chapter 17 spontaneity, entropy and free … 17 spontaneity, entropy and free energy.notebook 5...

Chapter 17 Spontaneity, Entropy and Free Energy.notebook

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September 10, 2015

Jan 291:54 PM

Chapter 17:

Spontaneity, Entropy and Free Energy

Jan 11:11 AM

Recall the first law of thermodynamics from chapter 6? It says that energy can neither be created nor destroyed just changed from one form to another. The law describes energy changes in chemical reactions but does not answer the more fundamental question, "Can we use thermodynamics to predict IF reactions will occur?" This chapter show such predictions can be done.

17.1 Spontaneous Processes and Entropy • Define entropy• Choose among alternatives which has the greatest entropy.


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September 10, 2015

Jan 11:11 AM

17.1 Spontaneous Processes & Entropy• Spontaneous means process occurring without outside intervention• Rate of reaction is irrelevant to spontaneity.

Thermodynamics tells us wether a reaction is spontaneous based only on the properties of the reactants and products. The predictions of thermodynamics do not require knowledge of the pathway between reactants and products.

Entropy S can be viewed as a measure of molecular randomness or disorder. The driving force for a spontaneous process is an increase of entropy in the universe• Entropy is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a give state Positional Entropy

Nature spontaneously proceeds towards the states that have the highest probabilities of existing.

Feb 78:46 AM

Which of the following pairs is likely to have the higher positional entropy per mole at a given temperature?a. Solid or gaseous phosphorusb. CH4 (g) or C3H8 (g)c. KOH (s) or KOH (aq)

Example 17.1B Positional Entropy

Practice Questions:Of the following, which would have a negative entropy change?1. A salt crystal dissolves in water.2. Dry ice (solid CO2) sublimes at room temperature.3. Individual amino acids bond together to make a protein.4. Liquid gasoline combusts to form CO2 and H2O gas molecules.

Choice 3 shows a change that decreases entropy. Individual amino acid molecules can be arranged in a variety of changing positions due to their independence. However, once bonded into a protein there is a set organized sequence. This would indicate a decrease (negative) in entropy.


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September 10, 2015

Jan 111:38 PM

For which process is ΔS negative?1) Evaporation of 1 mol of CCl4(l)2) Mixing 5 mL ethanol with 25 mL water3) Compressing 1 mol Ne at constant temperature from 0.5 atm to 1.5 atm4) Raising the temperature of 100 g Cu from 275 K to 295 K5) Grinding a large crystal of KCl to powder

Choice 3 The Ne atoms have less space to move, which decreases their randomness.

In which reaction is ΔS° expected to be positive?1) I2(g) → I2(s)2) H2O(1) → H2O(s)3) CH3OH(g) + (3/2)O2(g) → CO2(g) + 2H2O(l)4) 2O2(g) + 2SO(g) → 2SO3(g)5) None of these

Choice 5 Each of these reactions produces either a solid, liquid or fewer gas molecules.

Jan 11:11 AM

• Spontaneous processes increase the entropy of the universe. (The entropy of a system can decrease if that of the surroundings increase)

17.2 Entropy and the Second Law of ThermodynamicsSecond law of thermodynamics: in any spontaneous process, there is always an increase in the entropy of the universe.

ΔSuniverse = ΔSsystem + ΔSsurroundings

If the ΔSuniv is positive , the entropy of the universe increases and the process is spontaneous in the direction written. If ΔSuniv is negative, the process is spontaneous in the opposite direction. If ΔSuniv is zero, the process has no tendency to occur, and the system is at equilibrium.


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September 10, 2015

Jan 111:38 PM

Jan 111:38 PM

17.3 The Effect of Temperature on Spontaneity• State the importance of exothermic reactions to entropy• Calculate the change in entropy of the surroundings from the heat of reaction of the solution

Exothermic reactions give off energy to the surroundings. Therefore, random motions of particles in the surroundings increase. When random motions increase, positional probabilities increase. The key point from all this is that exothermic reactions increase the entropy of the surroundings (ΔSsurr).

The magnitude of the increase in ΔSsurr depends on the temperature. (See the moneyrelated discussion on page 781). Recall from Chapter 6 that we think of heat flow in terms of the system.

Exothermic reaction (at constant pressure) ΔH = Endothermic reaction (at constant pressure) ΔH = +

ΔSsurr = ΔH T (in Kelvin) Entropy Units: J/K or kJ/K


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September 10, 2015

Jan 212:16 AM

Example 17.3A ΔSsurr and the Heat of Reaction

Calculate ΔSsurr for each of the following reactions at 25˚C and 1 atm

a. C3H5 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) ΔH = 2045 kJb. (NH4)2Cr2O7 (s) → N2 (g) + Cr2O3 (s) + 4H2O (g) ΔH = 315 kJc. H2O (l) → H2O (g) ΔH = +44 kJ

The relationship between entropy change and reaction spontaneity is summarized in Table 17.3 in your textbook. Consider the information in that table, and try following that example.

Feb 77:16 AM

ΔSsys is related to the positional probabilities for each of the reactants, the number of moles of reactants and products in the reaction.

ΔSsurr is related to heat flow from the system.

ΔSuniv is related to the ΔSsurr and heat flow from the system to the environment.

Example 17.3B Reaction SpontaneityDetermine if the values for entropy in each of the following will produce a spontaneous process. Also, which of the following processes is endothermic (from the perspective of the system)?

a. ΔSsys = 30 J/K ΔSsurr = 50 J/Kb. ΔSsys = 27 J/K ΔSsurr = 40 J/Kc. ΔSsys = 140 J/K ΔSsurr = 85 J/Kd. ΔSsys = 60 J/K ΔSsurr = 85 J/K


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September 10, 2015

Jan 212:16 AM

17.4 Free Energy• Relate free energy to spontaneity

Free energy is a mathematical term that describes unequivocally whether a reaction will be spontaneous. It is experimentally useful because it reflects ΔSuniv.

ΔG = ΔH TΔS(When no subscript appears, it is assumed that we are referring to the system.)

This equation gives us an explicit way of calculating free energy. It further says that there are circumstances under which temperature will determine whether a reaction is spontaneous.

If ΔG is negative , the process is spontaneous in the direction written.

Jan 212:16 AM

The case of the temperature dependency of ice melting, as described in your textbook, is a perfect example of this.

You have two opposing entropy factors. On the one hand, the reaction is endothermic, which opposes the process (ΔSuniv = ). On the other hand, melting increases the positional probability of the system (ΔSuniv = +). The temperature will determine which process will dominate (whether ice melts).

Example 17.4A Free Energy and Spontaneity

Given the values for ΔH, ΔS, and T determine whether each of the following sets of data represent spontaneous or non spontaneous processes.

ΔH (kJ) ΔS (J/K) T (K)a. 40 300 130b. 40 300 150c. 40 300 150d. 40 300 130e. 40 300 150


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September 10, 2015

Feb 410:34 AM

Example 17.4B Free Energy and Temperature

You know that the boiling point of water is 373 K. See how this compares to the minimum temperature for reaction that you determine thermodynamically for the phase change;

H2O(l) → H2O(g)

where ΔH = 44 kJ and ΔS = 119 J/K

SolutionThe criterion for spontaneity is ΔG < 0. This means that ΔH TΔS < 0Adding TΔS to both sides

ΔH < TΔSDividing both sides by ΔS ΔH < T

ΔS

Using the data from our problem

T > 4.4 x 104 J or T > 370 K 119 J/K

Solve Q #10 in Exercise packet

Jan 23:40 AM

17.5 Entropy Changes in Chemical Reactions• Predict the sign of entropy changes for a given reaction• Calculate ΔS from thermodynamic data tables.

Reminder: ΔSsurr is related to heat flow from the system. BUT ΔSsys is related to the positional probabilities for each of the reactants. For example,

2H2(g) +O2(g) → 2H2O(g)

ΔS = 89 kJ (at 25°C). We lose entropy because 3 total moles of the gases on the left have more positional possibilities than 2 moles of vapor on the right.

• For a chemical reaction involving only the gas phase, entropy is related to the total number of moles on either side of the equation. A decrease means lower entropy, an increase means higher entropy.• For a chemical reaction involving different phases, the production of a gas will (in general) increase the entropy much more than an increase in the number of moles of a liquid or solid.


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September 10, 2015

Jan 23:40 AM

Example 17.5A The Sign of Entropy ChangesPredict the sign of ΔS° for each of the following reactionsa. (NH4)2Cr2O7(s) → Cr2O3(s) +4H2O(l) + N2(g)

b. Mg(OH)2(s) → MgO(s) + H2O(g)

c. PCl5(g) → PCl3(g) + Cl2(g)

For example,

2HNO3(aq) + Na2CO3(s) → 2NaNO3(aq) +H2O(l) + CO2(g)

ΔSsystem = +88 kJ (at 25°C)

Jan 23:40 AM

The third law of thermodynamics says that the entropy of a perfect crystal at 0 K is zero. This means that absolute entropy of substances can be explicitly measured. (See Appendix 4 in your textbook.)

As was true with ΔH° (a state function), ΔS° (also a state function) can be determined as the difference between the sum of the entropy of products minus the sum of the entropy of the reactants.

ΔS°reaction = Σ npS°products Σ nrS°reactants

Example 17.5B Entropy of ReactionCalculate ΔS° for each of the following reactions using data from Appendix 4 in your textbook.

a. N2O4(g) → 2NO2(g)

b. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

c. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)


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September 10, 2015

Jan 23:41 AM

17.6 Free Energy and Chemical Reactions• Calculate the standard free energy of formation and use it to predict spontaneity of chemical reactions.

Standard free energy change (ΔG°) is the free energy change that occurs if reactants in their standard states (1 atm, 25°C) are converted to products in theory standard states.

ΔG° cannot be measured directly. Three methods of calculating ΔG° are:1. ΔG° = ΔH° TΔS°2. By manipulating known equations, (as in Hess’s law problems for ΔH°)3. ΔG° = Σ np ΔG°products Σ nr ΔG°reactants

Example 17.6A Standard Free Energy from Entropy and EnthalpyUsing data for ΔH° and ΔS°, calculate ΔG° for the following reactions at 25°C and 1 atm.

a. Cr2O3(s) + 2Al(s) → Al2O3(s) + 2Cr(s)

b. C3H8(g) +5O2(g) → 3CO2(g) + 4H2O(g)

Jan 24:22 AM

Example 17.6B Standard Free Energy by Combining EquationsGiven the following data:

(Equation 1) S(s) +3/2O2(g) → SO3(g) ΔG° = 371 kJ(Equation 2) 2SO2(g) + O2(g) → 2SO3(g) ΔG° = 142 kJ

Calculate ΔG° for: S(s) +O2(g) → SO2(g)Is the reaction spontaneous?

Example 17.6C Standard Free Energy from “Products Reactants”Calculate ΔG° for the reaction

C3H8(g) + 5O2(g) → 3CO2(g) +4H2O(g)

using ΔG° data from Appendix 4. Compare the answer with that from our Example 17.6 A, part b.

chapter 17 spontaneity, entropy and free … 17 spontaneity, entropy and free energy.notebook 5...

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