Entropy, Free Energy, and Equilibrium Chapter 17.

Entropy, Free Energy, and Equilibrium Chapter 17

Thermodynamics Thermodynamics is the study of the relationship between heat and other forms of energy in a chemical or physical process. We have discussed the thermodynamic property of enthalpy, H. We noted that the change in enthalpy equals the heat of reaction at constant pressure. In this chapter we will define enthalpy more precisely, in terms of the energy of the system.

Thermodynamics Thermodynamics can be used to predict if a reaction will occur………. kinetics tell us how fast the reaction will occur.

First Law of Thermodynamics “energy can be converted from on form to another, but it cannot be created or destroyed” One measure of these changes is the amount of heat given off or absorbed by a system during a constant pressure process, which chemists define as a change in enthalpy (∆H)

Second Law of Thermodynamics “The entropy (S) of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.” Helps explain why chemical reactions favor a particular direction.

Third Law of Thermodynamics An extension of the Second Law stating: “the entropy of a perfect crystalline substance is zero at absolute 0 K” This allows us to determine the absolute entropies of substances.

Spontaneous Processes and Entropy A spontaneous process is a physical or chemical change that occurs by itself. Examples include: A rock at the top of a hill rolls down. Heat flows from a hot object to a cold one. An iron object rusts in moist air. These processes occur without requiring an outside force and continue until equilibrium is reached.

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

spontaneous nonspontaneous

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 = -890.4 kJ H + (aq) + OH - (aq) H 2 O (l)  H 0 = -56.2 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

Does a decrease in enthalpy mean a reaction proceeds spontaneously? We cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. To make predictions of spontaneity we need another thermodynamic quantity; ENTROPY

Entropy and the Second Law of Thermodynamics The second law of thermodynamics addresses questions about spontaneity in terms of a quantity called entropy. Entropy, S, is a thermodynamic quantity that is a measure of the randomness or disorder of a system. The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function.

Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process. The net change in entropy of the system,  S, equals the sum of the entropy created during the spontaneous process and the change in energy associated with the heat flow.

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 > 0

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

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

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)

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

First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe 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:

Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. Even without knowing the values for the entropies of substances, you can sometimes predict the sign of  S o for a reaction.

Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. 1.A reaction in which a molecule is broken into two or more smaller molecules. The entropy usually increases in the following situations:

Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. The entropy usually increases in the following situations: 2.A reaction in which there is an increase in the moles of gases.

Entropy Change for a Reaction You can calculate the entropy change for a reaction using a summation law, similar to the way you obtained  H o. The entropy usually increases in the following situations: 3.A process in which a solid changes to liquid or gas or a liquid changes to gas.

A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropy of NH 2 CONH 2 is 174 J/(mol. K). The calculation is similar to that used to obtain  H o from standard enthalpies of formation.

A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropy of NH 2 CONH 2 is 174 J/(mol. K). It is convenient to put the standard entropies (multiplied by their stoichiometric coefficients) below the formulas. So:So:2 x 19321417470

A Problem To Consider Calculate the change in entropy,  S o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropy of NH 2 CONH 2 is 174 J/(mol. K). We can now use the summation law to calculate the entropy change.

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 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) = 197.9 J/K mol S 0 (O 2 ) = 205.0 J/K mol S 0 (CO 2 ) = 213.6 J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K mol

Entropy Changes in the System (  S sys ) 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.

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

Entropy Changes of Surroundings ∆S surr = -∆H sys /T ∆S univ = ∆S sys + ∆S surr Entropy Changes of universe

Standard Entropies and the Third Law of Thermodynamics The third law of thermodynamics states that a substance that is perfectly crystalline at 0 K has an entropy of zero. When temperature is raised, however, the substance becomes more disordered as it absorbs heat. The entropy of a substance is determined by measuring how much heat is required to change its temperature per Kelvin degree.

Standard Entropies and the Third Law of Thermodynamics The standard entropy of a substance or ion, also called its absolute entropy, S o, is the entropy value for the standard state of the species. Standard state implies 25 o C, 1 atm pressure, and 1 M for dissolved substances.

Standard Entropies and the Third Law of Thermodynamics The standard entroy of a substance or ion, also called its absolute entropy, S o, is the entropy value for the standard state of the species. Note that the elements have nonzero values, unlike standard enthalpies of formation,  H f o, which by convention, are zero.

Standard Entropies and the Third Law of Thermodynamics The standard entropy of a substance or ion, also called its absolute entropy, S o, is the entropy value for the standard state of the species. The symbol S o, rather than  S o, is used for standard entropies to emphasize that they originate from the third law.

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

Entropy, Enthalpy, and Spontaneity Now you can see how thermodynamics is applied to the question of reaction spontaneity. Rearranging this equation, we find This inequality implies that for a reaction to be spontaneous,  H-T  S must be negative. If  H-T  S is positive, the reverse reaction is spontaneous. If  H-T  S=0, the reaction is at equilibrium (Spontaneous reaction, constant T and P)

Free Energy Concept The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation G=H-TS. This quantity gives a direct criterion for spontaneity of reaction.

Free Energy and Spontaneity Changes in H an S during a reaction result in a change in free energy,  G, given by the equation Thus, if you can show that  G is negative at a given temperature and pressure, you can predict that the reaction will be spontaneous.

Standard Free-Energy Change The standard free energy change,  G o, is the free energy change that occurs when reactants and products are in their standard states. The next example illustrates the calculation of the standard free energy change,  G o, from  H o and  S o.

 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.

A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? Use values of  H f o and S o, from Tables in book. Place below each formula the values of  H f o and S o multiplied by stoichiometric coefficients. So:So: 3 x 130.6191.52 x 193 J/K Hfo:Hfo: 002 x (-45.9) kJ

A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? Use values of  H f o and S o, from Tables in book. You can calculate  H o and  S o using their respective summation laws.

A Problem To Consider What is the standard free energy change,  G o, for the following reaction at 25 o C? Use values of  H f o and S o, from Tables in book. Now substitute into our equation for  G o. Note that  S o is converted to kJ/K.

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–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ < 0 spontaneous

Standard Free Energies of Formation The standard free energy of formation,  G f o, of a substance is the free energy change that occurs when 1 mol of a substance is formed from its elements in their stablest states at 1 atm pressure and 25 o C. By tabulating  G f o for substances, you can calculate the  G o for a reaction by using a summation law.

A Problem To Consider Calculate  G o for the combustion of 1 mol of ethanol, C 2 H 5 OH, at 25 o C. Use the standard free energies of formation given in the book. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ Place below each formula the values of  G f o multiplied by stoichiometric coefficients.

A Problem To Consider Calculate  G o for the combustion of 1 mol of ethanol, C 2 H 5 OH, at 25 o C. Use the standard free energies of formation given in the book. Gfo:Gfo: -174.802(-394.4)3(-228.6)kJ You can calculate  G o using the summation law.

 G o as a Criteria for Spontaneity The following rules are useful in judging the spontaneity of a reaction. 1.When  G o is a large negative number (more negative than about –10 kJ), the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached.

 G o as a Criteria for Spontaneity The following rules are useful in judging the spontaneity of a reaction. 2.When  G o is a large positive number (more positive than about +10 kJ), the reaction is nonspontaneous as written, and reactants do not give significant amounts of product at equilibrium.

 G o as a Criteria for Spontaneity The following rules are useful in judging the spontaneity of a reaction. 3.When  G o is a small negative or positive value (less than about 10 kJ), the reaction gives an equilibrium mixture with significant amounts of both reactants and products.

 G =  H - T  S

CaCO 3 (s) CaO (s) + CO 2 (g)  H 0 = 177.8 kJ  S 0 = 160.5 J/K  G 0 =  H 0 – T  S 0 At 25 0 C,  G 0 = 130.0 kJ  G 0 = 0 at 835 0 C 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 = 40.79 kJ 373 K = 109 J/K

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

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

 G 0 =  RT lnK

Relating  G o to the Equilibrium Constant The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation. Here Q is the thermodynamic form of the reaction quotient.

Relating  G o to the Equilibrium Constant The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation.  G represents an instantaneous change in free energy at some point in the reaction approaching equilibrium.

Relating  G o to the Equilibrium Constant The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change,  G o, by the following equation. At equilibrium,  G=0 and the reaction quotient Q becomes the equilibrium constant K.

Relating  G o to the Equilibrium Constant This result easily rearranges to give the basic equation relating the standard free-energy change to the equilibrium constant. When K > 1, the ln K is positive and  G o is negative. When K < 1, the ln K is negative and  G o is positive.

A Problem To Consider Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction. The standard free- energy change,  G o, at 25 o C equals –13.6 kJ. Rearrange the equation  G o =-RTlnK to give

A Problem To Consider Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction. The standard free- energy change,  G o, at 25 o C equals –13.6 kJ. Substituting numerical values into the equation,

A Problem To Consider Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction. The standard free- energy change,  G o, at 25 o C equals –13.6 kJ. Hence,

Spontaneity and Temperature Change All of the four possible choices of signs for  H o and  S o give different temperature behaviors for  G o. HoHo SoSo GoGo Description ––– Spontaneous at all T +++ Nonspontaneous at all T ––+ or – Spontaneous at low T; Nonspontaneous at high T +++ or – Nonspontaneous at low T; Spontaneous at high T

Calculation of  G o at Various Temperatures In this method you assume that  H o and  S o are essentially constant with respect to temperature. You get the value of  G T o at any temperature T by substituting values of  H o and  S o at 25 o C into the following equation.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Place below each formula the values of  H f o and S o multiplied by stoichiometric coefficients.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ You can calculate  H o and  S o using their respective summation laws.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now you substitute  H o,  S o (=0.1590 kJ/K), and T (=298K) into the equation for  G f o.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now you substitute  H o,  S o (=0.1590 kJ/K), and T (=298K) into the equation for  G f o. So the reaction is nonspontaneous at 25 o C.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now we’ll use 1000 o C (1273 K) along with our previous values for  H o and  S o.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Now we’ll use 1000 o C (1273 K) along with our previous values for  H o and  S o. So the reaction is spontaneous at 1000 o C.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ To determine the minimum temperature for spontaneity, we can set  G f o =0 and solve for T.

A Problem To Consider Find the  G o for the following reaction at 25 o C and 1000 o C. Relate this to reaction spontaneity. So:So: 38.292.9213.7 J/K Hfo:Hfo: -635.1-1206.9-393.5 kJ Thus, CaCO 3 should be thermally stable until its heated to approximately 848 o C.

a) Calculate ∆G° and K p for the following equilibrium reaction at 25 °C. The ∆G° values are 0 for Cl 2 (g), -286 kJ/mol for PCl 3 (g), and -325 kJ/mol for PCl 5 (g). b) Calculate ∆G for the reaction if the partial pressures of the initial mixture are P PCl = 0.0029 atm, P PCl = 0.27 atm, and P Cl = 0.40 atm. 2 3 PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g)

WORKED EXAMPLES

Worked Example 18.2

Worked Example 18.4

Worked Example 18.5

Worked Example 18.6

Worked Example 18.8

Entropy, Free Energy, and Equilibrium Chapter 17.

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Entropy, Free Energy, and Equilibrium Chapter 17.

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