Enthalpy, Entropy, and Free Energy

Mr. Bouyer
Day 1 - 2 | Day 3 - 4 | Day 5 | Lab | Skills Test


Describe the difference between enthalpy and entropy. Use the First Law of Thermodynamics to calculate system energy change. Use thermodynamic properties to calculate the enthalpy, free energy, and entropy of a substance. Use Hess's Law to determine the enthalpy change within a multiple-step reaction. Use Gibbs Free Energy calculations to determine the activation energy of a reaction.

Enthalpy, Entropy, and Free Energy

Review the basics about heat changes in matter.

All chemical changes are accompanied by changes in energy and the degree of disorder for the particles involved.

The study of matter and energy interactions is called Thermodynamics. This week we will examine the three variables of interest to chemists who study the thermodynamics of chemical reactions.

Enthalpy, H, (heat content of a substance): natural systems tend to go from a state of higher energy to a state of lower energy. For instance, a ball rolls down a hill spontaneously, but not up. The ball looses potential energy as it rolls downhill. At the bottom of the hill it has zero potential energy. This same idea can be applied to chemical potential.

Entropy, S, (disorder of a substance): natural processes tend to go from an orderly state to a disorderly state. For instance, what happens if you carefully place three layers of different colored marbles in a container and shake it? The marbles are no longer in an ordered state. No matter how much you shake the box, the probability of getting back to three layers of single colors is next to zero.

Free energy, G, (chemical potential of a substance): a comparison of the changes of enthalpy and entropy during a chemical reaction. All spontaneous processes move toward equilibrium. If the enthalpy and entropy for a chemical reaction have the same sign, there will be some temperature at which DH and TDS will be numerically equal and DG will be exactly zero. This state is the thermodynamic definition of a system at equilibrium.

While there are exceptions, exothermic reactions usually take place spontaneously and endothermic reactions do not. The products of an exothermic reaction have less energy than the reactants, therefore the reaction occurs naturally to reach a lower state of energy.

Since temperature and pressure affect enthalpy, you should be aware of two more reaction types. Isothermal reactions take place at a constant temperature. Isobaric reactions take place at a constant pressure.

Thermodynamics sounds pretty simple, right? Well ....... Professor Frank L. Lambert, of Occidental College in Los Angeles, will tell you that you can't understand the way the world works, to say nothing of chemistry, if you don't understand entropy and the second law of thermodynamics. He has three Internet resources that explain these using simple terms, not complicated equations. You should explore at least one of these resources before continuing.
detailed website
short and sweet website
magazine article

State functions depend only on the current state of the system.
The amount of change in each of these functions depends only on the beginning and the final states, not on the path followed during the change.

T - temperature
V - volume
P - pressure
U - internal energy
H - enthalpy
G - Gibbs free energy

Heat (q) and work (w) are not state functions. Their value depends on the path by which a system gets from the initial state to the final state.

Thermodynamics: the study of the flow of energy in matter.
There are two ways of transferring energy to a system: heat the system or do work on the system.

The first law of thermodynamics represents the energy change within a system with the equation:

DU = q + w

When energy is added to a system:

DU is positive.
q represents heat absorbed by the system, therefore is positive.
w represents work done on the system, therefore is positive.
This would be an endergonic reaction.

When energy is released by a system:

DU is negative.
q represents heat released by the system, therefore is negative.
w represents work done by the system, therefore is negative.
This would be an exergonic reaction.

Sample Thermodynamics Problems:

Problem #1 - A system receives 500 joules of heat from it surroundings and while those surroundings do 250 joules of work on the system. What is the change in the system's internal energy?

Problem solution:

DU = q + w
DU = 500 J + 250 J
DU = 750 J

Problem #2 - A system receives 325 kJ of heat from its surroundings and does 200 kJ of work on the surroundings. What is the change in its internal energy?

Problem solution:

DU = q + w
DU = 325 kJ + (- 200 kJ)
DU = 125 kJ
Notice that since work leaves the system,
it is given a negative value.

Problem #3 - The value for the DU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?

Problem solution:

DU = q + w
w = DU - q
w = - 120 J - 420 J
w = - 540 J
Again, the negative value indicates
work was done by the system.

Homework Assignment 312:
This assignment must be turned in by the beginning of class tomorrow to receive credit.
Scoring criteria
Do the following calculations using the first law of thermodynamics:

325 kJ of heat are added to a system and the system has 45 kJ of work done on it. What is the change the the internal energy of the system?
How would your answer above differ if the work had been done by the system instead of on it?
During a change, a system absorbs 1200 kJ. If the value of DU is -3.5 kJ, how much work did the system do?

Day 3 - 4

Enthalpy can be thought of as heat transfer in matter, but it is more than that:

Enthalpy equals internal energy plus the product of pressure and volume.
q_p = D(U + P V) ...... The _p represents an isobaric (constant pressure) process.
q_p = DH
In an exothermic reaction the products have less enthalpy than the reactants. DH < 0

In an endothermic reaction the products have more enthalpy than the reactants. DH > 0

The change in enthalpy that occurs in a chemical reaction is due to the energy required to break the chemical bonds in the reactants and the energy produced by forming the chemical bonds of the products.

Standard state:

Since temperature and pressure have an effect on enthalpy, these must be specified.
The standard states for enthalpy are 298 K (25^oC) and 100 kPa.
The enthalpy of a free element (an element not in a compound) is arbitrarily set equal to zero.

Enthalpy of formation - DH_f^o (the ^o indicates standard states) - the change in enthalpy when one mole of a compound is produced from the free elements.

The units will be kilojoules per mole.
A negative sign represents an exothermic reaction. The compound has less enthalpy than the elements from which it was formed.
A positive sign represents an endothermic reaction. The compound has more enthalpy than the elements from which it was formed.

If the reaction forming a compound is exothermic, the decomposition of the compound would be endothermic. Most enthalpies of formation are negative. Compounds that have large negative enthalpies of formation require large amounts of energy to decompose and are considered thermodynamically stable.

Enthaply of reaction: DH_r^o

The law of conservation of energy tells us that the sum of the enthalpies of the reactants must be equal to the sum of the enthalpies of the products plus any energy change during the reaction. Using the Greek letter sigma, S, to represent a sum, the total enthalpy for a reaction can be expressed mathematically: DH_r^o = SDH_f^o_(products) - SDH_f^o_(reactants)

A table with all the thermodynamic properties needed for the problems on this page.

Values on this table have been rounded for easy calculation.
Refer to page 859 in your textbook or the CRC Handbook for more information.
The NIST Chemistry WebBook has more information.

Thermodynamic Properties at (25^oC and 100 kPa)
Substance	Enthalpy DH^o_f kJ/mole	Free Energy DG^o_f kJ/mole	Entropy S^o J/mole^.K
BaCl₂_(aq)	-872	-823	123
BaSO₄_(cr)	-1473	-1362	132
Br₂_{( g )}	31	3	245
C₂H₄_{( g )}	52	68	220
C₂H₆_{( g )}	-85	-33	230
Cl₂_{( g )}	0	0	220
CO₂_{( g )}	-394	-394	214
HBr_{( g )}	-36	-53	200
HCl_{( g )}	-92	-95	190
HCl_(aq)	-167	-131	57
H₂O_{( l )}	-286	-237	70
H₂O_{( g )}	-242	-230	190
H₂O₂_{( l )}	-186	-120	110
H₃PO₄_(aq)	-1280	-1120	110
H₂SO₄_(aq)	-910	-745	20
NaCl_(s)	-411	-384	72
NaOH_(s)	-426	-	-
O₂_{( g )}	0	0	205
P₄O₁₀_(cr)	-2980	-2700	230

If the enthalpy of formation of each reactant and product is known, the amount of energy produced or absorbed can be calculated. With this information, you can predict whether a reaction will be exothermic or endothermic.

Sample Enthalpy Problem:

Calculate the DH_r^o for the reaction: P₄O₁₀ + 6H₂O_{( l )} 4H₃PO₄

The solution:

Use the equation DH_r^o = SDH_f^o_(products) - SDH_f^o_(reactants)
From the table:
Multiply each value by the coefficient from the balanced equation which represents the number of moles:
Plug the numbers into the equation and calculate:
DH_r^o = - 424 kJ

Hess's Law: The Enthalpy change for a reaction is the sum of the enthalpy changes for a series of reactions that add up to the overall reaction.

Sample Hess's Law Problem:

Use Hess's law to calculate the enthalpy of combustion for ethane.

The complete combustion of ethane is represented by the equation:
2C₂H₆_(g) + 7O₂_(g) 4CO₂_(g) + 6H₂O_(g)

This reaction is known to involve three distinct reactions:

C₂H₄_(g) + 3O₂_(g) 2CO₂_(g) + 2H₂O_(g)
C₂H₄_(g) + H₂_(g) C₂H₆_(g)
H₂_(g) + ¹/₂O₂_(g) 2H₂O_(g)
DH = - 1323 kJ/mole
DH = + 137 kJ/mole
DH = - 242 kJ/mole

The solution:
Hess's law states that the DH for a reaction is equal to the sum of the DH for the individual steps of the reaction.
Therefore: DH_(C₂H₆) = (- 1323 kJ/mole) + (137 kJ/mole) + (- 242 kJ/mole)

DH_(C₂H₆) = - 1428 kJ/mole

How can you use data from the Thermodynamic Properties Table
and this equation to verify Hess's law?
DH_r^o = SDH_f^o_(products) - SDH_f^o_(reactants)

Homework Assignment 314:
This assignment must be turned in by the beginning of class tomorrow to receive credit.
Scoring criteria

What is the enthalpy change for the gaseous reaction of chlorine with hydrogen bromide?

Day 5

Entropy (S) is the degree of disorder in a system.

States of matter are good examples of entropy.

Crystalline solids have a repeating pattern - a high degree of order.
The particles of a liquid are more disordered than those of a solid.
Gas particles, moving almost totally at random, have the largest entropy of all.

The symbol DS represents change in entropy.

A positive value means an increase in the degree of disorder. An example would be ice melting.
A negative value means a decrease in the degree of disorder. Water freezing would be an example.

Entropy changes for a reaction are calculated with the equation:

DS_r^o = SDS^o_(products) - SDS^o_(reactants)

Sample Entropy Problem:
Use the entropy equation to prove that water boiling increases the disorder in the system.
The solution:

Thermodynamics Calculator

Gibbs Free Energy:

Suppose the enthalpy change in a reaction tends to make it spontaneous, but the entropy change tends to prevent reaction. Is the reaction spontaneous or not? The next two equations help answer that. In the equations, G is Gibbs free energy, H is enthalpy, S is entropy, and T is the kelvin temperature.

G = H - TS
DG = DH - TDS
- When DG for a reaction is negative, the reaction will be spontaneous.
- When DG for a reaction is positive, the reaction will not be spontaneous.
- When DG is zero, the reaction is at equilibrium.

Gibbs free energy changes for a reaction are calculated with the equation:

DG_r^o = SDG_f^o_(products) - SDG_f^o_(reactants)

Sample Gibbs Free Energy Problem:

Calculate the Gibbs free energy change for the double displacement reaction between barium chloride and sulfuric acid. Does this reaction occur spontaneously?

The solution:

Write a balanced equation for the reaction.
Calculate SDG_f^o for the reactants and products.

Remember to multiply the data from the
thermodynamics properties table by the
coefficients from the balanced equation.
Calculate DG_r^o
Since DG_r^o is negative, the answer is yes.

Practice Problems:

Using a free energy calculation, determine if this reaction is spontaneous:
P₄O₁₀ + 6H₂O_{( l )} 4H₃PO₄
Does hydrogen peroxide spontaneously decomposed into water and oxygen?

Chemistry Class