# FreeEnergyLesson4

# 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∆ _{I}G^{o}/dT = -∆_{I}S^{o} ** (4.1)

since **∆ _{I}G^{o}** implies that all reactants and products are at standard pressure.

From the two expressions for **∆ _{I}G^{o}**:

**∆ _{I}G^{o} = ∆_{I}H^{o} - T∆_{I}S^{o}** from eq.(2.1)

and:

**∆ _{I}G^{o} = -RT ln K** from eq.(3.17)

we shall derive a relationship between **∆ _{I}H^{o}** and

**In K**. Differentiating eq. (3.17) with respect to

**T** and changing signs we obtain:

**∆S ^{o} = R In K + RT dIn K/dT**

Multiplying with **T**, and replacing **RT In K** with **∆ _{I}G^{o}**, give:

**T∆ _{I}S^{o} = -∆_{I}G^{o} + RT^{2} dIn K/dT**

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

**d ln K/dT = ∆ _{T}H^{o}/RT^{2}** (4.2)

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

Since **d ln K = - dT/T ^{2}**, eq. (4.2) can be given the form:

**d ln K/d(l/T) = -∆ _{T}H^{o}/R**(4.3)

When plotting **In K** as a function of **VT** we obtain **-∆ _{r}H^{o}/R** as the slope of the curve, see Fig. 4.1. The variation in

**∆**with temperature is given by eq (5.9)in 1st law of thermodynamics lesson 5

_{r}H^{o}∆_{r}H^{o}_{T}= ∆_{r}H^{o}_{298}+ ∆_{r}C_{P}^{o}dT

**∆ _{r}H^{o}_{T} = ∆_{r}H^{o}_{298} + ∆_{r}C_{p}^{o}dT**

The integral in the eq. **∆ _{r}H^{o}_{T} = ∆_{r}H^{o}_{298} + ∆_{r}C_{p}^{o}dT** will usually have a small value compared to

**∆**except when the temperature range is very large. This means that the curve in Fig. 4.1 is approximately a straight line.

_{r}H^{o}_{298}**∆**for a reaction is found from a plot similar to the one in Fig. 4.1. The experiment shows that the variation in

_{r}H^{o}**∆**with temperature can be ignored.

_{r}H^{o}

Fig. 4.1. The temperature dependence of the equilibrium constant for the reaction: **CO _{2}(g) + H_{2} (g) = CO (g) + H_{2}O (g). **

The variation with temperature of **∆ _{r}S^{o}** for a reaction is also usually very small, and therefore

**∆**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:

_{r}G^{o}

**∆ _{r}G^{o}_{T} = ∆_{r}H^{o}_{T} - T∆_{r}SO_{T}**

with the approximate equation:

**∆ _{r}G^{o}_{T} = ∆_{r}H^{o}_{298} - T∆_{r}S^{o}_{298}** (4.4)

Equation (6.45) represents a straight line with the slope **∆ _{r}S^{o}_{298}**. If the line is extrapolated to the absolute zero, the intercept with the ordinate gives

**∆**At temperatures of transition, there is a change in

_{r}H^{o}_{298}**∆**and a corresponding change in

_{r}H^{o}**∆**.

_{r}S^{o}

Equation (4.8)in Entropy lesson 4, gives the relation between **∆ _{fus}S^{o}** and

**∆**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 in

_{fus}H^{o}**∆**. There is a break in the curve for

_{r}G^{o}**∆**as a function of

_{s}G^{o}**T**, however, since the values of

**∆**and

_{r}H^{o}**∆**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.

_{r}S^{o}

### Example1

We shall calculate ** ∆ _{r}G^{o}** for NaCl as a function of temperature in the range 298 - 1400 K using the approximate eq. (4.4)

In a temperature range where there is no phase change we assume constant values for ** ∆ _{f}H^{o}** and

**∆**. A phase change leads to changes in

_{f}S^{o}**∆**and

_{f}H^{o}**∆**.

_{f}S^{o}

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

**∆ _{f}H^{o}** = - 411 kI,

**∆**= 72 - 51 - 112x223 = - 90.5 I K

_{f}S^{o}^{-1}(see eq. 2.2)

**∆ _{f}G^{o} = (- 411 + T 90.5/1000) kJ**

Temperature 371 K: **Na (s) = Na (1); ∆ _{fus}H/T_{fus} = 3000/371 = 8 I K-l = ∆_{fus}S** (see eq. (4.8)in Entropy lesson 4)

**∆ _{fus}G = ∆_{fus}H - T_{fus}∆_{fus}S = 0**

Temperature range 371 - 1074 K: **Na (I) + 1/2 Cl _{2} (g) = NaCI (s)**

**∆ _{f}H^{o} = - 411 - 3 = - 414 kJ; ∆_{f}S^{o} = - 90.5 - 8 = - 98.5 J K^{-1 }**

**∆ _{f}G^{o}** = (-414 + T 98.5/1000) kJ

Temperature 1074 K: **NaCl (s) = NaCl (l); _{fus}H/T_{fus} = 28000/1074 = 26.1 J K_{-1} = ∆_{f}S**

**∆ _{fus}G = 0**

Temperature range 1074 - 1165 K: **Na (I) + 1/2 Cl _{2} (g) = NaCl (I)**

**∆ _{f}H^{o} = - 414 + 28 = - 386 kJ; ∆_{f}S^{o} =- 98.5 + 26.1 = - 72.4 J K^{-1 }**

**∆ _{f}G^{o} = (-386 + T 72.4/1000) kJ **

Temperature 1165 K: **Na (l) = Na (g); ∆ _{vap}H/lT_{b} = 106000/1165 = 91 J K^{-1} = ∆_{vap}S = 0 **

**∆ _{vap}G =0**

Temperature range 1165 - 1400 K: **Na (g) + 1/2 Cl _{2} (g) = NaCI (I) **

**∆ _{f}H^{o} - 386 - 106 - = - 492 kJ; ∆_{f}S^{o} = - 72.4 - 91 = - 163.4 J K^{-1}**

**∆ _{f}G^{o} = (-492 + T 163.4/1000) kJ**