard enthalpy change for the combustion of propane:

\[ \mathrm{C_3H_8(g) + 5~O_2(g) \longrightarrow 3~CO_2(g) + 4~H_2O(l)} \]

Use the following standard enthalpies of formation:

  • ΔfH°[C3H8(g)] = −103.8 kJ mol−1
  • ΔfH°[O2(g)] = 0.0 kJ mol−1 (element in standard state)
  • ΔfH°[CO2(g)] = −393.5 kJ mol−1
  • ΔfH°[H2O(l)] = −285.8 kJ mol−1

Solution:

The standard enthalpy of reaction is calculated as:

\[ \Delta_\mathrm{r}H^\circ = \sum \nu_i \Delta_\mathrm{f}H^\circ(\text{products}) - \sum \nu_i \Delta_\mathrm{f}H^\circ(\text{reactants}) \]

where νi represents the stoichiometric coefficients from the balanced equation.

Step 1: Calculate the sum for products.

\[ \begin{align*} \sum \nu_i \Delta_\mathrm{f}H^\circ(\text{products}) &= 3 \times (-393.\bar{5}~\mathrm{kJ\,mol^{-1}}) + 4 \times (-285.\bar{8}~\mathrm{kJ\,mol^{-1}}) \\[1.5ex] &= -1180.\bar{5}~\mathrm{kJ\,mol^{-1}} + (-1143.\bar{2}~\mathrm{kJ\,mol^{-1}}) \\[1.5ex] &= -2323.\bar{7}~\mathrm{kJ\,mol^{-1}} \end{align*} \]

Step 2: Calculate the sum for reactants.

\[ \begin{align*} \sum \nu_i \Delta_\mathrm{f}H^\circ(\text{reactants}) &= 1 \times (-103.\bar{8}~\mathrm{kJ\,mol^{-1}}) + 5 \times (0.0~\mathrm{kJ\,mol^{-1}}) \\[1.5ex] &= -103.\bar{8}~\mathrm{kJ\,mol^{-1}} + 0.0~\mathrm{kJ\,mol^{-1}} \\[1.5ex] &= -103.\bar{8}~\mathrm{kJ\,mol^{-1}} \end{align*} \]

Step 3: Calculate ΔrH°.

\[ \begin{align*} \Delta_\mathrm{r}H^\circ &= -2323.\bar{7}~\mathrm{kJ\,mol^{-1}} - (-103.\bar{8}~\mathrm{kJ\,mol^{-1}}) \\[1.5ex] &= -2323.\bar{7}~\mathrm{kJ\,mol^{-1}} + 103.\bar{8}~\mathrm{kJ\,mol^{-1}} \\[1.5ex] &= -2219.\bar{9}~\mathrm{kJ\,mol^{-1}} \\[1.5ex] &= -2220~\mathrm{kJ\,mol^{-1}} \end{align*} \]

The standard molar enthalpy of combustion for propane is −2220 kJ mol−1.

This large negative value indicates a highly exothermic reaction, consistent with propane’s widespread use as a fuel.

Practice

Calculate the standard enthalpy of reaction for the synthesis of ammonia:

\[\mathrm{N_2(g) + 3~H_2(g) \longrightarrow 2~NH_3(g)}\]

Use the following standard enthalpy of formation: ΔfH°[NH3(g)] = −45.90 kJ mol−1

Solution

Apply the standard enthalpy of reaction formula:

\[ \Delta_{\mathrm{r}} H^{\circ} = \sum \nu_i \Delta_{\mathrm{f}}H^{\circ}(\text{products}) - \sum \nu_i \Delta_{\mathrm{f}}H^{\circ}(\text{reactants}) \]

Both N2(g) and H2(g) are elements in their standard states, so their ΔfH° = 0.

\[ \begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= \bigg[2 \times \Delta_{\mathrm{f}}H^{\circ}(\mathrm{NH_3, g})\bigg] - \bigg[\Delta_{\mathrm{f}}H^{\circ}(\mathrm{N_2, g}) + 3 \times \Delta_{\mathrm{f}}H^{\circ}(\mathrm{H_2, g})\bigg] \\[1.5ex] &= \bigg[2 \times (-45.9\bar{0}~\mathrm{kJ~mol^{-1}})\bigg] - \bigg[0 + 0\bigg] \\[1.5ex] &= -91.8\bar{0}~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -91.80~\mathrm{kJ~mol^{-1}} \end{align*} \]

The negative value indicates that ammonia synthesis is exothermic.

Summary

Enthalpy: Definition and Properties

Enthalpy is defined as:

\[ H = U + PV \]

At Constant Pressure (P = constant): Heat transfer equals enthalpy change: ΔH = qp

This makes enthalpy experimentally convenient because most laboratory reactions occur in open containers at atmospheric pressure.

Enthalpy is a state function (path-independent), while heat (qp) remains a path function but equals ΔH at constant pressure. For reactions involving gases, ΔH = ΔU + ΔngasRT (or ΔrH = ΔrU + ΔνgasRT for molar quantities). For reactions with only solids/liquids, ΔH ≈ ΔU.

Standard Enthalpies

Standard State (denoted °): 1 bar pressure, usually 298.15 K

Standard Enthalpy of Formation (ΔfH°): Enthalpy change when 1 mole of a compound forms from its elements in their standard states. Elements in their most stable form have ΔfH° = 0 by convention.

Standard Enthalpy of Reaction:

\[ \Delta_{\mathrm{r}} H^{\circ} = \sum \nu_i \Delta_{\mathrm{f}}H^{\circ}(\text{products}) - \sum \nu_i \Delta_{\mathrm{f}}H^{\circ}(\text{reactants}) \]

Sign conventions: ΔH < 0 is exothermic (heat released); ΔH > 0 is endothermic (heat absorbed).