Entropy and Gibbs free energy
1. Spontaneous Change
A spontaneous change is one that occurs
without a continuous input of energy from
outside the system.
All chemical processes require energy (activation energy) to
take place, but once a spontaneous process has begun, no
further input of energy is needed.
2
A nonspontaneous change occurs only if the surroundings
continuously supply energy to the system.
2H2(g) + O2(g) → 2H2O(l)
2H2O(l)→2H2(g) + O2(g)
If a change is spontaneous in one direction, it will be
nonspontaneous in the reverse direction.
3
DH Does Not Predict Spontaneous Change
Ba(OH)2 ·8H2O(s) + 2NH4NO3(s) → Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
DHθ = +62.3 kJ
water
This reaction occurs spontaneously when the solids are mixed. The
reaction mixture absorbs heat from the surroundings so quickly that
the beaker freezes to a wet block.
4
反应总是在一定条件下进行的。要讨论反应的自发性，也要指明反应条件。如果没有
交代反应条件，只讲“某某反应不自发”“某某反应自发”，实际上是指在常温常压下进
行的反应。“一定条件”指“一定的温度一定压强”。
例如，反应 CaCO3== CaO + CO2↑ 在常温常压是非自发的，而在高温下则是自发发
生的。
“外界帮助”这句话中，指的是为反应体系提供能量。“帮助”并不是指提供反应发生的
温度和压强等反应条件，也不是点燃、使用催化剂等等。一定的温度、一定的压强
是反应所需要的条件。很多场合下,提高温度,往往是为了加速反应。氢气氧气化合
成水的反应在常温下也是自发的。常温下将H2与O2混合，不能观察到有明显现象，
在氢氧燃料电池中，反应在常温下就能进行。在钯粉催化下H2与O2的混合气体也能
快速生成水。点燃氢气氧气，能迅速化合成水，也只是加速了反应速率。
“外界帮助”指的是对体系做“有用功”。例如，水的分解在常温常压下是不自发的，但
通电，为体系提供电能，可以使水分解成H2和O2.
5
Q1: Which of the following changes are likely to be spontaneous?
i sugar dissolving in water
spontaneous (because there are more ways of spreading out the sugar
molecules when they dissolve)
ii the smell from an open bottle of aqueous ammonia diffusing throughout a room
spontaneous (because there are more ways of spreading out the ammonia
molecules when they mix with the air molecules)
iii water turning to ice at 10℃
not spontaneous (because ice forms at 0 ℃ and the energy transferred from the
surroundings at 10℃ gives the molecules enough energy to move semi-independently of
each other and so there are a greater number of ways of arranging the energy in the
liquid than in the solid)
iv ethanol vaporising at 20℃
spontaneous (because there is enough energy given to the molecules at 20 ℃for some
of them to escape, thus allowing more spreading out of molecules in the vapour than in
the liquid)
v
water mixing completely with cooking oil
not spontaneous (because the strength of the bonding between water molecules alone
is greater than the strength between water and oil molecules; so the oil molecules
cannot mix and spread between the water molecules)
vi limestone decomposing at room temperature
not spontaneous (because the temperature is not high enough to break bonds in the
limestone (calcium carbonate); the ions in the calcium carbonate remain ordered)
6
law of thermodynamics
First law of thermoldynamics
Energy can be transformed from one form to another, but can be
neither created nor destroyed.
The total energy of the universe is constant:
DEsys = -DEsurr or DEsys + DEsurr = DEuniv = 0
Second law of thermoldynamics
A reaction is spontaneous if entropy change for the universe
or total is positive.
7
DH=0
When the stopcock opens, the
number of microstates is 2n,
where n is the number of particles.
8
2. Entropy
Entropy
is a measure of the dispersal of energy at a specific temperature, either from
systerm to the surroundings or from the surroundings to the system.
Entropy can also be thought of as a measure of the randomness or disorder of
a system. The higher randomness or disorder, ther greater entropy of the
system.
9
2.1 Comparing entropy values
Entropy can also be thought of as a measure of the randomness or disorder of a system.
To make any comparison of entropy values fair, we must use standard conditions.
 a pressure of 105Pa
 a temperature of 298K
 each substance involved in the
reaction is in its normal physical
state(solid, liquid or gas) at
105Pa and 298K
10
①Entropy and phase
Sθ (J/K•mol)
H2O(gas)
188.8
H2O(liq)
69.9
H2O (s)
47.9
11
The entropy change accompanying the dissolution of a salt.
pure solid
MIX
pure liquid
solution
The entropy of a salt solution is usually greater than that of the solid and
of water, but it is affected by the organization of the water molecules
around each ion.
12
②Entropy and temperature
The entropy of a substance increases with temperature.
Increasing the temperature of a solid makes the molecules, atoms or ions vibrate more.
Increasing the temperature of a liquid or gas increases the entropy.
Because it increases the disorder of particles.
13
The increase in entropy during phase changes from solid to liquid to gas.
The irregular arrangement of
particles in the liquid which are
close together and rotating
changes to an irregular
arrangment of particles, which
are free to move around rapidly
because they are far apart from
each other.
The regulary arranged lattice of
particles close together in the solid
changes to an irregular arrangement of
particles, which are close together bur
rotate and slide over each other in liquid.
14
③Entropy and structures
For similar types of substances, harder substance have a lower entropy
value.
θ
θ
S of graphite is 5.69 J/mol·K, whereas S of diamond is 2.44 J/mol·K.
ion pairs
Sθ (J/K•mol)
MgO
Mg2+ / O2-
26.9
NaF
Na+ / F-
51.5
These trends only hold for substances in the same physical state.
15
For compound, entropy increases with chemical complexity.
Simpler substances with fewer atoms have lower entropy values than more
complex substances with a greater number of atoms.
16
Q2: Explain the difference in the entropy of each ot the
following pairs of substances in terms of their state and
structure.
i Br2(l) Sθ= 151.6 JK-1mol-1 and I2(s) Sθ= 116.8 JK-1mol-1
Bromine is a liquid, so its molecules are able to rotate/slide over each other so
there is more disorder (higher entropy) than in iodine, which is a solid, in which the
molecules cannot move.
ii H2(g) Sθ= 130.6 JK-1mol-1 and CH4(g) Sθ= 186.2 JK-1mol-1
Both are gases but methane, CH4, is a more complex molecule with two different
types of atom. Complex molecules tend tohave greater entropy values than simpler
molecules.
iii Hg(l) Sθ= 76.00 JK-1mol-1 and Na(s)
Sθ= 51.20 JK-1mol-1
Mercury is a liquid, so its atoms are able to rotate/slide over each other so there is
more disorder (higher entropy) than in sodium, which is a solid, in which the atoms
cannot move.
iv SO2(g) Sθ= 248.1 JK-1mol-1 and SO3(l)
Sθ= 95.60 JK-1mol-1
Although SO3 has only one oxygen atom more than SO2, which makes SO3 a slightly
more complex molecule, SO2 has a much greater entropy because it is a gas, whereas
SO3 is a liquid. The particles in gases move freely from place to place and so there is
more disorder are many more ways of spreading out the energy than in liquids whose
particles can only rotate/slide over each other and vibrate.
17
2.2 Entropy change in reactions
We assume that gases have high entropy and solids have low entropy.
If there is a change in the number of gaseous molecules in a reaction,
there is likely be a significant entropy change.
There are greater number of molecules in products than in
reactants. In addition, there are two different types of molecules,
but only one type of reactant. This is also contributes to greater
disorder in products.
18
Q3: Suggest whether the entropy of the reactants or the
products will be greater or whether it is difficult to decide.
a The entropy of the reactants is greater.
Two gases (high entropy and two
molecules) are being converted to a solid
(low entropy and one molecule).
b Difficult to decide. The product SO2 is a
gas, which is likely to have a higher
entropy than oxygen because it is a more
complex molecule. However, there are
two molecules on the left and only one on
the right. The entropy of the solid sulfur,
however, is low.
cThe entropy of the reactants is greater. One
d The entropy of the reactants is greater. One
of the reactants, chlorine, is a gas, so has
a very high entropy. This outweighs the
entropies of the single product which is a
solid. Solids have low entropies.
of the reactants, carbon dioxide, is a gas, so
has a very high entropy. This outweighs the
entropies of the products which are both
solids. Solids have low entropies.
19
Q3: Suggest whether the entropy of the reactants or the
products will be greater or whether it is difficult to decide.
e
The entropy of the products is greater. Two gases are produced (2 moles of gases) both of which
have high entropy. The reactants have only one gas (1 mole of gas). The entropies of the solids are
much smaller than those of the gases and so can be ignored.
f Difficult to decide. There are equal numbers of moles of gases on each side of the equation. We
would have to know the individual values of the entropies of each molecule.
g The entropy of the products is greater.Reactants: potassium has low entropy
because it is a solid; water has medium entropy because it is a liquid.Products: aqueous potassium
hydroxide has medium entropy (not only is it a liquid, but potassium ions and hydroxide ions are
spread out randomly in the solution); hydrogen has high entropy because it is a gas.
20
3. Calculating entropy changes
3.1 Calculating DS for a Reaction/System
△Sθ = S Sθ (products) - S Sθ (reactants)
Note that:
1. We need to take account of the stoichiometry of the equation.
2. When looking up entropy values in tables of data, we need to choose the data
for the correct state.
21
22
Q4: Calculate the standard entropy change of the system in each
of the following reactions using the standard molar entropy
values given here:
(Values for Sθ in J K-1mol-1:
Cl2(g)=165.0
H2O(l)=69.90
Na(s)=51.20
a
Fe(s)= 27.30
H2O2(l) =109.6
NH4NO3(s)=151.1
Fe2O3(s)=87.40
Mg(s)=32.70
N2O(g)=219.7
H2(g)=130.6
MgO(s)=26.90
O2(g)=205.0 )
H2O2(l) → 2H2O(l)+O2(g)
b NH4NO3(S) → N2O(g) + 2H2O(g)
c
2Mg(s) +O2(g) → 2MgO(s)
d 2Na(s) + Cl2(g) → 2NaCl(s)
e
3Mg(s) + Fe2O3(s) → 3MgO(s) + 2Fe(s)
23
Q4:
If Entropy DECREASES,
Why is this a SPONTANEOUS REACTION??
—— Second law of thermoldynamics
24
Second law of thermoldynamics
A reaction is spontaneous if entropy change for the universe
or total is positive.
DSuniverse
=
DSsystem
+ DSsurroundings
DSuniverse
> 0 for spontaneous process
First calc. entropy created by matter dispersal (DSsystem)
Next, calc. entropy created by energy dispersal (DSsurround)
25
3.2 Calculating DS for the surroundings
Since entropy depends on temperature, DS°surr is also
affected by the temperature at which heat is transferred.
For any reaction, qsystem = -qsurroundings
The impact on the surroundings is larger when the
surroundings are at lower temperature, because there is
a greater relative change in Ssurroundings
26
Where
 △Hrθ is the standard enthalpy change of the reaction.
 T is the temperature in Kelvin. At standard temperature, the value
is 298K.
Note:
 When performing calcualtions to find △Sθ sorroundings the value of
△Hrθ in kJ/mol should be multiplied by 1000. This is because
entropy changes are measured in units of joules per kelvin per
mole.
 The negative sign in front of △Hrθ is part of equation and not the
sign of the enthalpy change.
27
28
29
Q5: Calculate the entropy change of the surroundings in each of
the following reactions. Assume that the value of △H does not
change with temperature.
a
C(s) + O2(g) → CO2(g)
carried out at 0℃
△Hθreaction = -393.5kJ/mol
b
2C(s) + N2(g) → C2N2(g)
carried out at 300℃
△Hθreaction = +307.9kJ/mol
c
H2(g) + F2(g) → 2HF(g)
△Hθreaction = -271.1kJ/mol
carried out at standard temperature
d
Si(s) + 2H2(g) → SiH4(g)
carried out at -3℃
△Hθreaction = +34.30kJ/mol
30
3.3 Calculating DS for the total
31
Q6: Calculate the the total standard entropy change in each of
the following reactions using the standard molar entropy values
given here.
( Values for Sθ in J K-1mol-1:
C(graphite)=5.700, C2N2(g)=242.1,
H2(g)=130.6,
H2O(l)=69.90,
O2(g)=205.0,
P(s)=41.10,
C3H8(g)=269.9, CO2(g)=213.6,
H2S(g)=205.7,
N2(g)=191.6,
P4O10(s)=228.0, S(s)=31.80)
a S(s) + H2(g) → H2S(g)
△Hθreaction = -20.6kJ/mol
b 2C(graphite) + N2(g) → C2N2(g)
△Hθreaction = +307.9kJ/mol
c 4P(s) + 5O2(g) → P4O10(s)
△Hθreaction = -2984.0kJ/mol
d C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) △Hθreaction = -2219.2kJ/mol
32
3.4 Entropy and temperature
Determining Reaction Spontaneity
Q7:
At 298 K, the formation of ammonia has a negative DS θ sys;
N2(g) + 3H2(g) → 2NH3(g); DSθ sys = -197 J/K
Calculate DS θ total, and state whether the reaction occurs spontaneously at
this temperature.
-91.8 kJ x
DSsurr = -
DHsys
T
1000 J
1 kJ
=-
= 308 J/K
298 K
DSθtotal = DSθsys + DSθsurr = -197 J/K + 308 J/K = +111 J/K
Since DSθtotal > 0, the reaction occurs spontaneously at 298 K.
Q8: Is this reaction occurs spontaneously at 600K?
33
Q9: The decomposition of calcium carbonate ,
CaCO3(s) → CaO(s) + CO2(g), does not take place at room temperature.
a Explain in terms of entropy changes why heating the calcium carbonate to
a high temperature increases the likelihood of this reaction taking place.
b In a closed system at high temperature, the reactants and products are in
equilibrium.
CaCO3(s) ⇌ CaO(s) + CO2(g)
i Explain the meaning of the term closed system.
ii Explain in terms of entropy changes what happens when the pressure on
this system is increased.
iii What is the value of the standard total entropy change at equilibrium?
34
4.1 Gibbs Free energy
DSuniv = DSsurr + DSsys
DSuniv =
 DH sys
T
+ DSsys
-TDSuniv =△G= DHsys - TDSsys
DG =
θ
θ
DH
reaction
-
θ
TDS
system
 a pressure of 105Pa
 a temperature of 298K
 each substance involved in the reaction is in its normal
physical state(solid, liquid or gas) at 105Pa and 298K
35
4.2 Calculating free energy
NOTE:
 the entropy chagne of the system in J K-1 mol-1
 the enthalpy change of the system in J mol-1 , we have to multiply the value
of the enthalpy change by 1000 because the entropy chang is in joules per
kelvin per mol
 the temperature; under standard conditions, this is 298K
Q10:
Calculate the Gibbs free energy change for the decomposition of zinc carbonate
at 298K.
ZnCO3(s) → ZnO(s) + CO2(g) △Hθr=+71.0kJ mol-1
(Values for Sθ in J K-1mol-1:
CO2(g)=+213.6, ZnCO3(s)=+82.4, ZnO(s)=+43.6)
36
Q11:
Calculate the standard Gibbs free energy of reaction in each of the following using
the standard molar entropy values given. Express your answers to 3 significant
figures in kJ mol-1, and in each case state whether the reaction is spontaneous or
not under standard conditions.
(Values for S in J K-1 mol-1: Ag2CO3(s)=167.4, Ag2O(s)=121.3, CH4(g)=186.2,
Cl2(g)=165, CO2(g)=213.6, H2(g)=130.6, HCl(g)=186.8, H2O(l)=69.9, Mg(s)=37.2,
MgCl2(s)=89.6, Na(s)=51.2, Na2O2(s)=95.0, O2(g)=205.0)
a H2(g) + Cl2(g) → 2HCl(g)
△Hθr= -184.6kJ mol-1
b CH4(g) + 2O2(g) → CO2(g) + 2HCl(l)
△Hθr= -890.3kJ mol-1
c 2Na(s) + O2(g) → Na2O2(s)
△Hθr= -510.9kJ mol-1
d Mg(s) + Cl2(g) → MgCl2(s)
△Hθr= -641.3kJ mol-1
e Ag2CO3(s) → Ag2O(s) + CO2(g)
△Hθr=+167.5kJ mol-1
37
4.3 standard Gibbs free energy
change of reaction
△Gθf
is the Gibbs free energy change whenthe amounts of the reactants shown in the
stiochiometric equation react under standard conditions to give products.
 △Gθf of an element is zero.
 Many compounds in the solid state have high negative values of Gibbs free energy
change of formation.
 Many gases and liquids have positive Gibbs free energy change of formation.
38
Q12: Draw a Gibbs free energy cycle to calculate the standard Gibbs free energy
change of decomposition of sodium hydrogencarbonate.
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)
The relevant Gibbs free energy values are:
△Gθf [NaHCO3(s)] =-851.0kJmol-1
△Gθf [Na2CO3(s)] =-1044.5kJmol-1
△Gθf [CO2(g)] =-394.4kJmol-1
△Gθf [H2O(l)] =-237.2kJmol-1
39
4.4 Free energy and spontaneous reactions
DG= DHreaction - T DSsys
DG < 0 for a spontaneous process
DG > 0 for a nonspontaneous process
DG = 0 for a process at equilibrium
DH
DS
-TDS
DG
–
+
–
–
Spontaneous at all T
+
–
+
+
Nonspontaneous at all T
+
+
–
+ or –
Spontaneous at higher T;
Description
nonspontaneous at lower T
–
–
+
+ or –
Spontaneous at lower T;
nonspontaneous at higher T
40
Q13:
The following scenes represent a familiar phase change for water.
(a) What are the signs of DH and DS for this process? Explain.
(b) Is the process spontaneous at all T, no T, low T, or high T? Explain.
41
Determining the Effect of Temperature on ΔG
Q14:
A key step in the production of sulfuric acid is the oxidation of SO2(g)
to SO3(g):
2SO2(g) + O2(g) → 2SO3(g)
At 298 K, DG = -141.6 kJ; DH = -198.4 kJ; and DS = -187.9 J/K
(a) Use the data to decide if this reaction is spontaneous at 25℃, and
predict how DG will change with increasing T.
(b) Assuming DH and DS are constant with increasing T, is the
reaction spontaneous at 900 ℃?
42
thermal stability of Group 2 carbonates and nitrates
XCO3 →XO + CO2
2X(NO3)2 →2XO + 4NO2 + O2
The temperature at which thermal decomposition takes place increases going
down Group 2.
43
5. DG, Equilibrium, and Reaction
Direction
 K is related to reaction favorability.
 If △G < 0, reaction is product-favored.
 But systems often reach a state of equilibrium in which
reactants have not converted completely to products.
44
DGθ = - RT lnK
45
 When a system is chemical equilibrium and the amounts of
products and reactants balance, the value of
46
47
48