The heatEnergy transferred as a result of a temperature difference; a form of energy stored in the movement of atomic-sized particles. changes which accompany a 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. are caused largely by changes in the electronic energyA system's capacity to do work. of the molecules. If we restrict our attention to gases, and hence to fairly simple molecules, we can go quite a long way toward predicting whether a reaction will be exothermicDescribes a process in which energy is transferred to the surroundings as a result of a temperature difference. by considering the bonds which are broken and made in the course of the reaction. In order to do this we must first become familiar with the idea of a bond enthalpyThe change in enthalpy when a mole of chemical bonds of a particular type are broken; the molecules whose bonds are broken must be in the gas phase. Closely related to bond energy..
In other sections we point out that when a chemical bond forms, negative charges move closer to positive charges than before, and so there is a lowering of the energy of the molecule relative to the atoms from which it was made. This means that energy is required to break a molecule into its constituent atoms. The bond enthalpy DX–Y of a diatomicContaining two atoms per molecule. molecule X—Y is the enthalpy change for the (usually hypothetical) process:
XY(g) → X(g) + Y(g) ΔH°(298 K) = DX―Y (1)
We have already used the term bond energyThe change in energy when a mole of chemical bonds of a particular type are broken; the molecules whose bonds are broken must be in the gas phase. Closely related to bond enthalpy. to describe this quantity, though strictly speaking the bond energy is a measure of ΔU rather than ΔH. As we have already seen, ΔU and ΔH are nearly equal, and so either term may be used.
As an example, let us consider the bond enthalpy for carbon monoxide. It is possible to establish the thermochemical equation
CO(g) → C(g) + O(g) ΔH°(298 K) = 1073 kJ mol–1 (2)
Accordingly we can write
- DC≡O = 1073 kJ mol–1
even though the process to which Eq. (2) corresponds is hypothetical: Neither carbon nor oxygen exists as a monatomic gas at 298 K. For triatomic and polyatomicContaining two or more atoms. molecules, the bond enthalpy is usually defined as a mean. In the case of water, for instance, we have
H2O(g) → 2H(g) + O(g) ΔH°(298 K) = 927.2 kJ mol–1
Since it requires 927.2 kJ to break open two O—H bonds, we take half this value as the mean bond enthalpy and write
- DO―H = 463.6 kJ mol–1
In methanol, CH3OH,however, a value of 427 kJ mol–1 for the O—H bond enthalpy fits the experimental data better. In other words the strength of the O—H varies somewhat from 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. to compound. Because of this fact, we must expect to obtain only approximate results, accurate only to about ± 50 kJ mol–1, from the use of bond enthalpies. Bond enthalpies for both single and multiple bonds are given in Table 1.
TABLE 1 Average Bond Energies/kJ mol–1.
As an example of how a table of bond enthalpies can he used to predict the ΔH value for a reaction, let us take the simple case
H2(g) + F2(g) → 2HF(g) 298 K, 1 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg. (3)
We can regard this reaction as occurring in two stages (Fig. 1). In the first stage all the reactantA substance consumed by a chemical reaction. molecules are broken up into atoms:
H2(g) + F2(g) → 2H(g) + F(g) 298 K, 1 atm (3a)
For this stage
- ΔHI = DH―H + DF―F
since 1 mol H2 and 1 mol F2 have been dissociated.
In the second stage the H and F atoms are reconstituted to form HF molecules:2H(g) + 2F(g) → 2HF(g) 298 K, 1 atm (3b)
- ΔHII = – 2DH―F
where a negative sign is necessary since this stage corresponds to the reverse of dissociationThe breaking apart of one species into two or more smaller species; often applied to ions in a crystal lattice, which dissociate when the ionic solid dissolves in water. Dissociation refers to separation of particles that already exist; ionization refers to the formation of ions from neutral species, as in the ionization of a weak acid in aqueous solutoin..
Since Eq. (3) corresponds to the sum of Eqs. (3a) and (3b), Hess' law allows us to add ΔH values:
- ΔH°reaction = ΔHI + ΔHII
- = DH―H + DF―F – 2DH―F
- = (436 + 159 – 2 × 566) kJ mol–1
- = –539 kJ mol –1
- ΔH°reaction = ΔHI + ΔHII
We can workA mechanical process in which energy is transferred to or from an object, changing the state of motion of the object. this same trick of subdividing a reaction into a bond-breaking stage followed by a bond-making stage for the general case of any gaseous reaction. In the first stage all the bonds joining the atoms in the reactant molecules are broken and a set of gaseous atoms results. For this stage
- ΔHI = ΣD (bond broken)
The enthalpy change is the sum of the bond enthalpies for all bonds broken. In the second stage these gaseous atoms are reconstituted into the productA substance produced by a chemical reaction. molecules. For this second stage therefore
- ΔHII = – ΣD (bond formed)
where the negative sign is necessary because the reverse of bond breaking is occurring in this stage. The total enthalpy change for the reaction at standard pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. is thus
ΔH° = ΔHI + ΔHII
or ΔH° = ΣD (bond broken) – ΣD (bond formed) (4)
The use of this equation is illustrated in the next example.
EXAMPLE 1 Using Table 1 calculate the value of ΔH°(298 K) for the reaction
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. It is best to sketch the molecules and their bonds in order to make sure that none are missed.
Thus ΔH° = ΣD (bond broken) – ΣD (bond formed)
= 4 DC―H + 2DO=O – 2DC=O – 4DO―H
= (4 × 416 + 2 × 498 – 2 × 803 – 4 × 467) kJ mol–1
= – 814 kJ mol–1
The experimental value for this enthalpy change can be calculated from standard enthalpies of formation. It is –802.4 kJ mol–1. The discrepancy is due to the unavoidable use of mean bond enthalpies in the calculation.