Chapter 17
Erin R

Objective: To understand the three laws of thermodynamics, spontaneous processes, entropy,
Gibbs free energy and Equilibrium.

Three Laws of Thermodynamics:

First Law: Energy can be converted from one form to another, but it cannot be created or destroyed. (Energy of the Universe is constant)

Second Law: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.

Third Law: Entropy of a substance approaches zero as its temperature approaches absolute zero.

Examples of Spontaneous Processes:

Ex 1: Perfume spreading in a room from its perfume bottle.

Ex. 2: Ice melting into Water

Ex. 3: A ball rolling down a hill.

Ex. 4: Iron Rusting

Non-Spontaneous:

Ex. 1: Burning Wood

Ex.
2: Combustion in a Combustion chamber.

Define Entropy:

Lack of order or predictability. (Gradual decline into disorder)

The universe moves towards disorder.

For example when you clean your room it eventually finds its way back to the beginning messy state.

Why is the entropy of a mole of steam larger than that of of a mole of water?:

Gases are inherently more random and have more disorder than a liquid and therefore would have higher entropy.

Predict the change of entropy (delta S) for common processes:
 PCl5 yields PCl3+Cl2…. There are more moles on the product side than on the reactant side the entropy will be positive.

When you freeze water that is negative entropy because it causes the atom to move about less randomly becoming more organized.
 2H2(g)+O2(g) yields 2H20 (g)…. There are 3 moles on the reactant side vs. only two on the product side causing the entropy to be negative.

When ice melts into water the entropy will be positive because the atoms are becoming more spread apart moving from a solid to a liquid.

Describe what delta S universe, delta S system, and delta S surroundings is:

Delta S Universe: The sum of the entropy changes in the system and the surroundings.

If it is positive than the process is spontaneous.

If it is negative than the process is nonspontaneous.

If it is zero than the process is at equilibrium.

Delta S System: The reactants and products of the reaction.

Delta S Surroundings: Everything not being reacted the surrounds the system… the environment that the reaction is happening in. (Ex. The lab area).

Relate Delta S Universe for spontaneous processes and for processes at equilibrium:

For a spontaneous process: Delta S Universe= Delta S System+ Delta S Surroundings is greater than 0.

For a equilibrium process: Delta S Universe= Delta S System + Delta S Surroundings =0

Use the thermodynamic table to determine Delta S reaction:

Zn(s)+2HCl(g) -> ZnCl2(s)+H2(g)

Zn(s) delta S is: 41.6

HCl(g) delta S is: 187.0

ZnCl2(s) delta S is: 108.37

H2(g) delta S is: 131.0

ΔH
+
+
-
-

The equation is Delta S reaction = sum of S products – sum of S reactants

The answer would be: -176.23 J/K or -176 J/K with correct sig figs.

Relate Delta S surroundings to Delta H system and also how temperature effects it:

Delta S is Entropy while Delta H is Enthalpy.

Delta G= Delta H- T Delta S
 Ms. Soto’s table helps explain this in the simplest terms and also most straight forward.
ΔS Spontaneous in forward direction?
+ Spontaneous at high temp “entropy driven”
(Exothermicity is unimportant)
-
+
-
ΔG is always positive;
It will never be spontaneous at any temp; reverse is spontaneous
ΔG is always negative;
It will always be spontaneous at all temperatures
Spontaneous at low temp “enthalpy driven”
(Exothermicity is important)
Practice Questions:
1.
Which process would be considered spontaneous? a) Burning Wood b) Freezing Water c) Charging a battery d) Ice melting into water e) A ball rolling up hill
Explanation:
The answer is D. The reason is because it doesn’t take any outside energy to cause this reaction.
Unlike burning wood, which causes someone or something to ignite it… it doesn’t just burst into flame.
Water has to have something force the molecules together for it to freeze it doesn’t just randomly do so.
Charging a battery is a no brainer because you have to put energy into it for it to hold a charge. And last but no least a ball does not roll up hill it would have to be pushed or pulled up hill.
2.
What is Delta S universe equal to when it is at equilibrium? a) 25 degrees Celsius b) 1 atm c) > 0 d) < 0 e) 0
Explanation:
The answer is E. The reason is yes 25 degrees Celsius is standard temperature for most experiments and also 1 atm is standard pressure but that has nothing to do with Delta S universe. Then C is for when
Delta S universe is spontaneous and D for when it is nonspontaneous. Equilibrium is = to 0 in this case.
Key Words: Equilibrium, Entropy, Enthalpy, Thermodynamic Laws 1, 2 and 3, Spontaneous, Nonspontaneous, Delta S System, Delta S Surroundings, Delta S universe, free energy, and Gibbs Free Energy.
Use pages 778-813 in the Chang Textbook
Helen N
Relating  S niverse
and  H system
and temperature
 S= H/T
Gibbs free energy
In general, nature tends to move toward two different and seemingly contradictory states— low energy and high disorder, so spontaneous processes must result in decreasing enthalpy or increasing entropy or both.
 G  =  H  -T  S  (T=absolute temperature in K)
Predicting a reaction from  G
 If  G is negative, the reaction is spontaneous.

 If  G is positive, the reaction is not spontaneous (or spontaneous in the reverse direction)
 If  G equals 0, the reaction is at equilibrium
Calculating  G
1.
Reference appendix 3 for thermodynamic values
2.
Using modified Hess’ Law, solve for variable  G  reaction
=  G products
 G reactants
3.
Standard free energy of formation for any element in its standard state equals zero
Predicting the direction of a spontaneous reaction given  H,  S and T
An exothermic process is favored because by giving up heat, the entropy of the surroundings increases. The sign of  S surr
depends on the direction of heat transfer and the magnitude of  S surr depends on the temperature: at constant temp/pressure. If  S is negative, the reaction is not spontaneous in the forward direction, but the reverse, and if  S is positive, it is spontaneous in the forward direction and not the reverse.
See table 17.3 on pg.795
Calculating  S for phase changes
 S
 S=  H/T equil
Given H fusion
or H vaporization
and melting or boiling point, you can determine
 G under non-ideal conditions
 G=  G  +RTln(Q) R=universal gas constant Q= reaction quotient
Q=[C] c [D] d /[A] a [B] b
Relating Gibb’s free energy and the equilibrium constant (K)
 G  = -RTln(K)
Calculating K
1.
Using thermodynamic values, solve for  G 
2.
R=universal gas constant (8.314 J/K  mol)
3.
T= temperature in Kelvin
G=0, K=1 G<0,K>0 G>0,K<0
Problems:
1.
Which of the following is false about standard free energy? a.
It is the free energy change that occurs when reactants in their standard state turn into products in their standard states b.
It is a state function c.
It cant be measured directly but can be calculated from other measurements d.
The standard free energy of formation for O
2 equals 205.0 J/K  mol
Answer: d (the standard free energy of formation for any element in its standard state is zero)
2.
Which shows a correct relationship between  H and  S? a.
 H is positive,  S positive = Spontaneous at low temperature b.
 H positive,  S negative = Never spontaneous c.
 H negative,  S positive = Never spontaneous d.
 H negative,  S negative = Spontaneous at high temperature
Answer: b (a positive minus a negative will always yield a positive indicating a spontaneous reaction)