Consider the reaction:
NH4NO3(s)
N2(g) + 2 H2O(g) + ½ O2(g)
Is ΔS0 expected to be positive or negative? Why?
ΔS0 = S0(N2) + 2 S0(H2O) + ½ S0(O2) – S0(NH4NO3)
= “big” + 2 x “big” + ½ x “big” – “small”
>0
Consider the reaction: 2 NO(g) + O2(g)
2 NO2(g)
Is ΔS0 expected to be positive or negative? Why?
ΔS0 = 2 S0(NO2) – 2 S0(NO) – S0(O2)
= 2 x “big” – 2 x “big” – “big”
< 0 (mixing of gases NO and O2)
1381
Gibbs Energy
1382
Gibbs Energy
To address the key question: Which reactions (or
processes) proceed spontaneously, we need a new
thermodynamic variable called the Gibbs energy or
sometimes called the Gibbs free energy.
1383
Gibbs Energy
To address the key question: Which reactions (or
processes) proceed spontaneously, we need a new
thermodynamic variable called the Gibbs energy or
sometimes called the Gibbs free energy.
The Gibbs energy combines two effects: changes in
ΔH and changes in ΔS. This covers both energy
changes and order-disorder changes.
1384
Gibbs defined the Gibbs energy, G, as:
G = H-TS
1385
Gibbs defined the Gibbs energy, G, as:
G = H-TS
This particular definition arises from a study of
entropy changes in both a system and the
surroundings (and requires some differential
calculus to fully appreciate what is behind the
definition).
1386
Gibbs defined the Gibbs energy, G, as:
G = H-TS
This particular definition arises from a study of
entropy changes in both a system and the
surroundings (and requires some differential
calculus to fully appreciate what is behind the
definition).
The Gibbs energy has units of energy, i.e. J, or when
specified for a mole of material, J mol-1.
1387
The Gibbs energy is the most important quantity in
this class, and is probably one of the three most
important quantities in Chemistry.
1388
The Gibbs energy is the most important quantity in
this class, and is probably one of the three most
important quantities in Chemistry.
Knowledge of the Gibbs energy change for a
reaction or process will answer the question: Will
the reaction or process occur spontaneously.
1389
From the definition of G, we obtain the following
result for a change in G when the temperature and
pressure are held constant:
ΔG  ΔH - T ΔS
1390
From the definition of G, we obtain the following
result for a change in G when the temperature and
pressure are held constant:
ΔG  ΔH - T ΔS
This will be a key equation from the point of view
of practical calculations, and is one of the most
important equations in the course.
1391
From the definition of G, we obtain the following
result for a change in G when the temperature and
pressure are held constant:
ΔG  ΔH - T ΔS
This will be a key equation from the point of view
of practical calculations, and is one of the most
important equations in the course.
Gibbs showed that this equation could be
employed to determine which reactions are
spontaneous.
1392
A change can only be spontaneous if it is
accompanied by a decrease in the Gibbs
energy, i.e.
ΔG 0
1393
Four general cases to consider
We can divide reactions (processes in general) into
four different groups. In two cases we can arrive at
a definitive answer as to whether the process is
spontaneous or not. For two of the cases, we can
add a qualifier to decide whether a reaction will be
spontaneous or not.
1394
Case 1. The reaction (or process) is exothermic and
the system becomes more disordered, i.e. the
entropy increases.
1395
Case 1. The reaction (or process) is exothermic and
the system becomes more disordered, i.e. the
entropy increases.
So ΔH  0 and ΔS  0
1396
Case 1. The reaction (or process) is exothermic and
the system becomes more disordered, i.e. the
entropy increases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (negative) – (positive)(positive)
therefore ΔG  0
1397
Case 1. The reaction (or process) is exothermic and
the system becomes more disordered, i.e. the
entropy increases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (negative) – (positive)(positive)
therefore ΔG  0
Conclusion: The reaction or process is spontaneous.
1398
Case 1. The reaction (or process) is exothermic and
the system becomes more disordered, i.e. the
entropy increases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (negative) – (positive)(positive)
therefore ΔG  0
Conclusion: The reaction or process is spontaneous.
Note that it does not matter what the temperature
is.
1399
Case 2. The reaction (or process) is endothermic
and the system becomes less disordered, i.e. the
entropy decreases.
1400
Case 2. The reaction (or process) is endothermic
and the system becomes less disordered, i.e. the
entropy decreases.
So ΔH  0 and ΔS  0
1401
Case 2. The reaction (or process) is endothermic
and the system becomes less disordered, i.e. the
entropy decreases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (positive) – (positive)(negative)
therefore ΔG  0
1402
Case 2. The reaction (or process) is endothermic
and the system becomes less disordered, i.e. the
entropy decreases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (positive) – (positive)(negative)
therefore ΔG  0
Conclusion: The reaction or process is NOT
spontaneous.
1403
Case 2. The reaction (or process) is endothermic
and the system becomes less disordered, i.e. the
entropy decreases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (positive) – (positive)(negative)
therefore ΔG  0
Conclusion: The reaction or process is NOT
spontaneous.
Note that it does not matter what the temperature
is.
1404
Case 3. The reaction (or process) is endothermic
and the system becomes more disordered, i.e. the
entropy increases.
1405
Case 3. The reaction (or process) is endothermic
and the system becomes more disordered, i.e. the
entropy increases.
So ΔH  0 and ΔS  0
1406
Case 3. The reaction (or process) is endothermic
and the system becomes more disordered, i.e. the
entropy increases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (positive) – (positive)(positive)
therefore ΔG  0 or ΔG  0 depending on T.
1407
Case 3. The reaction (or process) is endothermic
and the system becomes more disordered, i.e. the
entropy increases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (positive) – (positive)(positive)
therefore ΔG  0 or ΔG  0 depending on T.
Conclusion: The reaction or process could be
spontaneous depending on the value of T. The
reaction will be spontaneous only when T is high.
1408
Cu2O(s) + C(s)
2 Cu(s) + CO(g) ΔG375K  -3.8kJ
1409
Case 4. The reaction (or process) is exothermic and
the system becomes less disordered, i.e. the
entropy decreases.
1410
Case 4. The reaction (or process) is exothermic and
the system becomes less disordered, i.e. the
entropy decreases.
So ΔH  0 and ΔS  0
1411
Case 4. The reaction (or process) is exothermic and
the system becomes less disordered, i.e. the
entropy decreases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (negative) – (positive)(negative)
therefore ΔG  0 or ΔG  0 depending on T.
1412
Case 4. The reaction (or process) is exothermic and
the system becomes less disordered, i.e. the
entropy decreases.
So ΔH  0 and ΔS  0

T
ΔS
Hence from ΔG  ΔH
= (negative) – (positive)(negative)
therefore ΔG  0 or ΔG  0 depending on T.
Conclusion: The reaction or process could be
spontaneous depending on the value of T. The
reaction will be spontaneous only when T is low.
1413
Standard Gibbs Energies
When calculating Gibbs energy changes for chemical
reactions, it is most common to work with
tabulated values.
1414
Standard Gibbs Energies
When calculating Gibbs energy changes for chemical
reactions, it is most common to work with
tabulated values. Gibbs energies tabulated at 1 bar
are referred to as standard Gibbs energies, and the
symbol ΔG0 is employed.
1415
Standard Gibbs Energies
When calculating Gibbs energy changes for chemical
reactions, it is most common to work with
tabulated values. Gibbs energies tabulated at 1 bar
are referred to as standard Gibbs energies, and the
symbol ΔG0 is employed. A reference temperature
of 25 OC is employed, since the Gibbs energy
depends on the particular temperature.
1416
Note that it is impossible to calculate absolute
values of G0 (the same situation as for the enthalpy,
but different than the entropy).
1417
Note that it is impossible to calculate absolute
values of G0 (the same situation as for the enthalpy,
but different than the entropy).
Standard Gibbs energies can be calculated from
ΔG0  ΔH0 - T ΔS 0
1418
Note that it is impossible to calculate absolute
values of G0 (the same situation as for the enthalpy,
but different than the entropy).
Standard Gibbs energies can be calculated from
ΔG0  ΔH0 - T ΔS 0
For a chemical reaction the change in the Gibbs
energy can be calculated from
1419
Note that it is impossible to calculate absolute
values of G0 (the same situation as for the enthalpy,
but different than the entropy).
Standard Gibbs energies can be calculated from
ΔG0  ΔH0 - T ΔS 0
For a chemical reaction the change in the Gibbs
energy can be calculated from
ΔG0  Σ n ΔG0(products)  Σ n ΔG0(reactants)
1420
Example: For the reaction:
N2(g) + 3 H2(g) 
ΔH0 = – 92.22 kJ.
 2 NH3(g)
Determine ΔG0 and the direction for spontaneous
reaction at 25 oC.
1421
Example: For the reaction:
N2(g) + 3 H2(g) 
ΔH0 = – 92.22 kJ.
 2 NH3(g)
Determine ΔG0 and the direction for spontaneous
reaction at 25 oC. The following data are available
from standard tables (at 25 oC ):
S0(NH3) = 192.45 J K-1mol-1
S0(H2) = 130.684 J K-1mol-1
S0(N2) = 191.61 J K-1mol-1
1422
Example: For the reaction:
N2(g) + 3 H2(g) 
ΔH0 = – 92.22 kJ.
 2 NH3(g)
Determine ΔG0 and the direction for spontaneous
reaction at 25 oC. The following data are available
from standard tables (at 25 oC ):
S0(NH3) = 192.45 J K-1mol-1
S0(H2) = 130.684 J K-1mol-1
S0(N2) = 191.61 J K-1mol-1
First calculate ΔS0 for the reaction. Use
1423
Example: For the reaction:
N2(g) + 3 H2(g) 
ΔH0 = – 92.22 kJ.
 2 NH3(g)
Determine ΔG0 and the direction for spontaneous
reaction at 25 oC. The following data are available
from standard tables (at 25 oC ):
S0(NH3) = 192.45 J K-1mol-1
S0(H2) = 130.684 J K-1mol-1
S0(N2) = 191.61 J K-1mol-1
First calculate ΔS0 for the reaction. Use
ΔS0  Σ n S0(products)  Σ n S0(reactants)
1424
ΔS0 =
2 mol x 192.45 J K-1mol-1
– 1 mol x 191.61 J K-1mol-1
– 3 mol x 130.684 J K-1mol-1
1425
2 mol x 192.45 J K-1mol-1
– 1 mol x 191.61 J K-1mol-1
– 3 mol x 130.684 J K-1mol-1
= – 198.76 J K-1
= – 0.19876 kJ K-1
ΔS0 =
1426
2 mol x 192.45 J K-1mol-1
– 1 mol x 191.61 J K-1mol-1
– 3 mol x 130.684 J K-1mol-1
= – 198.76 J K-1
= – 0.19876 kJ K-1
ΔS0 =
Now use
ΔG0  ΔH0 - T ΔS 0
1427
2 mol x 192.45 J K-1mol-1
– 1 mol x 191.61 J K-1mol-1
– 3 mol x 130.684 J K-1mol-1
= – 198.76 J K-1
= – 0.19876 kJ K-1
ΔS0 =
Now use
ΔG0  ΔH0 - T ΔS 0
= – 92.22 kJ – (298.15 K)(– 0.19876 kJ K-1)
= – 92.22 kJ + 59.260 kJ
= – 32. 96 kJ
1428
2 mol x 192.45 J K-1mol-1
– 1 mol x 191.61 J K-1mol-1
– 3 mol x 130.684 J K-1mol-1
= – 198.76 J K-1
= – 0.19876 kJ K-1
ΔS0 =
Now use
ΔG0  ΔH0 - T ΔS 0
= – 92.22 kJ – (298.15 K)(– 0.19876 kJ K-1)
= – 92.22 kJ + 59.260 kJ
= – 32. 96 kJ
Since ΔG0  0 the reaction is spontaneous from left
to right at 25 oC.
1429
Standard Gibbs Energy of Formation
1430
Standard Gibbs Energy of Formation
Standard Gibbs Energy of Formation: The Gibbs
energy for the formation of 1 mol of a substance
from its component elements in their standard
states.
1431
Standard Gibbs Energy of Formation
Standard Gibbs Energy of Formation: The Gibbs
energy for the formation of 1 mol of a substance
from its component elements in their standard
states.
Symbol: ΔGf0
1432
Standard Gibbs Energy of Formation
Standard Gibbs Energy of Formation: The Gibbs
energy for the formation of 1 mol of a substance
from its component elements in their standard
states.
Symbol: ΔGf0
This quantity can be calculated in the same manner
as employed for ΔHf0 .
1433
Standard Gibbs Energy of Formation
Standard Gibbs Energy of Formation: The Gibbs
energy for the formation of 1 mol of a substance
from its component elements in their standard
states.
Symbol: ΔGf0
This quantity can be calculated in the same manner
as employed for ΔHf0 .
The standard Gibbs energies of formation of the
elements in their standard reference states are
zero by definition.
1434
The Gibbs Energy and maximum
Non-expansion Work
1435
The Gibbs Energy and maximum
Non-expansion Work
Reversible Process: A process that is essentially at
equilibrium at each stage of the path taken from
the initial to the final state.
1436
The Gibbs Energy and maximum
Non-expansion Work
Reversible Process: A process that is essentially at
equilibrium at each stage of the path taken from
the initial to the final state. At each stage, an
infinitesimal modification in a system variable can
reverse the direction of the process.
1437
The Gibbs Energy and maximum
Non-expansion Work
Reversible Process: A process that is essentially at
equilibrium at each stage of the path taken from
the initial to the final state. At each stage, an
infinitesimal modification in a system variable can
reverse the direction of the process.
In different words: A process is reversible if its
driving force is opposed by another force that is just
infinitesimally weaker.
1438
Example: Compression of a gas in a piston by the
gradual addition of very tiny masses.
An irreversible process is one that occurs rapidly
and does not resemble an equilibrium state.
1439
1440