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
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2CH3OH(g) + H2(g) → C2H6(g) + 2H2O(g)
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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.
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To Do:
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Categorise the following as either spontaneous or non-spontaneous:
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Cooking an egg, acid reacting with metal, the reaction in a battery, electrolysis, respiration,
photosynthesis
Think of an example of each of the following:
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A spontaneous exothermic reaction
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A non-spontaneous endothermic reaction
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A spontaneous endothermic reaction
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A non-spontaneous exothermic reaction (difficult!)
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Gibbs free energy is a useful accounting tool
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For chemical reactions neither ∆H(system) nor ∆S(system) alone can reliably be used
to predict the feasibility of a reaction.
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To calculate feasibility we use this expression:
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Tidying up and moving T to the other side:
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Multiplying by -1 and reversing the inequality:
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This gives the function known as Gibbs free energy (∆G(system)):
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∆G(system) must be negative for a spontaneous process. Units: kJ or J mol-1
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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)
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Using ∆G(system) to predict the feasibility
of a change
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We generally assume that both the enthalpy and entropy changes of the
system do not change with temperature.
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Temperature, T, is alike a tap which adjusts the significance of the term
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At low temperature:
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This shows that all exothermic reactions can occur at low
temperatures.
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At high temperature:
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This means all reactions which have a positive value of ∆S(system)
can be feasible at high temperatures even if they are endothermic.
∆S(system) in determining the value of ∆G(system).
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What does the Gibb’s free energy value tell us about a
reaction?
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.
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Solutions
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Summarising the Effect Of ∆H, ∆S and T on Spontaneity
of Reaction
∆G = ∆H - T∆S
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Task: Use the equation to help yourself reason these through
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Spontaneity
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We can calculate Gibbs free energy in two ways:
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Method 1:
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Directly from (appropriate) data using ∆G = ∆H - T∆S < 0
Method 2:
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Indirectly from standard ∆Gfo values in the data booklet, using a Hess Cycle
Note: Similar to ∆Hfo values, ∆Gfo for any element in its standard state is zero.
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Method 1: From ∆G = ∆H - T∆S
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Is the reaction of ethene with hydrogen (to form ethane), spontaneous at
room temperature (298K)?
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C2H4(g) +H2(g)  C2H6(g)
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Calculate Gibbs Free Energy
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∆Ho = -137 kJ mol-1
∆So = -121 J K-1 mol-1
∆G = ∆H - T∆S
∆G = -137 – (298 x -121)/1000
∆G = -149 – (-36) = -101 kJ mol-1
Divide by 1000 to convert to kJ
Evaluate the answer
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The answer is negative, which means the reaction is spontaneous at room temperature
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Method 2: Using standard ∆Gfo values
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Is the reaction of ethene with hydrogen (to form ethane), spontaneous at room temperature (298K)?
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C2H4(g) +H2(g)  C2H6(g)
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∆Gfo (C2H4) = 68 kJ mol-1, ∆Gfo (H2) = 0 kJ mol-1, ∆Gfo (C2H6) = -33 kJ mol-1
Construct a Hess Cycle
C2H4(g) + H2(g)
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∆G
C2H6(g)
∆G1 = 68 + 0 = 68
∆G2 = -33
2C(s) + 3H2(g)
∆Go = ∑ ∆Gf(products) - ∑ ∆Gf(reactants)
= -68 + -33
= -101 kJ mol-1
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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.
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...
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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
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Why 1132K?
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This slide just explains the maths behind working out when the reaction becomes
spontaneous.
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
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:
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∆G = ∆H - T∆S
0 = ∆H - T∆S
T∆S = ∆H
T = ∆H/∆S
T = -137 / (-121/1000)
T = 1132 K
The initial equation
Set ∆G = 0
Rearrange to make T subject
Sub in values
Evaluate
∆G = -101 kJ mol-1,
-1, ∆So = -121 J K-1 mol-1, T = 298 K
∆Ho = -137
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kJ mol
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Solutions
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Solutions
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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.
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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 nonspontaneous
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)
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Substance
∆Hof kJ mol-1
HCl(g)
-92
Substance
So J K-1 mol-1
SO2(g)
248
O2(g)
205
HCl(g)
187
Cl2(g)
223
S(g)
32
Substance
∆Gof kJ mol-1
SO2(g)
-300
S(s) + O2(g)  SO2(g)
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Method 1:
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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:
∆G
S(s) + O2(g)
∆G1 = 0 + 0 = 0
SO2(g)
∆G2 = -300
S(s) + O2(g)
∆Go = -0 + -300
= -300 kJ mol-1
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At what temperature
does it become non
spontaneous?
This reaction is always
spontaneous, since H is
negative and S is positive.
Gibbs free energy and equilibrium
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So far we have considered reactions in which it is assumed that all reactants are converted
into products.
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Many reactions do not go to completion, but instead reach equilibrium.
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The extent of the reaction can be quantified by the ratio of concentrations:
[products]/[reactants].
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As ∆G becomes more negative, the reaction will favour the products. (More spontaneous
reaction)
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Summary of ∆G(reaction) and the extent
of the reaction
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Less than 30 kJmol-1
but more than -30
kJmol-1 = Partial
reaction
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Above 30 kJmol-1
= No reaction
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Below -30 kJmol-1
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= Complete reaction
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Solutions
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Summarising
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∆G = ∆H - T∆S
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The mathematics means that:
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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
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