University of Malta CH237 - Chemical Thermodynamics and Kinetics

Dr. Joseph N. Grima, Department of Chemistry
University of Malta, Msida, MSD 06, MALTA
http://staff.um.edu.mt/jgri1/teaching/ch237
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Statistical Thermodynamics (vi)
 

The equilibrium constant 

(a) Derivation of the equilibrium constant in terms of the partition function
(b) The physical basis for equilibrium constants
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The equilibrium constant 

(a) Derivation of the equilibrium constant in terms of the partition function

We have already seen from classical thermodynamics that the equilibrium constant for a reaction is related to the Gibbs free energy of the forward process through:

Thus, if we were to consider a gas phase reaction where the equilibrium constant can be expressed in terms of the partial pressures. 

In this derivation we shall make use of the standard molar Gibbs free energy, defined as:

for each of the species J. To calculate this, we need to the value of the standard molar partition function, qm0, that is, the molar partition function when p = p0 = 1 bar. 

Consider the reaction:

The standard Gibbs energy is given by:
where:
as derived in the previous section (ideal gasses).

Thus we have:

But G(0) = U(0), the first term on the right equates to the difference in the molar energies of the ground states of the products and reactants and is calculated from the bond dissociation enthalpies (seefig. 1), i.e.:
whilst since a ln(x) + b ln(y) - c ln(z) = ln(xa) + ln(yb) - ln(zc) = ln(xayb/zc), the second term is given by:

But we know that 
i.e.:
i.e. by taking antilogarithms, we obtain the equilibrium constant:
This equation may be summarised by:
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Fig. 1: The definition of DrE0 for the calculation of equilibrium constants. 
 

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(b) The physical basis for equilibrium constants

In this section we shall look into the physical basis of equilibrium constants, vis-à-vis the simple R <=>P gas-phase equilibrium (R for reactants, P for products).

In each of the situations in the fig. 2 below, we have two sets of energy levels: one set of states belongs to R other belongs to P. The populations of the states are given by the Boltzmann distribution and are independent of whether any given state happens to belong to R or to P. We can therefore imagine a single Boltzmann distribution spreading, without distinction over the two sets of states. Then:

  • If the spacings of R and Pare similar (as in Fig. 2a), and P lies above R, then R will dominate in the equilibrium mixture. 
  • If P has a high density of states (a large number of states in a given energy range, as in Fig. 2b) then, even though its zero-point energy lies above that of R, the species P might still dominate at equilibrium.
Let us now attempt to quantify these deductions.

The population in a state i of the composite system (i.e. R, P) is given by the Boltzmann distribution, i.e.:

where N is the total number of molecules.

The total number of R molecules is the sum of these populations taken over the states belonging to R; these states we label r with energies er. The total number of P molecules is the sum over the states belonging toP; these states we label p with energies e`p:

where the last step is the conversion of molecular energies to molar energies.

The equilibrium constant is hence given by:


 

(a)

(b)

Fig. 2: The array of R(eactants) and P(roducts) energy levels:(a) At equilibrium all are accessible (to differing extents, depending on the temperature), and the equilibrium composition of the system reflects the overall Boltzmann distribution of populations. As DrE0 increases, R becomes dominant. (b) It is important to take into account the densities of states of the molecules. Even though P might lie well above R in energy (that is, (DrE0  is large and positive), P might have so many states that its total population dominates in the mixture. In classical thermodynamic terms, we have to take entropies into account as well as enthalpies when considering equilibria. 
 
 
 

NOTE:

The significance of this last equation can be seen most clearly by exaggerating the molecular features that contribute to it (see Fig. 3). We shall suppose that R has only a single accessible level, which implies that qR = 1. We also suppose that P has a large number of evenly, closely spaced levels. The partition function of P is then qP = kT/e. In this model system, the equilibrium constant is:

  • When DrE0 is very large, the exponential term dominates and K < 1, which implies that very little P is present at equilibrium. When DrE0 is small but still positive, K can exceed 1 because the factor kT/e may be large enough to overcome the small size of the exponent term. The size of K then reflects the predominance of P at equilibrium on account of its high density of states. 
  • At low temperatures K << 1, and the system consists entirely of R. At high temperatures the exponential function approaches 1 and the pre-exponential factor is large. Hence P becomes dominant. We see that, in this endothermic reaction (endothermic because P lies above R), a rise in temperature favours P, because its states become accessible. 
  • The model also shows why the Gibbs energy, G, and not just the enthalpy, determines the position of equilibrium. It shows that the density of states (and hence the entropy) of each species as well as their relative energies controls the distribution of populations and hence the value of the equilibrium constant.
Fig. 3: The exageration used for exploring the effects of energy separations and densities of states on equilibria - The products P can dominate provided DrE0 is not too large and that P has an appreciable density of states. 
 

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