APSI 2018 AP Chemistry 2 Chapter 3.1

3.1 Entropy — A Quantitative Treatment Warm Up In the chapter on Entropy Change versus Enthalpy Change) we dealt with the impact of ΔS and ΔH on spontaneity. we also know that heat flows from a region of high energy to a region of low energy. This might lead to the conclusion that spontaneous change must be exothermic.

1. Consider the following change occurring at room temperature:

time ice cube (crystal structure)

puddle of water (no organized structure)

(a) Is this change spontaneous (did it occur without outside intervention)? _______ (b) Is this process endothermic or exothermic? ________________ (c) What is the sign of ΔH? _______ 2. Consider the change at a molecular level: Order

Disorder

ΔS > 0

ice

water

(a) What is the sign for ΔS? _______ (b) Is this process enthalpy driven or entropy driven? ______________ Explain. ____________________________________________________________________________________________ ___________________________________________________________________________________________________

The First Law of Thermodynamics

Energy is conserved — never created or destroyed. When energy is released by an exothermic change, the potential energy stored in chemical bonds is converted to kinetic energy, which is the energy of motion of atoms and molecules. As the kinetic energy of the atoms and molecules in the system and the surroundings increases, the temperature rises. Sometimes we speak of exothermic changes as “giving off heat.” Heat is the energy transferred between a system and its surroundings.

When thermodynamicists say that a reaction is spontaneous, they mean that the reaction will occur

Spontaneous Change without any outside intervention. They do not mean the reaction occurs rapidly. We studied the rate or speed of a reaction in Unit 1: Reaction Kinetics.

In contrast, a non-spontaneous event can continue only as long as it receives some sort of outside assistance. All non-spontaneous events occur at the expense of spontaneous ones. They can occur only when some spontaneous change occurs first.

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Enthalpy and Spontaneity

Figure 3.1.1 Every

When a change lowers the potential energy of a system, it tends to occur spontaneously. Fuels evolve heat as they combust to form H2O and CO2 and convert the potential energy in the bonds of the reactants to heat. Since a change that lowers the potential energy of a system is exothermic, we can say that exothermic changes have a tendency to occur spontaneously.

While many spontaneous changes are indeed exothermic, there are common examples of spontaneous endothermic changes including the melting of ice, the vaporization of water, and the dissolving of most salts in water to form aqueous solutions. Since each of these events is both endothermic AND spontaneous, they must have some other factor in common that allows them to overcome the unfavorable energy change. Careful examination shows that in each of these events, there is an increase in the randomness or disorder of the system. These observations led to the development of the second law of thermodynamics.

Second law of thermodynamics: The entropy of the universe increases with every spontaneous process.

spontaneous process increases the entropy of the universe.

Quick Check 1. Use qualitative reasoning to decide which of the two options in each of these examples will occur spontaneously: (a) A mole of ice at 0°C or a mole of liquid water at the same temperature

(b) Solid sodium chloride in the bottom of a bucket of water or a solution of the salt in water

(c) A collection of jigsaw pieces or the completed puzzle

(d) One kilogram of raw rubber or 1 kg of vulcanized rubber (Vulcanized rubber is rubber that is polymerized into long chains; vulcanized rubber is used to make tires.)

(e) A pack of cards arranged in suits or a pack of cards that is randomly shuffled

The Boltzmann Equation

Ludwig Boltzmann, a famous physicist, studied fluid mechanics in the late 1800s. During this time, Boltzmann determined the probability that a group of particles might occupy a particular set of positions in space at any particular time. These sets of positions are referred to as microstates. It has been suggested that this term, microstates (states), may have led to the symbol S being selected to represent entropy. Boltzmann noted that entropy is directly proportional to the natural log of the number of microstates available. In 1872, Boltzmann derived his famous equation to represent entropy: S = kB ln W where S = entropy kB = Boltzmann’s constant = R/NA = 8.314 L·kPa/mol·K = 1.38 × 10–23 J/K 6.02 × 1023 mol–1

W = number of microstates

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In a pure crystal at absolute zero (0 K), the particles orient themselves in only one possible microstate. At absolute zero, the entropy is S = 1.38 x 10–23 J/K (ln 1). Since the natural log of 1 = 0, the entropy of a pure crystal at 0 K equals zero. This is the third law of thermodynamics.

Third law of thermodynamics: The entropy of a pure crystal at absolute zero is zero (S = 1.38 x 10–23 J/K × ln (1) = 0).

We know that increasing the number of electrons in an element or increasing the number of atoms in a molecule leads to an increase in entropy. More electrons or atoms mean more possible microstates and therefore a higher entropy according to the Boltzmann equation.

Dependence of Entropy on Temperature

The second law of thermodynamics states “the entropy of the universe increases with every spontaneous change.” The “universe” is the system plus the surroundings and the entropy of a system is a quantitative measure of the disorder or randomness within a system. There are two kinds of disorder in a system: positional disorder, depending on the arrangement of particles (molecules and atoms) in space, and thermal disorder, depending on how the total energy is distributed among all the particles. The more random the distribution of particles in space, the higher the entropy. The more randomly energy is distributed among the particles, the greater the entropy.

Table 3.1.1 shows all possible arrangements for two identical particles among three relative energy levels. The most likely total energy is 2 kJ. Maximizing thermal disorder does not necessarily lead to maximum overall energy.

Table 3.1.1 All possible two-particle arrangements among three relative energy levels

Energy Level

Particle Arrangements

2 kJ

X

1 kJ

X

0 kJ

XX

X

X

Total energy

0 kJ

1 kJ

2 kJ

X XX

X

2 kJ

3 kJ

XX

4 kJ

Adding heat to a system results in greater disorder and, therefore, greater entropy. The entropy increase is directly proportional to the amount of heat added. However, the increase in entropy is more significant when we add heat to a system at a lower temperature where little disorder exists, rather than at high temperature where there is already substantial disorder. For a given quantity of heat, the entropy change is inversely proportional to the temperature at which we add the heat. Thus entropy can be determined as: ∆Ssurroundings = q = –ΔH T T

tem pe ratu re

hi gh

Energy

The equation above shows that entropy has units of energy divided by temperature (J/K). If an entropy change is being calculated for a particular quantity of material undergoing a physical or chemical change, the units may be J/K·mol.

re atu r e p tem low

Entropy Figure 3.1.2 The change in entropy (ΔS) is most noticeable at a low temperature.

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Quick Check 1. When 1.0 mole of steam condenses to form 1.0 mole of liquid water at 100.0° C, 40.7 kJ of heat is released. What is ΔS for the surroundings?

2. Select the example with greater entropy in each case. (a) F2(g) or Cl2(g)

(b) CO2(g) or CO(g) (c)

A

Entropy Is a State Function

or

We know the third law of thermodynamics states that at absolute zero, the entropy of a pure crystal is equal to zero joules. Mathematically, we can state: S = 0 J when W = 1 at T = 0 K Because we know the point at which entropy has a value of 0 J, it is possible to determine the actual amount of entropy that a substance possesses at temperatures above 0 K. If we determine the entropy of 1 mole of a substance at a temperature of 298 K (25°C) and a pressure of 1 atm, we call it the standard entropy or S°. Tables of standard entropies are available, similar to the tables of standard enthalpies of formation used in AP 1. These values are listed in Table A8 in the appendix at the back of this book. (Notice that the standard entropy values of pure elements in their standard states are not equal to zero because standard entropy values are not recorded at 0 K.) Once we know the entropies of a variety of substances we can calculate the standard entropy change, or ΔS °, for chemical reactions in the same way as we calculated ΔH ° earlier.

Entropy is a state function — it is not pathway dependent. Therefore, we can use the same logic as that of a Hess’s law type of calculation to determine the entropy change for a given chemical reaction. It is important to note that like enthalpy, entropy is an extensive property — it depends on the amount of substance present. This means that the number of moles of all species in the reaction must be taken into account. Hence:

ΔS °reaction = ∑ nS °products – ∑ nS °reactants

Recall that for a process to be spontaneous, there must be an overall increase in the entropy of the universe. If the entropy change of the universe is negative, the process is non-spontaneous. A spontaneous entropy change is positive or in the reverse direction of non-spontaneous change.

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Sample Problem 3.1.1 — Quantifying Entropy Change Urea (from urine) hydrolyzes slowly in the presence of water to produce ammonia and carbon dioxide.

CO(NH2)2(aq) + H2O(l) → CO2(g) + 2 NH3(g) What is the standard entropy change, for this reaction when 1 mol of urea reacts with water? Is the process spontaneous? The standard entropies for reactants and products are:

Substance S° (J mol–1 K–1) Substance S° (J mol–1 K–1) CO(NH2)2(aq) 173.7 CO2(g) 213.8 H2O(l) 70.0 NH3(g) 192.8

What to Think About 1. Calculate the change in entropy using the equation: ΔS°rxn = ∑nS°products – ∑nS°reactants 2. After determining the stoichiometrically correct entropy values, substitute and solve.

3. Does the answer make sense?

One mole of aqueous substance and 1 mole of liquid form 3 moles of gas.

4. What does the answer mean?

How to Do It

Determine So values for all reactants and products. These are listed in the table given above in this example.

[mol rxn mol K mol rxn mol K] [ mol rxn mol K mol rxn mol K ] 1 mol CO2 x 213.8 J + 2 mol NH3 x 192.8 J

_

1 mol CO(NH2)2 x 173.7 J + 1 mol H2O x 70.0 J

= 599.4 – 243.7 = 355.7 J/molrxn·K The number of microstates (W) has increased.

So we expect a positive ΔS value. ✓ A positive ΔS value is indicative of a spontaneous change

Practice Problems 3.1.1 — Quantifying Entropy Change 1. Use qualitative reasoning (no calculations are necessary) to predict whether the entropy change in each of the following processes is positive or negative. Does the entropy change of the system favor spontaneous change or not? (a) The evaporation of 1 mole of rubbing alcohol (b) 2 Ca(s) + O2(g) → 2 CaO(s) (c) XeF4(g) → Xe(g) + 2 F2(g) (d) The dilution of an aqueous solution of lime Kool-Aid

2. Use the standard entropy table in Table A8 in the appendix at the back of this book to calculate the standard entropy change for each of the following reactions. Is the reaction thermodynamically favorable (spontaneous) based on the ΔS o value? (a) CaO(s) + 2 HCl(g) → CaCl2(s) + H2O(l)

(b) C2H4(g) + H2(g) → C2H6(g)

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3.1 Review Questions 1. Predict the sign you would expect for ΔS° in each of the following processes. Give a short explanation of why you think it is (+) or (–). Do NOT do any calculations. (a) Blowing up a building

(b) Organizing a stack of files in alphabetical order

(c) AgCl(s) → Ag+(aq) + Cl–(aq)

(d) 2 H2(g) + O2 (g) ➝ 2 H2O(l)

(e) Na(s) + ½ Cl2(g) ➝ NaCl(s)

(f) Removing medication from bottles and organizing the pills by the day in “blister packs”

2. Predict the sign of ΔS° you would expect for each of the following changes. Give a short explanation of why you think it is (+) or (–). Do NOT do any calculations.

hydrogen

oxygen

water

(a)

physical change of water into ice

(b)

(c)

HH

H

Cl

H

Cl

condense

Cl

(d)

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Cl

(e) (f)

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3. The boiling point of chloroform (CHCl3) is 61.7° C. The enthalpy of vaporization is 31.4 kJ/mol. Calculate the entropy of vaporization. Does the sign for entropy of vaporization make sense? Explain.

4. (a) What is the normal boiling point for formic acid, HCOOH?

Given: ΔH°f = –424.72 kJ/mol for liquid HCOOH and –379.1 kJ/mol for gaseous HCOOH

S ° = 128.95 J/mol K for liquid HCOOH and 251.0 J/mol K for gaseous HCOOH

(b) Based on the calculated boiling point, in what state does formic acid exist at room temperature? Explain.

5. The normal boiling point of water is 100.0°C , and its molar enthalpy of vaporization is 40.7 kJ/mol. What is the change in entropy in the system in J/K when 39.3 g of steam at 1 atm condenses to a liquid at the normal boiling point? Does the sign for the value you calculate make sense? Explain.

6. Iodine is an unusual element that sublimes under standard conditions. If the ΔS °subl of iodine is 145 J/mol K and the ΔH °subl of iodine is 62 kJ/mol, what is the standard sublimation temperature for iodine?

7. Consider the following addition reaction: C2H2(g) + 2 H2(g) → C2H6(g) The following data is known for this reaction: Substance

S° (J/mol K)

ΔHf (kJ/mol)

C2H2(g)

200.9

226.7

H2(g)

130.7

0

C2H6(g)

????

–84.7

(a) If the value of ΔS ° for the reaction is –232.7 J/mol K, calculate the value of S ° for C2H6 gas.

(b) Calculate the enthalpy change ΔH ° for the reaction.

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(c) Assume that Boltzmann’s equation applies when the system is in equilibrium. What is the temperature at this point (use the non-SI unit of °C)?

8. Consider the following reaction: 2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g) (a) Predict the sign of ΔS° for the reaction at 298 K. Explain the basis of your prediction.

(b) Calculate the actual value of ΔS° for the reaction. Assume the sulfur produced is the rhombic form. Does your calculation justify your prediction? Explain.

9. The combustion of methanol is described by the following equation:

2 CH3OH(l) + 3 O2(g) → 4 H2O(l) + 2 CO2(g) The value of ∆S° for this reaction is –38.6 cal/mol K at 25°C. One calorie is the energy required to warm 1 g of water by 1°C. One calorie is equivalent to 4.18 J. Data collected for this reaction: Reaction Species

∆H° (kcal/mol) at SATP

S° (cal/mol K) at SATP

CH3OH(l)

–57.0

30.3

H2O(l)

–68.3

16.7

CO2(g)

–94.0

51.1

f

Calculate the standard absolute entropy S° per mole of O2(g). Convert this value to units of J/mol K.

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10. Use the values in Table A8 in the appendix to calculate ΔS ° for each of the following equations. (It is important to use the values in the appendix because entropy values may differ slightly from one source to another. It is important to refer to the recommended source for reference values to ensure consistency in answers.) (a) H2(g) + ½ O2(g) → H2O(g)

(b) 3 O2(g) → 2 O3(g)

(c) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

(d) P4O10(s) + 6 H2O(l) → 4 H3PO4(s) (Given S° for solid phosphoric acid = 110.54 J/mol K.)

(e) The complete combustion of octane in gasoline

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