Standard Enthalpies of Formation

Submitted by jwmoore on Sat, 03/26/2011 - 10:37

By now chemists have measured the enthalpyA thermodynamic state function, symbol H, that equals internal energy plus pressure x volume; the change in enthalpy corresponds to the energy transferred as a result of a temperature difference (heat transfer) when a reaction occurs at constant pressure. changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Fortunately Hess' law makes it possible to list a single value, the standard enthalpy of formationThe enthalpy change when one mole of a compound is produced from its elements, each in their most stable form at a given temperature and standard pressure (1 bar). Also called standard heat of formation. ΔHf, for each compoundA substance made up of two or more elements and having those elements present in definite proportions; a compound can be decomposed into two or more different substances.. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. and a specified temperatureA physical property that indicates whether one object can transfer thermal energy to another object. (usually 25°C).

For example, if we know that ΔHf[H2O(l)] = –285.8 kJ mol–1, we can immediately write the thermochemical equation


H2(g) + ½O2(g) → H2O(l)      ΔHm = –285.8 kJ mol–1      (1)


The elements H and O appear as diatomicContaining two atoms per molecule. molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per moleThat chemical amount of a substance containing the same number of units as 12 g of carbon-12. of H2O(l) formed. Equation (1) must specify formation of 1 mol H2O(l), and so the coefficient of O2 must be ½.

In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heatEnergy transferred as a result of a temperature difference; a form of energy stored in the movement of atomic-sized particles. absorbed serves to determine ΔHf. Quite often, however, elements do not react directly with each other to form the desired compound, and ΔHf must be calculated by combining the enthalpy changes for other reactions. A case in point is the gas acetylene, C2H2. In Example 1 from the Hess' law section we showed that the thermochemical equation


2C(s) + H2(g) → C2H2(g) ΔHm = 227.0 kJ mol–1

was valid. Since it involves 1 mol C2H2 and the elements are in their most stable forms, we can say that ΔHf[C2H2(g)] = 227.0 kJ mol–1.

One further point arises from the definition of ΔHf. The standard enthalpy of formation for an element in its most stable state must be zero. If we form mercury from its elements, for example, we are talking about the reaction


Hg(l) → Hg(l)


Since the mercury is unchanged, there can be no enthalpy change, and ΔHf = 0 kJ mol–1.

Some Standard Enthalpies of Formation at 25°C.

Compound ΔHf/kJ

mol–1

ΔHf/kcal

mol–1

Compound ΔHf/kJ

mol–1

ΔHf/kcal

mol–1

AgCl(s)
–127.068
–30.35
H2O(g)
–241.818
–57.79
AgN3(s)
+620.6
+148.3
H2O(l)
–285.8
–68.3
Ag2O(s)
–31.0
–7.41
H2O2(l)
–187.78
–44.86
Al2O3(s)
–1675.7
–400.40
H2S(g)
–20.63
–4.93
Br2(l)
0.0
0.00
HgO(s)
–90.83
–21.70
Br2(g)
+30.907
+7.385
I2(s)
0.0
0.0
C(s), graphite
0.0
0.00
I2(g)
+62.438
+14.92
C(s), diamond
+1.895
+0.453
KCl(s)
–436.747
–104.36
CH4(g)
–74.81
–17.88
KBr(s)
–393.798
–94.097
CO(g)
–110.525
–26.41
MgO(s)
–601.7
–143.77
CO2(g)
–393.509
–94.05
NH3(g)
–46.11
–11.02
C2H2(g)
+226.73
+54.18
NO(g)
+90.25
+21.57
C2H4(g)
+52.26
+12.49
NO2(g)
+33.18
+7.93
C2H6(g)
–84.68
–20.23
N2O4(g)
+9.16
+2.19
C6H6(l)
+49.03
+11.72
NF3(g)
–124.7
–29.80
CaO(s)
–635.09
–151.75
NaBr(s)
–361.062
–86.28
CaCO3(s)
–1206.92
–288.39
NaCl(s)
–411.153
–98.24
CuO(s)
–157.3
–37.59
O3(g)
+142.7
+34.11
Fe2O3(s)
–824.2
–196.9
SO2(g)
–296.83
–70.93
HBr(g)
–36.4
–8.70
SO3(g)
–395.72
–94.56
HCl(g)
–92.307
–22.06
ZnO(s)
–348.28
–83.22
HI(g)
+26.48
+6.33


Standard enthalpies of formation for some common compounds are given in the table above. These values may be used to calculate ΔHm for any chemical reactionA process in which one or more substances, the reactant or reactants, change into one or more different substances, the products; chemical change involves rearrangement, combination, or separation of atoms. Also called chemical change. so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example.



EXAMPLE 1 Use standard enthalpies of formation to calculate ΔHm for the reaction

2CO(g) + O2(g) → 2CO2(g)


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. We can imagine that the reaction takes place in two steps, each of which involves only a standard enthalpy of formation. In the first step CO (carbon monoxide) is decomposed to its elements:


2CO(g) → 2C(s) + O2(g)      ΔHm = ΔH1      (2)


Since this is the reverse of formation of 2 mol CO from its elements, the enthalpy change is


ΔH1 = 2 × {–ΔHf [CO(g)]} = 2 × [– (–110.5 kJ mol–1)] = +221.0 kJ mol–1


In the second step the elements are combined to give 2 mol CO2(carbon dioxide):


2C(s) + 2O2(g) → 2CO2(g)      ΔHm = ΔH2      (3)


In this case


ΔH2 = 2 × ΔHf [CO2(g)] = 2 × (– 393.5 kJ mol–1) = – 787.0 kJ mol–1


You can easily verify that the sum of Eqs. (2) and (3) is


2CO(g) + 2 O2(g) → 2CO2(g)      ΔHm = ΔHnet


Therefore


ΔHnet = ΔH1 + ΔH2 = 221.0 kJ mol–1 – 787.0 kJ mol–1 = – 566.0 kJ mol–1


Note carefully how Example 1 was solved. In step 1 the reactantA substance consumed by a chemical reaction. compound CO(g) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so ΔH1 was opposite in sign from ΔHf. Step 1 also involved 2 mol CO(g) and so the enthalpy change had to be doubled. In step 2 we had the hypothetical formation of the productA substance produced by a chemical reaction. CO2(g) from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same.

Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients. This can he summarized by the equation


ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)      (4)


The symbol Σ means “the sum of.” Since ΔHf values are given per mole of compound, you must be sure to multiply each ΔHf by an appropriate coefficient derived from the equation for which ΔHm is being calculated.



EXAMPLE 2 Use the table of standard enthalpies of formation at 25°C to calculate ΔHm for the reaction

4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g)


Solution Using Eq. (4), we have


ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)

       = [6 ΔHf (H2O) + 4 ΔHf (NO)] – [4 ΔHf (NH3) + 5 ΔHf (O2)]

       = 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0

       = –1450.8 kJ mol–1 + 361.2 kJ mol–1 + 184.4 kJ mol–1

       = –905.2 kJ mol–1

Note that we were careful to use ΔHf [H2O(g)] not ΔHf [H2O(l)]. Even though water vaporThe gaseous state of a substance that typically exists as a liquid or solid; a gas at a temperature near or below the boiling point of the corresponding liquid. is not the most stable form of water at 25°C, we can still use its ΔHf value. Also the standard enthalpy of formation of the element O2(g) is zero by definition. Obviously it would be a waste of space to include it in the table above.