Ch. 18: Free Energy and Thermodynamics

Dr. Namphol Sinkaset Chem 201: General Chemistry II

I. Chapter Outline

I. Introduction II. Spontaneity III. Entropy IV. The 2nd Law of Thermodynamics V. Predicting Entropy Changes VI. Entropy and Spontaneity VII. Gibbs Free Energy

I. Introduction

• In this chapter, we look more closely at what causes chemical change.

• We will see that everything comes back to entropy, and we seek to understand what it is.

• Most simply, it can be described as “energy spreads out.”

I. Energy Spreads Out

II. Spontaneity

• Thermodynamics allows us to predict which processes will occur spontaneously.

• spontaneous process: a process that occurs without ongoing outside intervention e.g. objects falling towards the center of a

gravitational field.

II. Kinetics vs. Thermodynamics • A spontaneous process is NOT equivalent to

a fast process. • Thermodynamics tells us…* • Kinetics tells us…*

II. Spontaneous, But Slow

II. Nonspontaneity

• Note that a nonspontaneous process is not impossible!

• e.g. Iron metal in an iron ore won’t just come out on its own.

• A nonspontaneous process can be made to occur by using external energy or coupling it to a spontaneous process.

III. What Causes Something to Happen?

• Up until this point, what have you been taught to say why something occurs?

• At first, we might consider enthalpy and postulate: Systems proceed in direction of lowest

enthalpy (in direction of lowest PE).

III. Maybe Enthalpy?

• If lower enthalpy was the determining factor in spontaneity, then we’d expect: All exothermic processes to be

spontaneous. All endothermic processes to be

nonspontaneous. • Is this true?*

III. Spontaneous, But Endo!

• Enthalpy is not the whole story because there are endothermic processes that are spontaneous. Evaporation of liquid water to gaseous water NaCl(s) dissolving in water.

III. Evaporation of Water

III. NaCl(s) Dissolving in Water

III. Entropy

• Entropy is the criterion for spontaneity in all systems, including chemical ones.

• entropy (S): a thermodynamic function that increases w/ the number of energetically equivalent ways to arrange the components of a system to achieve a particular state

III. Equation for Entropy

• Developed by Boltzmann, so k is Boltzmann’s constant, (k = R/NA = 1.38 x 10-23 J/K).

• W is the # of energetically equivalent ways to arrange the components of the system.

• Since W is unitless, entropy has units of J/K. As W increases, entropy increases.

III. Understanding W • Imagine a system of particles at a

certain T, P, and V. This system will have a set amount of energy. This set of conditions (or energy) defines

the macrostate of the system. • The distribution of that energy is

constantly changing (particles exchange energy via collisions). The exact internal energy distribution at an

instant is called a microstate.

III. How Many Microstates in a Macrostate?

• This is the fundamental question, and its answer gives us W.

• Consider 2 systems composed of 2 particles with E = 4 J.

III. Microstates and E Dispersal

• The macrostate with the highest entropy has the greatest dispersal of E.

• In the previous example, System B disperses 4 J over two levels whereas System A confines it to just one.

• State that has a higher dispersal (or randomization) of energy has more entropy than a state that concentrates it.

IV. 2nd Law of Thermodynamics

• 2nd Law of Thermodynamics: for any spontaneous process, the entropy of the universe increases (ΔSuniverse > 0).

• The criterion for spontaneity is thus the entropy of the universe.

• A system, chemical or otherwise, proceeds in the direction that will increase the entropy of the universe.

IV. S Is a State Function

• Typically, we will be interested in changes of entropy that accompany a chemical or physical change.

• Luckily, entropy is a state function, so ΔS is easily calculated.

IV. 2nd Law In Action • To understand the 2nd law more, we consider

the expansion of 4 ideal gas molecules into a vacuum.

• This process is spontaneous, has no change in enthalpy, and has no associated work.

• If it’s spontaneous, ΔS > 0. But how?

IV. Allowing Expansion

• When we open the stopcock, there are several energetically-equivalent final arrangements possible.

• For simplicity, we consider only 3 out of the possible 5 arrangements. (What are the other 2?*)

IV. Macrostates & Microstates

• To identify microstates, we label each atom.

IV. Microstates for State C

IV. Probability and Entropy

• Comparing the # of microstates for each macrostate, we see that the probability of finding State C is 6x greater.

• State C has the highest # of energetically-equal ways of distributing the components of the system.

• What happens statistically when we increase the number of atoms?*

IV. ΔS > 0

• When the system goes from all particles on the left to equally distributed, there’s an increase in entropy.

• Therefore, it’s spontaneous.

IV. Units on Entropy • What do the units J/K mean? Numerator is amount of energy dispersed. Denominator can be read as “per kelvin.”

• Thus, energy dispersal at different temperatures results in different entropies!

• Energy dispersal at low T is more entropic than at high T.

IV. Sample Problem 18.1

• Which of the following diagrams represents a positive ΔS?

V. Predicting/Quantifying S

• Now that we have an idea about what entropy is, we can try to analyze systems from an entropy point of view. Entropy differences between states of matter. Entropy changes associated with phase

changes. Entropy differences between compounds. Entropy changes associated with reactions.

V. Entropy and State

• The structure of a substance influences its entropy.

• Consider how many energetically equivalent ways you can arrange each phase.

V. Entropy and State

• As a substance goes from solid to gas, its entropy increases. Consider how each phase can “hold” energy.

V. ΔS of a Phase Change

• If we know how the states of matter are related to each entropically, we can predict ΔS’s for phase changes.

• Which phase changes have positive ΔS’s? Freezing, condensation, deposition,

evaporation, melting, sublimation.

V. S of Elements/Compounds

• Every element and compound has a certain amount of entropy.

• These values are tabulated from experimental measurements.

• The reference point for these measurements is given by the 3rd Law of Thermodynamics.

V. 3rd Law of Thermodynamics • 3rd Law of

Thermodynamics: the entropy of a perfect crystal at absolute zero (0 K) is zero.

• Standard molar entropies can thus be tabulated based on this zero point.

V. Standard Molar Entropies

V. Factors Influencing S°

• The standard entropy of a substance is the amount of energy dispersed into one mole of that substance at 25 °C.

• Thus, the more “places” the compound can put the energy, the more “entropic” that compound will be. Places depends on structure. State, molar mass, allotrope, and

molecular complexity.

V. S° and Physical State

• As mentioned before, entropy increases as a substance goes from solid to liquid, to gas. Liquid H2O has S° = 70.0 J/mole.K. Gaseous H2O has S° = 188.8 J/mole.K.

V. S° and Molar Mass

• As the molar mass increases, the entropy increases.

• The “why” is outside the scope of this course, but it has to do with spacing of energy levels.

V. S° and Allotropes

• Recall that allotropes are different forms of the same element.

• Less constrained allotropes have more entropy than more constrained ones.

V. S° and Molecular Complexity • The more complex the compound, the more

places it can put energy due to its different modes of motion.

• e.g. Ar(g) vs. NO(g).

V. S° and Molecular Complexity

• Similar molar mass, but more complex (bonds):

• Increasing molar mass and increasing complexity:

V. S° and Dissolution

• Dissolved ionic solids have more entropy than the nondissolved solid.

• Energy that was concentrated in the crystalline solid becomes dispersed when dissolved in a solution.

• e.g. KClO3(s) with S° = 143.1 J/mole.K vs. KClO3(aq) with S° = 265.7 J/mole.K.

V. Calculating ΔS°rxn • Recall that the “not” symbol means standard

conditions, i.e. 25 °C and 1 atm. Standard state would be the state the substance

exists as at these conditions. For a solution, [ ] must be exactly 1 M.

• The standard entropy change for a reaction (ΔS°rxn) is the change in entropy for a process in which all reactants and products are in their standard states.

V. Calculating ΔS°rxn

• Calculating changes in standard entropies for a reaction is done via a Hess’s Law type of calculation.

VI. Predicting Spontaneity • We know processes that decrease

entropy aren’t spontaneous. • So why does water spontaneously

freeze at temperatures below 0 °C? • Consider the phrasing of the 2nd Law –

it’s all about the entropy of the universe. • So even though entropy of the water

has decreased, somehow entropy of the universe must have increased.

VI. System & Surroundings • We need to recall the terms system and

surroundings from thermodynamics. • To satisfy the 2nd Law, ΔSuniv > 0. But, the

universe can be split into the system and the surroundings.

• For water spontaneously freezing at 0 °C:

VI. Increasing S of Surroundings

• How do we increase the entropy of the surroundings?

• Remember that entropy is higher dispersal of energy.

• Thus, if a system releases energy to the surroundings, then ΔSsurr > 0.

• Why then does water freeze below 0 °C?*

VI. ΔSsurr Depends on T • Above 0 °C, water does not freeze

spontaneously. Why? • Remember that entropy is E dispersed per

kelvin (J/K), so as temperature increases, ΔS decreases (for a constant amount of J).

• So, above 0 °C, the joules released from freezing are not enough to make ΔSsurr > ΔSsys.

VI. Calculating ΔSsurr

• An exothermic process increases the entropy of the surroundings.

• An endothermic process decreases the entropy of the surroundings.

• The size of ΔSsurr is proportional to the size of qsys.

VI. Calculating ΔSsurr

• Additionally, the higher the temperature, the smaller the change in entropy.

• We combine the proportionalities into an equality.

VI. Sample Problem 18.2

• Consider the reaction 2N2(g) + O2(g) 2N2O(g) with a ΔHrxn = 163.2 kJ.

a) Calculate the entropy change in the surroundings associated w/ this reaction at 25 °C.

b) What is the sign of the entropy change for the system?

c) Will the reaction be spontaneous?

VI. Sample Problem 18.3

• A certain reaction has ΔHrxn = -107 kJ and ΔSrxn = 285 J/K. At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?

VII. Determining Spontaneity

• It would be easier if we could determine spontaneity by just considering changes in the system.

• We derive an equation from the relationships we have learned and define a new quantity – the Gibbs free energy.

VII. Gibbs Free Energy Equation

VII. Gibbs Free Energy

• This equation calculates the change in the Gibbs free energy.

• Gibbs free energy (G) is formally defined as G = H – TS.

• From the derivation, ΔG = -TΔSuniv. Since ΔSuniv is the criterion for spontaneity, ΔG

can be used as a criterion for spontaneity.

VII. Using ΔG

• The change in free energy can be calculated w/ respect to the system, and the result can be used to determine spontaneity. If ΔG is negative, the process is

spontaneous. If ΔG is positive, the process is

nonspontaneous.

VII. Possible Combos in ΔG

• There are 3 factors that determine the outcome of the sign on ΔG.

• Sometimes ΔH and ΔS work together; sometimes they don’t.

VII. “To Lower Energy”

• We’ve always explained why things happen by explaining how they lower energy.

• To be technically correct, things happen to lower the Gibbs free energy (which is equal to increasing Suniverse).

VII. Sample Problem 18.4

• The reaction C2H4(g) + H2(g) C2H6(g) has ΔH = -137.5 kJ and ΔS = -120.5 J/K. Calculate ΔG at 25 °C and determine whether the reaction is spontaneous at this temperature. Does ΔG become more negative or more positive as temperature increases?

VII. Rxn Free Energy Changes

• So, to determine whether or not a process is spontaneous, we can calculate ΔG°rxn.

• Of course, this is the standard free energy change of a reaction.

• There are three ways to calculate ΔG°rxn, depending on what information is given.

VII. Calculating ΔG°rxn with ΔH°rxn and ΔS°rxn

• The first method involves using the Gibbs free energy equation. ΔG°rxn = ΔH°rxn - TΔS°rxn

• To use this method, typically need to do Hess’s Law type calculations for enthalpy and entropy.

VII. Sample Problem 18.5 • Determine whether the reaction NO(g) +

½ O2(g) NO2(g) is spontaneous at 25 °C. Note that the enthalpies of formation for NO(g) and NO2(g) are 91.3 kJ/mole and 33.2 kJ/mole, respectively and the standard entropies for NO(g), O2(g), and NO2(g) are 210.8 J/mole.K, 205.2 J/mole.K, and 240.1 J/mole.K, respectively.

VII. Calculating ΔG°rxn Via Hess’s Law

• standard free energy of formation, ΔG°f: change in free energy when 1 mole of a compound forms from its constituent elements in their standard states.

• If a table of ΔG°f’s is available, ΔG°rxn can be calculated using a Hess’s Law type of calculation.

VII. Sample Problem 18.6

• Calculate ΔG°rxn for the reaction 2CO(g) + 2NO(g) 2CO2(g) + N2(g) given that the standard free energies of formation for CO(g), NO(g), and CO2(g) are -137.2 kJ/mole, 87.6 kJ/mole, and -394.4 kJ/mole, respectively.

VII. Calculating ΔG°rxn Stepwise • Just like thermochemical equations,

reactions w/ associated free energies can be manipulated with the same rules applying. 1) If the equation is multiplied by a factor, then

ΔGrxn is multiplied by the same factor. 2) If an equation is reversed, then ΔGrxn

changes sign. 3) If a series of reactions adds up to an overall

reaction, the ΔGrxn for the overall process is the sum of ΔGrxn for each step.

VII. Sample Problem 18.7

• Using the equations below, find the ΔG°rxn for the reaction N2O(g) + NO2(g) 3NO(g).

2NO(g) + O2(g) 2NO2(g) ΔG°rxn = -71.2 kJ N2(g) + O2(g) 2NO(g) ΔG°rxn = 175.2 kJ 2N2O(g) 2N2(g) + O2(g) ΔG°rxn = -207.4 kJ

VII. Why Is It “Free” Energy? • The energy of a reaction

available to do work is the free energy.

• The free energy is the theoretical maximum; usually, it will be less.

• The only way to get the maximum is by using an infinitesimally slow reaction known as a reversible reaction.

VII. A Puzzling System

• Consider the following: H2O(l) H2O(g) ΔG° = +8.59 kJ/mole

• Is this process spontaneous?* • When you spill water, does it

evaporate? • How do we explain this discrepancy?*

VII. Nonstandard Free Energies

• Of course, most of the time, a reaction will not be under standard conditions.

• We need a way to calculate spontaneity under a variety of conditions.

VII. Sample Problem 18.8

• The reaction 2H2S(g) + SO2(g) 3S(s,

rhombic) + 2H2O(g) has a standard free energy change of -102 kJ. Calculate ΔGrxn when the partial pressures of H2S, SO2, and H2O are 2.00 atm, 1.50 atm, and 0.0100 atm. Is the reaction more or less spontaneous under these conditions?

VII. Relationship Between ΔG°rxn and K

• Recall that K is a measure of how far a reaction goes towards products; the value of K depends on the value of free energy change.

• Thus, there is a relationship between free energy and equil. constants.

• If ΔGrxn < 0, reaction is spontaneous. If ΔGrxn > 0, reaction is nonspontaneous. If ΔGrxn = 0, reaction is at equilibrium.

VII. Relationship Between ΔG°rxn and K

VII. ΔG°rxn and K

VII. Temp. Dependence of K

• Previously, we discovered that equilibrium constants depend on temperature.

• With the relationship between standard free energy and the equilibrium constant, we can see how temperature is affected by temperature.

VII. Temp. Dependence of K

VII. Two-Point Form

• The previous equation can expressed in two-point form, allowing calculation of the change in enthalpy if equilibrium constants at 2 different temperatures are known.

ln𝐾𝐾2𝐾𝐾1

= −∆𝐻𝐻°𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅

1𝑇𝑇2−

1𝑇𝑇1