# LESSON 4: TEMPERATURE DEPENDENCE OF GIBBS ENERGY

The temperature dependence of Gibbs energy, (G/T)p = -S, was given by eq (3.5). For a chaemical reaction with all species in the standard state we can write:

d∆IGo/dT = -∆ISo (4.1)

since IGo implies that all reactants and products are at standard pressure.

From the two expressions for IGo:

IGo = ∆IHo - T∆ISo from eq.(2.1)

and:

IGo = -RT ln K from eq.(3.17)

we shall derive a relationship between IHo and In K. Differentiating eq. (3.17) with respect to

T and changing signs we obtain:

∆So = R In K + RT dIn K/dT

Multiplying with T, and replacing RT In K with IGo, give:

T∆ISo = -∆IGo + RT2 dIn K/dT

Using eq. (2.1) and rearranging, we obtain:

d ln K/dT = ∆THo/RT2 (4.2)

This equation is often called the van t'Hoff equation.

Since d ln K = - dT/T2, eq. (4.2) can be given the form:

d ln K/d(l/T) = -∆THo/R(4.3)

When plotting In K as a function of VT we obtain -∆rHo/R as the slope of the curve, see Fig. 4.1. The variation in rHo with temperature is given by eq (5.9)in 1st law of thermodynamics lesson 5

∆rHoT = ∆rHo298 + ∆rCPodT


rHoT = ∆rHo298 + rCpodT

The integral in the eq. rHoT = ∆rHo298 + rCpodT will usually have a small value compared to rHo298 except when the temperature range is very large. This means that the curve in Fig. 4.1 is approximately a straight line. rHo for a reaction is found from a plot similar to the one in Fig. 4.1. The experiment shows that the variation in rHo with temperature can be ignored.

Fig. 4.1. The temperature dependence of the equilibrium constant for the reaction: CO2(g) + H2 (g) = CO (g) + H2O (g).

The variation with temperature of rSo for a reaction is also usually very small, and therefore rGo for a reaction is usually a practically linear function of temperature. For many purposes one will obtain results with sufficient accuracy by replacing the equation:

rGoT = ∆rHoT - T∆rSOT

with the approximate equation:

rGoT = ∆rHo298 - T∆rSo298 (4.4)

Equation (6.45) represents a straight line with the slope rSo298. If the line is extrapolated to the absolute zero, the intercept with the ordinate gives rHo298 At temperatures of transition, there is a change in rHo and a corresponding change in rSo.

Equation (4.8)in Entropy lesson 4, gives the relation between fusSo and fusHo at the melting point. When a transition takes place reversibly, i.e., the system and the surroundings are at the same temperature, the transition temperature, there is no change inrGo. There is a break in the curve for sGo as a function of T, however, since the values of rHo and rSo have changed by the transition. An increased slope corresponds to a phase transition in a reactant, whereas a decreased slope corresponds to a phase transition in a product. The change in slope corresponds to the entropy of transition. It is small for melting and large for boiling.

### Example1

We shall calculate rGo for NaCl as a function of temperature in the range 298 - 1400 K using the approximate eq. (4.4)

Table4.1 NaCl as a function of temperature

In a temperature range where there is no phase change we assume constant values for fHo and fSo. A phase change leads to changes in fHo and fSo.

Temperature range 298 - 371 K: Na (s) + 112 Cl2 (g) = NaCI (s)

fHo = - 411 kI, fSo = 72 - 51 - 112x223 = - 90.5 I K-1 (see eq. 2.2)

fGo = (- 411 + T 90.5/1000) kJ

Temperature 371 K: Na (s) = Na (1); ∆fusH/Tfus = 3000/371 = 8 I K-l = ∆fusS (see eq. (4.8)in Entropy lesson 4)

fusG = ∆fusH - TfusfusS = 0

Temperature range 371 - 1074 K: Na (I) + 1/2 Cl2 (g) = NaCI (s)

fHo = - 411 - 3 = - 414 kJ; ∆fSo = - 90.5 - 8 = - 98.5 J K-1

fGo = (-414 + T 98.5/1000) kJ

Temperature 1074 K: NaCl (s) = NaCl (l); fusH/Tfus = 28000/1074 = 26.1 J K-1 = ∆fS

fusG = 0

Temperature range 1074 - 1165 K: Na (I) + 1/2 Cl2 (g) = NaCl (I)

fHo = - 414 + 28 = - 386 kJ; ∆fSo =- 98.5 + 26.1 = - 72.4 J K-1

fGo = (-386 + T 72.4/1000) kJ

Temperature 1165 K: Na (l) = Na (g); ∆vapH/lTb = 106000/1165 = 91 J K-1 = ∆vapS = 0

vapG =0

Temperature range 1165 - 1400 K: Na (g) + 1/2 Cl2 (g) = NaCI (I)

fHo - 386 - 106 - = - 492 kJ; ∆fSo = - 72.4 - 91 = - 163.4 J K-1

fGo = (-492 + T 163.4/1000) kJ

Fig. 4.2 The Gibbs energy of formation for one mole NaCI as a function of temperature