Chapter 16: Spontaneity, Entropy, and Free Energy
16.1 Spontaneous Processes and Entropy
* First law of thermodynamics: energy can be neither created nor destroyed; the energy of the universe is
constant.
* Spontaneous process can be either fast or slow but it must occur without outside intervention.
*Entropy: a thermodynamic function that measures randomness or disorder, the larger the entropy value (S)
the more disorder is present.
*Positional probability: a type of probability that depends on the number of arrangements in space that
yield a particular state.
If it has a high positional probability, this indicates that the substance is a gas while if it has a low
positional probability, it must be a solid. If in between, the substance is in the form of a liquid.
*S solid < S liquid < S gas (entropy)
16.2 Entro, y and the Second Law of Thermodynamics
*Second law of thermodynamics: In any spontaneous process there is always an increases in the entropy of
the universe. (in other words, the entropy of the universe is increasing)
∆S universe = ∆S system + ∆S surrounding
∆S system and ∆S surrounding represent the changes in entropy that occur (in the system and surrounding)
If ∆S universe was….
(+) = the entropy of the universe is increases and its spontaneous in the direction written
( - ) = the process is spontaneous, the process is spontaneous in the opposite written
( 0 ) = the process does not occur and the system is at equilibrium
16.3 The Effect of Temperature on Spontaneity
*Effect of Temperature: The impact of the transfer of a given quantity of energy as heat to or from the
surroundings will be greater at lower temperatures.
*Two characteristics of the changes of S in the surrounding:
1. ∆S surrounding depends on the direction of the heat flow.
Constant temperature: Exothermic reaction occurs in the system causing more amounts of random
motion so the ∆S surrounding is positive. If it was an endothermic reaction, this will cause less
random motion since heat is being absorbed, causing the ∆S surrounding would be negative.
2. The magnitude of ∆S surrounding depends on the temperature.
Driving force provided by the energy flow (heat) = magnitude of the entropy change of the
surroundings = quantity of heat (J)/ temperature (K)
 Exothermic process: ∆S surrounding = + quantity of heat (J)/ temperature (K)
 Endothermic process: ∆S surrounding = - quantity of heat (J)/ temperature (K)
16.4 Free Energy
Free energy: a thermodynamic function equal to the enthalpy (H) minus the product of the entropy (S) and
the Kelivn temperature. Under certain conditions the change in free energy for a process is equal to the
maximum useful work.
G = H – TS  G= free energy H = enthalpy T= temperature S= entropy
*Eventually, the formula above can be derived into the formula below;
∆S universe = - ∆G/ T at constant T and P
*This formula shows that a process can only be carried out with constant temperature and pressure and be
spontaneous if ∆G is negative.
Possible Combinations of ∆H and ∆S
Case
Result
∆S positive, ∆H negative
Spontaneous at all temperatures
∆S positive, ∆H positive
Spontaneous at high temperatures
∆S negative, ∆H negative
Spontaneous at low temperatures
∆S negative, ∆H positive
Process not spontaneous, temperature does not matter
16.5 Entropy Changes in Chemical Reactions
*In order to determine the changes in entropy by the positional probability.
*Again, the smaller number of molecules, the smaller the number of possibilities that it can be arranged
differently which leads to a lower value of positional probability.
In a chemical reaction:
4 mol gas + 5 mole gas  4 mol gas + 6 mol gas
In this generic reaction, 9 gas molecules reacts to yield 10 molecules of gas, when the molecules of product
is greater than the reactants, the positional entropy increases.
*Third law of thermodynamics: the entropy of a perfect crystal at 0 K is zero
*∆So reaction = ∑npSo products - ∑nrSo reactants
∑ represents the sum of the terms and “n” represents the moles of products or reactants
*Also, the more complex the molecule, the higher the standard entropy value. This is due to its atomic
structure. Ex. H20 vs. H2, one is triatomic while the other is diatomic respectively. H20’s structure must be
3-D, which allows it more rotational and vibrational motions while H2’s structure is linear, limiting its
motions.
16.6 Free Energy and Chemical Reactions
*Standard free energy change: the change in the free energy that will occur for one unit of reaction if the
reactants in their standard states are converted to products in their standard states.
1st way of solving ∆G for a reaction: ∆Go = ∆Ho - T∆So
2nd way of solving ∆G for a reaction: add the products and reactants value of 1 st ∆Go if given,
you can manipulate the ∆Go just like enthalpy.
*Standard free energy of formation: the change in free energy that accompanies the formation of one mole
of a substance form its constituents elements with all reactants and products in their standard states.
*∆Go reaction = ∑npGo products - ∑nrGo reactants
*Note that the standard free energy of formation of an element in its standard stat is zero!
16.7 The Dependence of Free Energy on Pressure
*Entropy depends on pressure because it also depends on volume.
If there were 1 mol of an ideal gas at a given temperature, one is in a 10L container while another
one is in a 1L container. Obviously, the positional entropy is greater in the 10L container since there is
move volume, allowing more possibilities of different positions.
S large volume > S small volume
S low pressure > S high pressure
G = Go + RTln(P)  G is the free energy of the gas at P atm, Go stands for free energy at 1 atm
Also, there is another formula that can be derived from the equation above:
G = ∆Go + RT ln (Q)  Q is the reaction quotient (products/reactants)
*A system can achieve the lowest possible free energy by going to equilibrium, which is when the forward
and the reverse reaction are equal, causing the net reaction to be zero, not when the reaction undergoes
completion.
16.8 Free Energy and Equilibrium
Equilibrium point: the position where the free energy of a reaction system has its lowest possible value.
Another formula for finding free energy at 1 atm when provided the K value;
∆Go = -RT ln(K) ∆Go is the free energy at 1 atm
3 different cases for the value of ∆Go:
Case #1: ∆Go is equal to zero. This happens when the free energies of the products and reactants
are equal, causing the net amount of energy to be zero. This shows that the reaction is at
equilibrium and at STP so K must equal 1.
Case #2: ∆Go is negative. This can only occur when the free energy of the reactants is greater in
value than the free energy of the products. This means that the system will shift to the right to the
product side to reach equilibrium so the value for K will be greater than 1.
Case #3: ∆Go is positive. This occurs when the free energy of the products is greater than the free
energy of the reactants. This will cause the system to shift to the left towards the reactants to
achieve equilibrium because there is less reactants than products. Because of this, the value of K
will be less than one.
16.9 Free Energy and Work
∆G = w maximum  this equation shows that the maximum possible useful work obtainable from
a process at constant temperature and pressure is equal to the change in free energy
*Reversible process: a cyclic process carried out by a hypothetical pathway, which leaves the universe
exactly the same as it was before the process. No real process is reversible.
Irreversible process: Any real process. When a system undergoes the changes State 1  State 2  State 1
by any real pathway, the universe id different than before the cyclic process took place in the system.
A real cyclic process: This process occurs in a system where the surroundings’ work is limited so it has
more thermal energy.
Multiple Choice Questions
1) Which of the following processes are spontaneous?
A. A clear solution becomes a uniform color after a few drops of dye are added.
B. You clean your bedroom.
C. Salt dissolves in H20
D. Iron rusts. (decomposes)
2)
I.
II
III
If ∆G for a certain reaction has a negative value at 298 K, which of the following must be true?
The reaction is exothermic.
The reaction occurs spontaneously at 298 K.
The rate of reaction is fast at 298 K.
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
3) Calculate ∆S surrounding for these reactions at 25o C and 1 atm.
SbS3 (s) + 3Fe (s)  2Sb (s) + 3FeS (s)
∆H = -125 kJ
Sb4O6 (s) + 4Sb (s) + 6CO(g)
∆H = 778 kJ
A. –2.16 x 10 (3) J/K
B. –2.16 J/K
C. –4.32 x 10 (-3) J/K
D. +4.32 x 10(-3) J/K
4) Which statement is true for this following information?
∆S negative and ∆H positive
A. Spontaneous at all temperatures
B. Spontaneous at high temperatures
C. Spontaneous at low temperatures
D. Process not spontaneous at any temperature.
5) Calculate ∆G using the following data.
2SO2 (g) + O2 (g)  2SO3 (g) at 25 degrees Celsius and 1 atm
Substance
SO2 (g)
SO3 (g)
O2 (g)
A.
B.
C.
D.
–198 kJ
142 kJ
–142 kJ
198 kJ
∆Hof (kJ/mol)
-297
-396
0
So (J/K x mol)
248
257
205
Free Response Question
Nitrogen monoxide, NO (g), and carbon monoxide, C0 (g), are air pollutants generated by
automobiles. It has been proposed that under suitable conditions these two gases could react to form
N2 (g) and CO2 (g), which are components on unpolluted air.
a.Write a balanced equation for the reaction described above. Indicate whether the carbon in Co is
oxidized or whether it is reduced in the reaction. Justify your answer.
b.Write the expression for the equilibrium constant, Kp for the reaction.
c.Consider the following thermodynamic data.
∆Gof (kJ mol (-1)
NO/ 86.55
CO/ -137.15
CO2/ -394.36
I.Calculate the value of ∆Go for the reaction at 298 K.
II.Given the ∆Ho for the reaction at 298 K is -746 kJ per mol of N2 (g) formed, calculate the value of
∆So for the reaction at 298 K. Include units with your answer.
d. For the reaction at 298 K, the value jof Kp is 3.3 x 10 (120). In an urban area, typical pressures of
the gases in the reaction are P (NO) = 5.0 x 10(-7) atm, P (N2) = .781 atm, \
and P (CO2) = 3.1 x 10 (-4) atm.
I.
Calculate the value of ∆G for the reaction at 298 K when the gases are at the partial pressures
given above.
II.
In which direction will the reaction be spontaneous at 298 K with these partial pressures?
Explain.
Answers
Multiple choice
1.A, C, and D. Spontaneous reaction favors disorder.
2.B, because not enough information are given to assume the other possibilities.
3.A
4.D
5.C
Free Response
A.
B.
C
C.
D.
E.
2NO (g) + 2CO (g)  N2 + 2CO2
Kp = [P CO2](2) [P N2] / [P NO] (2) [P CO]2
I. –687.52 kJ
II. –196 J/K
–607.24 kJ
To the right because the reaction is spontaneous at 298K and it tries to relieve the stress by
forming more products.