Entropy, Free Energy, and Equilibrium Chapter 18

Thermodynamics is the study of heat and other forms of energy involved in chemical or physical processes.

First Law of Thermodynamics The first law of thermodynamics is essentially the law of conservation of energy applied to a thermodynamic system. Recall that the internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system:  U = U f – U i

Exchanges of energy between the system and its surroundings are of two types: heat, q, and work, w. Putting this in an equation gives us the first law of thermodynamics.  U = q + w

Sign Convention for q When heat is evolved by the system, q is negative. This decreases the internal energy of the system. When heat is absorbed by the system, q is positive. This increases the internal energy of the system.

Sign Convention for w Recall from Chapter 6 that w = – P  V. When the system expands,  V is positive, so w is negative. The system does work on the surroundings, which decreases the internal energy of the system. When the system contracts,  V is negative, so w is positive. The surroundings do work on the system, which increases the internal energy of the system.

Here the system expands and evolves heat from A to B. Zn 2+ (aq) + 2Cl - (aq) + H 2 (g)  V is positive, so work is negative.

At constant pressure: q P =  H The first law of thermodynamics can now be expressed as follows:  U =  H – P  V

To understand why a chemical reaction goes in a particular direction, we need to study spontaneous processes. A spontaneous process is a physical or chemical change that occurs by itself. It does not require any outside force, and it continues until equilibrium is reached.

A ball will roll downhill spontaneously. It will eventually reach equilibrium at the bottom. A ball will not roll uphill spontaneously. It requires work.

Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2

Examining some spontaneous processes will clarify this definition. First, heat energy from a cup of hot coffee spontaneously flows to its surroundings—the table top, the air around the cup, or your hand holding the cup. The entropy of the system (the cup of hot coffee) and its surroundings has increased.

The rock rolling down the hill is a bit more complicated. As it rolls down, the rock’s potential energy is converted to kinetic energy. As it collides with other rocks on the way down, it transfers energy to them. The entropy of the system (the rock) and its surroundings has increased.

Now consider a gas in a flask connected to an equal-sized flask that is evacuated. When the stopcock is open, the gas will flow into the evacuated flask. The kinetic energy has spread out, and the entropy of the system has increased.

spontaneous nonspontaneous 18.2

Does a decrease in enthalpy mean a reaction proceeds spontaneously? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H 0 = kJ H + (aq) + OH - (aq) H 2 O (l)  H 0 = kJ H 2 O (s) H 2 O (l)  H 0 = 6.01 kJ NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H 0 = 25 kJ H2OH2O Spontaneous reactions 18.2

The first law of thermodynamics cannot help us to determine whether a reaction is spontaneous as written. We need a new quantity—entropy. Entropy, S, is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy.

Examining some spontaneous processes will clarify this definition. First, heat energy from a cup of hot coffee spontaneously flows to its surroundings—the table top, the air around the cup, or your hand holding the cup. The entropy of the system (the cup of hot coffee) and its surroundings has increased.

Entropy (S) is a measure of the randomness or disorder of a system. orderS disorder S  S = S f - S i If the change from initial to final results in an increase in randomness S f > S i  S > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l)  S >

W = 1 W = 4 W = 6 W = number of microstates S = k ln W  S = S f - S i  S = k ln WfWf WiWi W f > W i then  S > 0 W f < W i then  S < 0 Entropy 18.3

Processes that lead to an increase in entropy (  S > 0) 18.2

How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (  S < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (  S < 0) (c) Heating hydrogen gas from 60 0 C to 80 0 C Randomness increases Entropy increases (  S > 0) (d) Subliming dry ice Randomness increases Entropy increases (  S > 0) 18.3

In the case of the cup of hot coffee, as heat moves from the hot coffee, molecular motion becomes more disordered. In becoming more disordered, the energy is more dispersed.

Likewise, when the gas moves from one container into the evacuated container, molecules become more disordered because they are dispersed over a larger volume. The energy of the system is dispersed over a larger volume.

When ice melts, the molecules become more disordered, again dispersing energy more widely. When one molecule decomposes to give two, as in N 2 O 4 (g)  2NO 2 (g) more disorder is created because the two molecules produced can move independently of each other. Energy is more dispersed.

In each of these cases, molecular disorder increases, as does entropy. Note: This understanding of entropy as disorder applies only to molecular situations in which increasing disorder increases the dispersion of energy. It cannot be applied to situations that are not molecular—such as a desk.

Entropy State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, enthalpy, pressure, volume, temperature, entropy 18.3

First Law of Thermodynamics Energy can be converted from one form to another Second Law of Thermodynamics The entropy of the universe (i.e system+surrounding) increases in a spontaneous process and remains unchanged in an equilibrium process.  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: 18.4

Third Law of Thermodynamics A substance that is perfectly crystalline at zero Kelvin (0 K) has an entropy of zero.

The standard entropy of a substance—its absolute entropy, S°—is the entropy value for the standard state of the species. The standard state is indicated with the superscript degree sign. For a pure substance, its standard state is 1 atm pressure. For a substance in solution, its standard state is a 1 M solution.

Entropy Changes in the System (  S sys ) aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - The standard entropy of reaction (  S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn 18.4 What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = J/K mol S 0 (O 2 ) = J/K mol S 0 (CO 2 ) = J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = – [ ] = J/K mol

When wine is exposed to air in the presence of certain bacteria, the ethyl alcohol is oxidized to acetic acid, giving vinegar. Calculate the entropy change at 25°C for the following similar reaction: CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l) The standard entropies, S°, of the substances in J/(K  mol) at 25°C are CH 3 CH 2 OH(l),161; O 2 (g), 205; CH 3 COOH(l), 160; H 2 O(l), 69.9.

CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l)  S = J/K – 366 J/K  S = –136 J/K S° J/(mol  K) n mol1111 nS° J/K

Entropy Changes in the System (  S sys ) 18.4 When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes,  S 0 > 0. If the total number of gas molecules diminishes,  S 0 < 0. If there is no net change in the total number of gas molecules, then  S 0 may be positive or negative but  S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s) The total number of gas molecules goes down,  S is negative.

The opening of Chapter 6, on thermo- chemistry, describes the endothermic reaction of solid barium hydroxide octahydrate and solid ammonium nitrate: Ba(OH) 2  8H 2 O(s) + 2NH 4 NO 3 (s)  2NH 3 (g) + 10H 2 O(l) + Ba(NO 3 ) 2 (aq) Predict the sign of  S° for this reaction. 3 moles of reactants produces 13 moles of products. Solid reactants produce gaseous, liquid, and aqueous products.  S° is positive.

Entropy Changes in the Surroundings (  S surr ) Exothermic Process  S surr > 0 Endothermic Process  S surr <

Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature S = k ln W W = 1 S = 0

 S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: Gibbs Free Energy For a constant-temperature process:  G =  H sys -T  S sys Gibbs free energy (G)  G < 0 The reaction is spontaneous in the forward direction.  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.  G = 0 The reaction is at equilibrium. 18.5

aA + bB cC + dD G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - The standard free-energy of reaction (  G 0 ) is the free- energy change for a reaction when it occurs under standard- state conditions. rxn Standard free energy of formation (  G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f

2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x– x–237.2 ] – [ 2x124.5 ] = kJ Is the reaction spontaneous at 25 0 C?  G 0 = kJ < 0 spontaneous 18.5

Calculate the free-energy change,  G°, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation. CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l)

CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l)  G f °, kJ/mol– –392.5 –237.2 n, mol1111 n  G f °, kJ–174.80–392.5–237.2 –174.8 kJ–629.7 kJ  G° = –454.9 kJ  G° = –629.7 – (–174.8)

 G =  H - T  S 18.5

CaCO 3 (s) CaO (s) + CO 2 (g)  H 0 = kJ  S 0 = J/K  G 0 =  H 0 – T  S 0 At 25 0 C,  G 0 = kJ  G 0 = 0 at C 18.5 Temperature and Spontaneity of Chemical Reactions Equilibrium Pressure of CO 2

Gibbs Free Energy and Phase Transitions H 2 O (l) H 2 O (g)  G 0 = 0=  H 0 – T  S 0  S = T HH = kJ 373 K = 109 J/K 18.5

Using standard enthalpies of formation,  H f ° and the value of  S° from the previous problem, calculate  G° for the oxidation of ethyl alcohol to acetic acid. CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l)

CH 3 CH 2 OH(l) + O 2 (g)  CH 3 COOH(l) + H 2 O(l)  H f ° kJ/mol – –487.0 –285.8 n mol1111 n  H f ° kJ –277.60–487.0 –285.8 –277.6 kJ –772.8 kJ  H° = –495.2 kJ  S° = –136 J/K T = 298 K

 H° = –495.2 kJ  S° = –136 J/K = –0.136 kJ/K T = 298 K  G° =  H° – T  S°  G° = –495.2 kJ – (298 K)(–0.136 kJ/K)  G° = –495.2 kJ kJ  G° = –454.7 kJ The reaction is spontaneous.

Gibbs Free Energy and Chemical Equilibrium  G =  G 0 + RT lnQ R is the gas constant (8.314 J/K mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium  G = 0 Q = K 0 =  G 0 + RT lnK  G 0 =  RT lnK 18.6

Free Energy Versus Extent of Reaction  G 0 < 0  G 0 > 0

 G 0 =  RT lnK 18.6

18.7 ATP + H 2 O + Alanine + Glycine ADP + H 3 PO 4 + Alanylglycine Alanine + Glycine Alanylglycine  G 0 = +29 kJ  G 0 = -2 kJ K < 1 K > 1

18.7 The Structure of ATP and ADP in Ionized Forms