Title: Lesson 9 Entropy and Gibb’s Free Energy

Learning Objectives:– Calculate Gibb’s Free Energy from ∆G = ∆H - T∆S

– Calculate Gibb’s Free Energy from Hess Cycles

– Predict and explain the effect of temperature changes on Gibb’s Free Energy

Main Menu

Refresh

2CH3OH(g) + H2(g) → C2H6(g) + 2H2O(g)

The standard entropy for CH3OH(g) at 298 K is 238 J K–1 mol–1, for H2(g) is 131 J K–1 mol–1 and for H2O(g) is 189 J K–1 mol–1.

Using information from Table 12 of the Data Booklet, determine the entropy change for this reaction.

Main Menu

To Do: Categorise the following as either spontaneous or non-spontaneous:

Cooking an egg, acid reacting with metal, the reaction in a battery, electrolysis, respiration, photosynthesis

Think of an example of each of the following:

A spontaneous exothermic reaction

A non-spontaneous endothermic reaction

A spontaneous endothermic reaction

A non-spontaneous exothermic reaction (difficult!)

Main Menu

Gibbs free energy is a useful accounting tool

For chemical reactions neither ∆H(system) nor ∆S(system) alone can reliably be used to predict the feasibility of a reaction.

To calculate feasibility we use this expression:

Tidying up and moving T to the other side:

Multiplying by -1 and reversing the inequality:

This gives the function known as Gibbs free energy (∆G(system)):

∆G(system) must be negative for a spontaneous process. Units: kJ or J mol-1

Whereas ∆H(system) is a measure of the quantity of heat change, ∆G(system) gives a measure of the quality of the energy available. (Ability of energy free to do useful work rather than leave the system as heat)

Main Menu

Using ∆G(system) to predict the feasibility of a change

We generally assume that both the enthalpy and entropy changes of the system do not change with temperature.

Temperature, T, is alike a tap which adjusts the significance of the term ∆S(system) in determining the value of ∆G(system).

At low temperature:

This shows that all exothermic reactions can occur at low temperatures.

At high temperature:

This means all reactions which have a positive value of ∆S(system) can be feasible at high temperatures even if they are endothermic.

Main Menu

1. If G is negative, the reaction is spontaneous in the forward direction.

2. If G is equal to zero, the reaction is at equilibrium.

3. If G is positive, then the reaction is non-spontaneous in the forward direction, but the reverse reaction will be spontaneous.

4.       for elements at standard state (pure elements at 25ºC and 1 atm are assigned a value of zero).

The Gibb’s free energy equation can be used to calculate the phase change temperature of a substance. During a phase change, equilibrium exists between phases, so if the G is zero, we know that the reaction is in equilibrium.

What does the Gibb’s free energy value tell us about a reaction?

Main Menu

Main Menu

Main Menu

Solutions

Main Menu

Main Menu

Summarising the Effect Of ∆H, ∆S and T on Spontaneity of Reaction

∆G = ∆H - T∆S

Task: Use the equation to help yourself reason these through

Main Menu

Spontaneity We can calculate Gibbs free energy in two ways:

Method 1: Directly from (appropriate) data using ∆G = ∆H - T∆S < 0

Method 2: Indirectly from standard ∆Gf

o values in the data booklet, using a Hess Cycle

Note: Similar to ∆Hfo values, ∆Gf

o for any element in its standard state is zero.

Main Menu

Method 1: From ∆G = ∆H - T∆S Is the reaction of ethene with hydrogen (to form ethane),

spontaneous at room temperature (298K)? C2H4(g) +H2(g) C2H6(g)

∆Ho = -137 kJ mol-1

∆So = -121 J K-1 mol-1

Calculate Gibbs Free Energy ∆G = ∆H - T∆S ∆G = -137 – (298 x -121)/1000 Divide by 1000 to convert to kJ ∆G = -149 – (-36) = -101 kJ mol-1

Evaluate the answer The answer is negative, which means the reaction is spontaneous at room

temperature

Main Menu

Method 2: Using standard ∆Gfo values

Is the reaction of ethene with hydrogen (to form ethane), spontaneous at room temperature (298K)? C2H4(g) +H2(g) C2H6(g)

∆Gfo (C2H4) = 68 kJ mol-1, ∆Gf

o (H2) = 0 kJ mol-1, ∆Gfo (C2H6) = -33 kJ mol-1

Construct a Hess CycleC2H4(g) + H2(g) C2H6(g)

2C(s) + 3H2(g)

∆G

∆G1 = 68 + 0 = 68

∆Go = ∑ ∆Gf(products) - ∑ ∆Gf(reactants)

= -68 + -33= -101 kJ mol-1

∆G2 = -33

Evaluate Answer The value is

negative so the reaction is spontaneous

Note this is the same as the previous answer, which is to be expected.

Main Menu

The Effect of Changing the Temperature

In the previous example: ∆G = -101 kJ mol-1, ∆Ho = -137 kJ mol-1, ∆So = -121 J K-1 mol-1, T = 298 K

∆G = ∆H - T∆S

What happens if we increase the temperature? Since ∆H and both ∆S are both negative, ∆G can be either

positive or negative depending on the temperature...

A high temperature (in this case 1132 K) will make the ‘T∆S’ term large enough to counter-balance the ∆H term. This will mean that the reaction is not spontaneous.

At any temperature below 1132 K, the reaction will be spontaneous This does not mean that the reaction will be fast, just that it will (in

theory) happen

Main Menu

Why 1132K? This slide just explains the maths behind working out when the

reaction becomes spontaneous. This is interesting but not needed!

The transition from spontaneous to non-spontaneous happens at ∆G = 0, so if we use this we can determine the temperature necessary for this as follows: ∆G = ∆H - T∆S The initial equation 0 = ∆H - T∆S Set ∆G = 0 T∆S = ∆H Rearrange to make T subject T = ∆H/∆S T = -137 / (-121/1000) Sub in values T = 1132 K Evaluate

∆G = -101 kJ mol-1, ∆Ho = -137 kJ mol-1, ∆So = -121 J K-1 mol-1, T = 298 K

Main Menu

Main Menu

Main Menu

Solutions

Main Menu

Main Menu

Main Menu

Solutions

Main Menu

Main Menu

Practice Questions Using data from the data booklet, calculate

∆Go for the following reactions using both methods, and comment on whether the reaction is spontaneous. Assume 298 K. Note: in some cases you will first need to calculate

∆Ho and ∆So. Extension: for non-spontaneous reactions, find the

minimum temperature necessary to make them spontaneous, and for spontaneous reactions, find the maximum temperature where they are non-spontaneous

1. S(s) + O2(g) SO2(g) – see next slide for example

2. C5H12(l) CH4(g) + 2C2H4(g)

3. CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

Substance So J K-1 mol-1

SO2(g) 248

O2(g) 205

HCl(g) 187

Cl2(g) 223

S(g) 32

Substance ∆Hof kJ mol-1

HCl(g) -92

Substance ∆Gof kJ mol-1

SO2(g) -300

Main Menu

S(s) + O2(g) SO2(g) Method 1:

H = -297 kJ mol-1 This is in the data booklet as Hc for sulfur S = 248 – (32 + 205) = 11 J K-1 mol-1 Data is on previous slide ∆G = ∆H - T∆S

= -297 – (298 x 11)/1000

= -300 kJ mol-1

Method 2:

S(s) + O2(g) SO2(g)

S(s) + O2(g)

∆G

∆G1 = 0 + 0 = 0

∆Go = -0 + -300= -300 kJ mol-1

∆G2 = -300

At what temperature does it become non spontaneous?

This reaction is always spontaneous, since H is negative and S is positive.

Main Menu

Gibbs free energy and equilibrium

So far we have considered reactions in which it is assumed that all reactants are converted into products.

Many reactions do not go to completion, but instead reach equilibrium.

The extent of the reaction can be quantified by the ratio of concentrations: [products]/[reactants].

As ∆G becomes more negative, the reaction will favour the products. (More spontaneous reaction)

Main Menu

Summary of ∆G(reaction) and the extent of the reaction

Above 30 kJmol-1 = No reaction

Less than 30 kJmol-1 but more than -30 kJmol-1 = Partial reaction

Below -30 kJmol-1 = Complete reaction

Main Menu

Main Menu

Solutions

Main Menu

Main Menu

Summarising ∆G = ∆H - T∆S

The mathematics means that: Some reactions are spontaneous at all temperatures Some reactions are only spontaneous above certain temperatures Some reactions are only spontaneous below certain temperatures

∆S can be calculated using Hess cycles in much the same way as ∆G and ∆H