THERMOCHEMISTRY
I would like to revise a few facts about enthalpy;
- It is a pain in the neck to try to evaluate absolute H value for any system.
- ΔH = q when the process is conducted at constant pressure.
- When heat energy is transferred into the system from the surrounding it is an exothermic process and q > 0. When it is from the system to the surrounding it is an endothermic process and q < 0.
Chemists in general has accepted that there is not much point in seeking for the absolute value of H. Instead we focus on the change in enthalpy (ΔH) of one mole of matter at 760 torr pressure and referred to it as the standard enthalpy, ΔHΘ. It is important to note that the standard enthalpy varies with temperature.
ENTHALPY is a STATE FUNCTION
PHYSICAL PROCESSES
Let us consider some examples.
- The standard enthalpy of fusion (or melting) ΔfusHΘ is the change in enthalpy when one mole of solid is transformed into a liquid at its fusion temperature at 760 torr of pressure.
- The standard enthalpy of vapourisation ΔvapHΘ is the change in enthalpy when one mole of liquid is transformed into vapour at its boiling temperature at 760 torr of pressure.
Let us try to compute the transformation of 1 mole of ice water at 0�C into one mole of steam at 100�C.
Since enthalpy is a state function it does not matter how you execute the process, you should get the same value at the end.
Let us do the straight forward method:
| ΔH | = | ΔfusHΘ + Cp,mΔT + ΔvapHΘ | |
| = | 6.01 kJ mol‾� + (0.075.3 kJ K‾� mol‾� x 100 K) + 40.7 kJ mol‾� | ||
| = | 54.2 kJ mol‾� |
This is referred to as the Born-Haber cycle, which states that the enthalpy of a physical process is the sum enthalpies of its composite processes.
Similarly we can define the standard enthalpy of solution (for dissolving a substance in a solvent); the standard enthalpy of hydration, sublimation, etc. We can apply the Born-Haber cycle in all these processes.
CHEMICAL REACTIONS
Let us move to processes that are more challenging. How much energy would be produced by the reaction;
Let us use the series of thermochemical processes;
If the enthalpies of the following thermochemical processes are known;
| NH3(g) | � | � { N2(g) + 3H2(g) } | 46,11 kJ mol‾� | ||
| HCl(g) | � | � { Cl2(g) + H2(g) } | 92.31 kJ mol‾� | ||
| � { N2(g) + 4H2(g) + Cl2(g) } | � | NH4Cl(s) | −314.4 kJ mol‾� | ||
| SUM: | NH3(g) + HCl(g) | � | NH4Cl(s) | −176.0 kJ mol‾� |
This is commonly referred to as Hess' Law. The enthalpy of a reaction is the sum enthalpies of its composite thermochemical reactions. You can actually make up thermochemical reactions regardless of whether they can actually take place, as long as the sum of the thermochemical equations are balanced. This is the power of a state function.
 TransitionState              |
| Energy profile for a general reaction |
It must be mindful that just because the products are energetically more stable than the reagents it does not mean that the reaction can proceed. The other factor is the activation energy. The reagents must have sufficient energy (that is by heating the system) for them to climb the energy barrier to become the products. The summit of the barrier is known as the transition state and the species found in this state is known as the intermediate species. Every steps in the reaction mechanism has its own energy barrier, but the step with the highest barrier determines whether the reaction will occur or not. If it can conquer this barrier the temperature is enough for it to climb over all the other barriers.
ENTHALPY of FORMATION
This illustrated the fact that if we can compile a comprehensive data bank we will be able to compute the enthalpy for any reaction. One of the most diverse group of covalent compounds in the chemical industry, for which the heat of reactions are of interest, are the organic compounds. Let us consider the reaction;
This become known as the enthalpy of formation of propane. The enthalpy of formation is stated as the enthalpy of the reaction at 760 torr pressure that forms the compound from its elements in their natural states at the specified temperature. By and large most tabulated data are for 298 K (that is SATP). The natural state being the state the element is most stable. For example the natural state at SATP of hydrogen is diatomic hydrogen gas. Where it is not so obvious we will have to specify it. For example the natural state for carbon is taken as graphite and for sulphur it is rhombic sulphur.
Tutorial 1
- What are the natural state of oxygen, chlorine, nitrogen, and sulphur at SATP?
- What is the enthalpy of formation of H2(g) at SATP? Answer
The natural state of the element at SATP is the same for all reactions, but the natural state of the product can be different. One important illustration is water;
H2(g) + O2(g) � H2O(l) ΔfHΘ{H2O(l)} = −285.8 kJ mol‾�
This is because water can exist as a liquid or a vapour at SATP. For compounds where such possibilities do not exist than this problem will not arise. However you will have noticed by now that it is common practice to denote the state of the compounds involved in thermochemical equations.
ENTHALPY of COMBUSTION
This is the amount of heat energy from the burning of a compound in excess oxygen. For the point of compiling data this is about the simplest process. Just burn the compound in a bomb with excess of oxygen, in a calorimeter, and determine the heat given out. For example propane;
Such data are very important for the energy industry. Household cooking is a by-product of the petrochemical industry. It is mainly propane with slight amount of butane. If you assumed it to be just propane you can try computing the amount of energy your household uses daily for cooking, from the weight of gas you used.
Another application of the heat of combustion is that it can be used to compute for the enthalpy of formation of almost all compounds.
Tutorial 2
Given ΔcHΘ{propane(g)} = −2.1229 x 106 J mol‾� and ΔfHΘ{CO2(g)} = −393.5 kJ mol‾�, compute the enthalpy of formation of propane(g)? Answer
For any class of reaction that interest you, you can state a standard enthalpy for the reaction.
Tutorial 3
The reaction
CH3(CH2)7-CH=CH-(CH2)7COOH (oil) + H2(g) � CH3(CH2)16COOH (s) ΔHΘ{oleic acid(l)} is used to prepare margarine from vegetable oil. It belongs to a class of reaction known as hydrogenation. How would you state the standard enthalpy of hydrogenation? Answer
Other class of reactions are ionisation, electron affinity, etc.
INTERPRETATION of STANDARD ENTHALPY
All standard enthalpies are based on the natural (or stable) state of the elements and the compounds at SATP. Certainly not all reactions occur at SATP, so what does this mean? By setting the condition at SATP we are saying that they are every time we refer to a particular element or compound its particular internal energy is a constant, and so the change in enthalpy reflects (not equal) the sum total of the energies involve in the breaking and forming of bonds in the reaction. If it is an exothermic enthalpy than the sum total of all the bonds formed in the products is less then that of the bonds broken in the reagents.
If there is a need to compute the enthalpy at a particular temperature, like the temperature of the reaction, this can be easily done by using a combination of Hess's Law and the Born-Haber cycle. Example;
Let the molar heat capacity at constant pressure for A, B, C, and D be Cp,m(A), Cp,m(B), Cp,m
at temperature T by using the thermochemical process;
So ΔH(T) = ΔHΘ + [3 ΔCp,m(C) + 4 ΔCp,m(D) - ΔCp,m(A) - 2 ΔCp,m(B)}SATP→T.
Assuming that Cp,m does not vary with temperature, then:
where ΔCp(298.15K→T) = [ΔCp(products) - ΔCp(reagents)] (T - 298.15K)
In general Kirchhoff's Law applies to conversions of ΔH between any two temperatures, not necessary SATP.
Tutorial 4
The standard enthalpy of formation at 298 K for the following compounds are;
Compounds n-Hexane Propane Propene Neo-hexane ΔfH� / kJ mol‾� −167 −104 20.4 −186 Temp. / K Cp / J mol‾� * 200
300
400
500
600
700
800111
143
181
217
248
274
29656.1
73.9
94.0
113
129
143
15550.2
64.6
80.5
95.2
108
119
129101
142
183
220
253
282
307*Data taken from National Institute of Standard and Technology.
Compute ΔfH� of the compounds at 700 K. Answer
ENTHALPY of BOND DISSOCIATION
Since the standard enthalpy is an indication of the change in bond energies we will do the ultimate and compile data for the energy for each and every covalent bond. Example;
| CH4(g) | � | CH3(g) + H(g) | ΔHΘ(CH3-H) = 435 kJ mol‾� |
This is known as the enthalpy of bond dissociation of bond dissociation energy.
When the thermochemical process is the dissociation of diatomic elements it is given the special name of standard enthalpy of atomisation, ΔatHΘ.
| H2(g) | � | H(g) + H(g) | ΔHΘ{ H2(g) } = −218.0 kJ mol‾� |
However this is not exclusive. The term standard enthalpy of atomisation is also used for the enthalpy of the reaction where solid elements in its natural state at SATP are atomised to gaseous atoms at SATP. Example;
| C (graphite, solid) | � | C(g) | ΔatHΘ{ C(graphite) } = −716.7 kJ mol‾� |
The next question to ask will be: "Will similar type of bond between two similar atoms have the same bond dissociation energy?" For example is ΔHΘ{CH3-H} = ΔHΘ{CH3CH2-H}? The answer is no, they are close but not exactly the same. However a mean value can sometimes be useful for a preliminary computation. So we computed the statistical mean of the bond dissociation energies of a particular type of bond between two particular atoms. For example the mean bond energy (or bond energy) of C-H bond is 348 kJ mol‾�. This will allow us to save time to look for the specific value if we only wanted an estimation.
Tutorial 5
Given the following; ΔatHΘ{ C(graphite) }= 716.7 kJ mol‾�, ΔatHΘ{ H(g) } = 218.0 kJ mol‾�, ΔH(C-C) = −331.0 kJ mol‾�, and ΔH(C-H) = −416.0 kJ mol‾�; estimate the standard enthalpy of formation of ethane (CH3CH3(g)). Answer
