Standard Enthalpy of Reaction

Changes in enthalpy are normally reported for processes taking place under a set of standard conditions. The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25 °C). The standard enthalpy change for a reaction or a physical process is the difference between the products in their standard states and the reactants in their standard states, all at the same specified temperature.

The standard enthalpy change, ΔH°, is the change in enthalpy for a process in which the initial and final substances are in their standard states:

For example, the standard state of liquid water at 298.15 K is pure water at 2981.15 K and 1 bar. As an example of a standard enthalpy change, the standard enthalpy of vaporization, ∆_vap H°, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar

Unless otherwise mentioned, all thermodynamic tabuled data will refer to the conventional temperature of 298.15 K.

For the transition from solid to liquid, we have

Standard Enthalpy of Formation

The standard enthalpy of formation (or standard heat of formation) Δ_fH°_T of a pure substance at temperature T is the standard reaction enthalpy for the formation of the substance from its elements in theor reference state. The reference state of an element is the stablest form (physical state and allotrope) of the element under standard thermodynamic conditions. The reference form of oxygen at 25 °C is O₂ (g); the reference form of carbon at 25 °C is graphite. By convention the standard enthalpies of formation of elements in their reference states are arbitrarily assigned a zero value.

Some elements exist in the same physical state (gas, liquid, or solid) in two or more distinct forms. For example, oxygen in any of the physical states occurs both as dioxygen (commonly called simply oxygen), with O₂ molecules, and as ozone, with O₃ molecules. Dioxygen gas is odorless; ozone gas has a characteristic pungent odor. Solid carbon has two principal crystalline forms: graphite and diamond. Graphite is a soft, black, crystalline substance; diamond is a hard, usually colorless crystal. The elements oxygen and carbon are said to exist in different allotropic forms. An allotrope is one of two or more distinct forms of an element in the same physical state.

To understand this definition, consider the standard enthalpy of formation of liquid water. Note that the stablest forms of hydrogen and oxygen at 1 atm and 25°C are H₂ (g) and O₂ (g), respectively. These are therefore the reference forms of the elements. You write the formation reaction for 1 mol of liquid water as follows:

The standard enthalpy change for this reaction is −285.8 kJ per mole of H₂O. Therefore, the thermochemical equation is H₂ (g) + 12 O₂ (g) ⟶ H₂O(l); ΔH_f° = −285.8 kJ. The values of standard enthalpies of formation listed in the table and in other tables are determined by direct measurement in some cases and by applying Hess’s law in others. Oxides, such as water, can often be determined by direct calorimetric measurement of the combustion reaction. Note that the standard enthalpy of formation of an element will depend on the form of the element. For example, the ΔH_f° for diamond equals the enthalpy change from the stablest form of carbon (graphite) to diamond. The thermochemical equation is

On the other hand, the ΔH_f° for graphite equals zero. Note the values of ΔH_f° for the elements listed in Table 6.2; the reference forms have zero values.

Now let us see how to use standard enthalpies of formation (listed in the Table) to find the standard enthalpy change for a reaction. We will first look at this problem from the point of view of Hess’s law. But when we are finished, we will note a pattern in the result, which will allow us to state a simple formula for solving this type of problem. Consider the equation

From Table 6.2 you find the enthalpies of formation for CH₄ (g), CCl₄ (l), and HCl(g). You can then write the following thermochemical equations:

C(graphite) + 2H₂ (g) ⟶ CH₄ (g); ΔH_f° = −74.9 kJ (1)

C(graphite) + 2Cl₂ (g) ⟶ CCl₄ (l); ΔH_f° = −135.4 kJ (2)

2H₂ (g) + 12Cl₂ (g) ⟶ HCl(g); ΔH_f° = −92.3 kJ (3)

You now apply Hess’s law. Since you want CH₄ to appear on the left, and CCl₄ and 4HCl on the right, you reverse Equation 1 and add Equation 2 and 4 x Equation 3.

CH₄ (g) ⟶ C(graphite) + 2H₂ (g)  (−74.9 kJ) x (−1)

C(graphite) + 2Cl₂ (g) ⟶ CCl₄ (l)   (−135.4 kJ) x (1)

2H₂ (g) + 2Cl₂ (g) ⟶ 4HCl(g)   (−92.3 kJ) x (4)

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CH₄ (g) + 4Cl₂ (g) ⟶ CCl₄ (l) + 4HCl(g)   ΔH° = −429.7 kJ

The setup of this calculation can be greatly simplified once you closely examine what you are doing. Note that the ΔH_f° for each compound is multiplied by its coefficient in the chemical equation whose ΔH° you are calculating. Moreover, the ΔH_f° for each reactant is multiplied by a negative sign. You can symbolize the enthalpy of formation of a substance by writing the formula in parentheses following ΔH_f°. Then your calculation can be written as follows:

ΔH° = [ΔH_f° (CCl₄) + 4 ΔH_f° (HCl)] − [ΔH_f°(CH₄) + 4 ΔH_f° (Cl₂)] = [(−135.4) + 4(-92.3)] kJ − [(−74.9) + 4(0)] kJ) = −429.7 kJ

In general, you can calculate the ΔH° for a reaction by the equation

ΔH° = Σ n ΔH°_f (products) − Σ m Δ_f° (reactants).

m and n are the coefficients of the substances in the chemical equation.

Example 1. Find the standard-state enthalpy change of the reaction of N₂O₄(g) → 2NO₂(g) at 298.15 K, using values of enthalpy changes of formation tabulated.

Solution.

∆H°= 2∆_fH° (NO₂) + (−1)∆_fH° (N₂O₄) = 2(33.095 kJ mol⁻¹) − (9.079 kJ mol⁻¹) = 57.11 kJ mol⁻¹

The standard enthalpy of formation of ions in solutions poses a special problem because it is impossible to prepare a solution of cations alone or of anions alone. This problem is solved by defining one ion, conventionally the hydrogen ion, to have zero standard entalphy of formation at all temperature

Δ°_f(H⁺, aq) = 0

Thus, if the enthalpy of formation of HBr(aq) is found to be -122 kJ mol^-1, then the whole of that value is ascribed to the formation of Br^-(aq), and we write Δ°_f(Br^-, aq) = -122 kJ/mol. That value may then be combined with, for instance, the enthalpy formation of AgBr(aq) to determine the value of Δ°_f(Ag⁺, aq) and so on.

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