So far we know that for a reaction with ΔG < 0 ; the reaction can proceed spontaneously.
A + B � C + D ΔG < 0
What about a reaction with ΔG > 0 ; for example A + B � C + D ΔG > 0
No problem just reverse the reaction C + D � A + B ΔG < 0 . Now ΔG < 0 and thermodynamics says that this reaction is spontaneous.
However a spontaneous reaction do not implied that all the reactants would be converted completely into the product. In most cases it would not. It can safely be said that "100%" reaction is an exception than a norm.
CHEMICAL POTENTIAL, μ
To study the effect of concentration on a reaction we have to define a new parameter known as the chemical potential, μ, where;
| μ | = | ( | ∂G ∂n |
) | p,T |
n is the number of moles. Let us now consider a system of A, B, C and D;
A + B � C + D
such that ξ moles of A, a very small amount, has just reacted, then;
| ( | ∂G ∂ξ |
) | p,T | = | μC + μD − μA − μB | (1) |
Mathematics! mathematics! mathematics!
To make this concept practical we have to ask our mathematician friends for help.
| (1) (2) (3) So |
G dG H dH dU dG |
= = = = = = = |
H − TS dH − TdS − SdT U + PV dU + pdV + Vdp dq + dw TdS − pdV ; when system only do expansion work. Vdp − SdT (2) |
Consider the effect of the variation of p and T on G, i.e. G = f(p,T), we can write the expression for the partial derivatives as;
| dG | = | ( | ∂G ∂p |
) | T | dp | + | ( | ∂G ∂T | ) | p | dT | (3) |
Comparing Eqn (2) and (3), mathematics says;
| ( | ∂G ∂p | ) | T | = | V | ( | ∂G ∂T | ) | p | = | − S |
Tutorial 1
Using the same approach show that;
V = ( ∂U
∂S) V and p = − ( ∂U
∂T) p Answer
IDEAL GAS
| The only equation that interest us at the moment is; | ( | ∂G ∂p |
) | T | = | V |
The free energy of a liquid or solid is only weakly influence by pressure. So let us take a simple process where one mole of ideal gas at pressure p˚ is changed to pressure p. Remember to understand complex systems we always start with the simpliest system possible and develop from there. Note that free energy per mole (also known as molar Gibb's energy) = chemical potential.
| The change in free energy will then be given by; | μ − μ� | = | ∫ | Vdp | = | RT ln | ( | p p� |
) |
Of course we are more interested in solutions rather than gases, as most reactions take place in the liquid state. If we have a pure liquid it will vapourise. At equilibrium the vapour pressure above the pure liquid is p*. Let us now define a standard state for the liquid such that the chemical potential is μ� and the vaporu pressure is p�. The chemical potential of the liquid would be given by;
| μ*(l) − μ�(l) | = | ∫ | Vdp | = | RT ln | ( | p* p� | ) |
Let us now consider this liquid is mixed with another liquid. Notice we are working towards A + B � C + D. Then the chemical potential of the liquid is now μ and the partial vapour pressure is p. The situation becomes;
| μ(l) − μ�(l) | = | ∫ | Vdp | = | RT ln | ( | p p� |
) |
Note that the reference state used in both expressions is the standard state of the compound.
| Combining the two expressions; | μ(l) − μ*(l) | = | ∫ | Vdp | = | RT ln | ( | p p* |
) |
Francois Raoult found that for certain systems p = Χp* ; where Χ is the mole fraction of the compound in the solution. This is known as Raoult's law and such systems are known as ideal solution.
| So for an ideal solution; | μ(l) − μ*(l) | = | ∫ | Vdp | = | RT ln | ( | Χ | ) |
Now we can return to Equation (1) above. The free energy at any time is given by;
| ΔG − ΔG� | = | ∫ | Vdp | = | RT ln | ( | ΧCΧD ΧAΧB | ) |
For ΔG < 0 the reaction will proceed to products C + D. For ΔG > 0 the reaction will proceed from C +D to A + B. Then if ΔG = 0 it means that no more net reaction occur. That is the reaction has arrived at the point of equilibrium and the mole composition of A, B, C and D remains unchanged. So at ΔG = 0;
| RT ln(K) | = | − ΔG� | where | K | = | ( | ΧCΧD ΧAΧB |
) | the equilibrium constant |
Tutorial 2
Compute the standard free energy and the equilibrium constant for the reaction;
PbO(s) + CO(g) � Pb(s) + CO2(g) at 298 K and 400 K
Compound ΔH˚ (298K)
/ kJ mol‾�S˚ (298K)
/ JK‾� mol‾�ΔG˚ (298K)
/ kJ mol‾�Cp
/ JK‾� mol‾�Lead
Lead oxide
Carbon monoxide
Carbon dioxide0
− 219.0
− 110.5
− 393.564.8
66.5
197.7
213.70
− 188.9
− 137.2
− 394.426.4
45.8
29.1
37.1What is the standard free energy and the equilibrium constant for the reaction;
Pb(s) + CO2(g) � PbO(s) + CO(g) at 298 K and 400 K Answer
REAL GASES and REAL SOLUTIONS
For real gases and real solutions we define a term, activity, a, as;
a = φ p/p� (for real gases) and a = γ Χ (for real solutions)
p� is then defined as the pressure of the standard state of the real gas. φ is the fugacity coefficient and γ the activity coefficient.
The equilibrium constant computed from ΔG� = 0 will then be known as the thermodynamic equilibrium constant. It is related to the practical equilibrium constants, Kp and KΧ by;
K = Kφ x Kp (for real gases) and K = Kγ x KΧ (for real gases)
One expression used to determine the value for φ is;
ln φ = B'p + � C'p� + -----------
Another method is to express φ = f / p ; where f is known as the fugacity of the gas. It has the same unit as pressure.
Considering reduced pressure as the most significant condition for an ideal gas (high pressure introduces interaction between gas particles), one simple approach is to introduce a compression coefficient, Z, to the ideal gas equation for one mole of gas.
| Vm | = | Z | RT p |
So that as p tends to zero ; Z will tend to 1. The fugacity coefficient can then be expressed as (from p=0 to p);
| ln φ | = | ∫ | ( | Z − 1 p |
) | dp |
So that as Z tends to 1 ; φ will tend to 1.
If we consider the van der Waals equation for real gas, and assuming that a = 0, then;
| Z | = | 1 + | bp RT |
| Or | ln φ | = | ∫ | ( | b RT |
) | = | bp RT |
Then f = p exp(bp/RT).
Tutorial 3
Estimate the fugacity for ammonia at 10 atm. pressure and 298 K. (b for van der Waals gas expression is 2.72 x 10‾� dm� mol‾� (Answer: 10.15 atm)
It is more difficult to determine the activity coefficient. Measurements must be obtained when the mixture is at equilibrium. In general as a first approximation we will assume that the gas and the solution are ideal.
