CHEM 3310 Thermodynamics Change in Gibbs Free Energy

CHEM 3310 Thermodynamics Change in Gibbs Free Energy, G

Review According to the 2 nd law of Thermodynamics, a process is spontaneous if Suniverse = Ssystem + Ssurroundings > 0 Even though Ssystem<0, if there is enough changes to the surroundings such that Ssurroundings>0, such that Ssystem + Ssurroundings >0 the process will be spontaneous. As chemists, we are interested in the criteria of spontaneity as they apply to chemical reactions. Determining Suniverse is not always easy. We need to modify the method of determining the spontaneity of a process by only looking within the system! CHEM 3310 2

Review Recall how H and S favour spontaneity. 1. Spontaneous processes are: • favoured by a decrease in H (exothermic) • favoured by an increase in S 2. Nonspontaneous process, • favoured by an increase in H (endothermic) • favoured by an decrease in S We will expand on this. CHEM 3310 3

Gibbs Free Energy, G At this point, we introduce a new function, G, such that G = H – TS where G is called the Gibbs free energy. For a process that is carried out under constant T and P, G = (H – TS) = H – (TS) at constant T G = H – T S G is the change in Gibbs free energy. G is a state function and has the properties of a state function. The Gibbs energy is the maximum useful work that a system can do on the surroundings when the process is carried out under constant temperature and pressure. At constant P, qsurr = -qsys = - Hsys T Ssurr = -qsurr = - Hsys T Suniv = T Ssurr + T Ssys = - Hsys + T Ssys G = – T Suniv for constant T and P CHEM 3310 4

Change in Gibbs Free Energy, G Process carried out under constant T and P G Irreversible and Spontaneous <0 Reversible =0 Irreversible and Nonspontaneous >0 G = H – T S Enthalpy term Entropy term The change in the free energy of a system that occurs during a reaction can be measured under any set of conditions. If the data are collected under standard-state conditions, the result is the standard-state free energy of reaction (Go). Go = Ho - T So CHEM 3310 5

Change in Gibbs Free Energy, G G = H – T S Enthalpy term Summary: H S G Case I <0 >0 Always < 0 Spontaneous at all T Case II >0 <0 Always > 0 Nonspontaneous at all T Case III >0 >0 > 0 at low T < 0 at high T Entropy term Nonspontaneous at low T and becomes spontaneous as T is raised Case IV Factors favouring a spontaneous process, making G < 0 <0 <0 < 0 at low T Spontaneous at low T and becomes > 0 at high T nonspontaneous as T is raised 25 o. C is considered “low T”. As a result, exothermic reactions are in general usually spontaneous at room temperature and 1 atmosphere. [Note: H is a fairly reliable indicator of a reaction’s spontaneity at room temperature, but it is NOT a general criterion. ] CHEM 3310 6

Change in Gibbs Free Energy, G Summary: G = H – T S H S G Case I <0 >0 Always < 0 Spontaneous at all T Case II >0 <0 Aways > 0 Nonspontaneous at all T Case III >0 >0 > 0 at low T < 0 at high T Nonspontaneous at low T and becomes spontaneous as T is raised Case IV <0 <0 < 0 at low T > 0 at high T Spontaneous at low T and becomes nonspontaneous as T is raised Enthalpy term Entropy term G versus T graphs at constant pressure CHEM 3310 7

G = H – T S Change in Gibbs Free Energy, G Example: (a) Sketch G versus T for the following reaction. (b) Calculate the cross-over temperature. 2 H 2 S (g) + SO 2 (g) 3 S (s) + 2 H 2 O (g) Enthalpy term Entropy term Ho = -35. 00 kcal Go = -21. 73 kcal Answers: (a) Expect S < 0 Go = Ho – T So -21. 73 = -35. 00 – 298( So) So = -44 cal K-1 Ho < 0 and So < 0 CHEM 3310 (b) Go = Ho – T So 0 = -35. 00 – T(-0. 044) T = 795 K = 522 o. C Process is spontaneous below 522 o. C. 8

Change in Gibbs Free Energy, G G = H – T S Enthalpy term Entropy term Example: Would you invest your money and time in developing a glue which is endothermic as it sets? Answer: Glue (l) Glue (s) H > 0 S < 0 CHEM 3310 Expect G > 0 over all temperature. This is a nonspontaneous process. The glue will never set. 9

Standard Gibbs Free Energy of formation, Gfo • Gf of a substance refers to the reaction in which that substance is formed from the elements as they exist in their most stable forms at 1 atm pressure and (usually) 298 K. • Standard free energies of formation, Gf are analogous to standard enthalpies of formation, Hf. • Like the enthalpies of elements in their standard states (1 atm, 25 o. C) is assigned zero, we assign a free energy of zero to each free element in its standard state. • Gf can be looked up in tables or calculated from So and Hf. CHEM 3310 10

Standard Gibbs Free Energy of formation, Gfo Example: What is the standard free energy for the burning of n-pentane? C 5 H 12 (g) + 8 O 2 (g) 5 CO 2 (g) + 6 H 2 O (g) Given Gf (CO 2 (g)) = -394. 4 k. J mole-1 Gf (H 2 O (g)) = -228. 6 k. J mole-1 Gf (C 5 H 12 (g)) = -8. 37 k. J mole-1 Goreaction = 5 Gf (CO 2 (g)) + 6 Gf (H 2 O (g ) - Gf (C 5 H 12 (g)) – 8 Gf (O 2 (g)) = 5(-394. 4) + 6(-228. 6) – (-8. 37) – 8(0) = -3335. 2 k. J G < 0 Process is spontaneous. CHEM 3310 11

G° Examples Given that at 400°C the reaction 4 Cu(s) + O 2(g) 2 Cu 2 O(s) G° = -250 k. J/mol 1. What is G° at 400°C for the following reactions a) 2 Cu(s) + ½O 2(g) Cu 2 O(s) G° = ½(-250 k. J/mol) = -125 k. J/mol b) Cu 2 O(s) 2 Cu(s) + ½O 2(g) G° = -½(-250 k. J/mol) = 125 k. J/mol 2. Will Cu 2 O(s) spontaneously decompose at 400°C with the compounds in their standard states? G° = >0, ∴ no 3. What happens if the Cu 2 O(s) decomposition reaction is made to run simultaneously with C(s) + ½O 2(g) CO(g) G° = -175 k. J/mol C(s) + ½O 2(g) CO(g) Cu 2 O(s) 2 Cu(s)+ ½O 2(g) Cu 2 O(s) + C(s) 2 Cu(s) + CO(g) G° < 0, ∴ reaction is spontaneous G° = -175 k. J/mol G° = 125 k. J/mol G° = -50 k. J/mol 12

G° Examples Given for water at 25°C H°vap = 42 k. J/mol S°vap = 119 J mol-1 K-1 What is G°vap at 25°C? Will water spontaneously evaporate at 25°C? G°vap = H° - T S° = 42 k. J/mol –(298 K)(0. 119 k. J mol-1 K-1) G°vap = 7 k. J/mol Since G°vap > 0, water @ 25°C will not evaporate under standard conditions Does this agree with your intuition? 13

Change in Gibbs Free Energy, G G = H – T S • For many reactions, the rates at 25 o. C are too low to be useful. We need to find G at other temperatures. Enthalpy term Entropy term • We will assume that Ho and So to be constant and use them to estimate G at temperatures other then 25 o. C. Example: Estimate the G of the formation and decomposition of ammonia at 25 o. C and at 427 o. C. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Given Ho = -92. 22 k. J and So = -0. 1989 k. J K-1 Answer: Go = -92. 22 – (298)(-0. 1989) = - 32. 96 k. J G 427°C -92. 22 – (427+273)(-0. 1989) = 47 k. J Go therefore describes this reaction when all three components are present at 1 atm pressure. Go < 0, therefore, expect the process to be spontaneous. Under standard condition, the equilibrium favours the products. G 427°C > 0 Expect that the process is nonspontaneous. Therefore, we can produce ammonia at 25 o. C, but not at 427 o. C spontaneously. But commercial processes synthesize ammonia at near 500 o. C. Why? CHEM 3310 14

G = Go + RT ln Q Gibbs Free Energy and Keq N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) ΔG = ΔH - TΔS ΔG° = ΔH° - TΔS° For ideal gases ΔH = ΔH° ΔG = ΔH° - TΔS For 1 mole of gas ΔS = Sf – Si = R ln Pi Pf = -R ln Pf Pi Let i be the standard state Sf = S° - R ln CHEM 3310 P P° = S° - R ln P 1 = S° - R ln P 15

G = Go + RT ln Q Gibbs Free Energy and Keq N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) ΔG = ΔH° - TΔS SN 2 = S°N 2 – Rln PN 2 SH 2 = S°H 2 – Rln PH 2 SNH 3 = S°NH 3 – Rln PNH 3 ΔSrxn = 2(S°NH 3 – Rln PNH 3) – (S°N 2 – Rln PN 2) – 3(S°H 2 – Rln PH 2) ΔSrxn = 2 S°NH 3 – S°N 2 – 3 S°H 2+ Rln ΔSrxn = ΔS°rxn + Rln CHEM 3310 PN 2 PH 2 3 PN 2 PH 23 2 PNH 3 16

G = Go + RT ln Q Gibbs Free Energy and Keq N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) ΔG = ΔH° - TΔS ΔSrxn = ΔS°rxn + R ln ΔG = ΔH° - TΔS°rxn – TR ln ΔG = ΔG° + RT ln PN 2 PH 23 2 PNH 3 PN 2 PH 23 ΔG = ΔG° + RT ln Q CHEM 3310 17

Gibbs Free Energy and Keq G = Go + RT ln Q Go = -RT ln Keq Remember if G = 0, the system is at equilibrium. So G must be related to the equilibrium constant, Keq. Standard free energy, G°, is directly related to Keq G = Go + RT ln Q at equilibrium G = 0, and Q = Keq. 0 = Go + RT ln Keq Go = -RT ln Keq where Go is the standard free energy change R is the gas constant T is the temperature in Kelvin Keq is the equilibrium constant Q is the reaction quotient CHEM 3310 18

Change in Gibbs Free Energy, G G = Go + RT ln Q Go = -RT ln Keq Example: Consider the equilibrium reaction at 427 o. C N 2 (g) + 3 H 2 (g) 2 NH 3 (g) + heat According to Le Chatelier’s Principle there are two ways to increase the yield of ammonia. 1. Increase the pressure causes the equilibrium position to shift to the right resulting in a higher yield of ammonia (since there are more gas molecules on the left hand side of the equation). 2. Decrease the temperature causes the equilibrium position to shift to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). The rate of the reaction at lower temperatures is extremely slow making the production of ammonia is negligible. Even though increasing the temperature shifts the equilibrium to the left, a higher temperature must be used to speed up the reaction. At the same time, pressure is also increased in order to increase the yield of ammonia. CHEM 3310 19

G = Go + RT ln Q Change in Gibbs Free Energy, G Example: Consider the equilibrium reaction at 427 o. C Go = -RT ln Keq Ptotal = 134 atm N 2 (g) + 3 H 2 (g) 2 NH 3 (g) If p(N 2) = 33. 0 atm, p(H 2) = 99. 0 atm, p(NH 3) = 2. 0 atm, calculate G. Evaluate Q, Recall G 427 o. C = H – T S G 427 o. C -92. 22 – (427+273)(-0. 1989) 47 k. J G G 427 o. C + RT ln Q G 47 + 0. 008314(700 K) ln (1. 3 x 10 -7) = - 45 k. J As a result of the new pressure conditions at 427 o. C, G 427 o. C < 0. Usually, a catalyst (eg Fe) is used to speed up the reaction by lowering the activation energy so that N 2 bonds and H 2 bonds can be broken more readily. CHEM 3310 20

G = Go + RT ln Q Change in Gibbs Free Energy, G Go = -RT ln Keq Example: Consider the equilibrium reaction at 427 o. C N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Ptotal = 134 atm Under this condition, p(N 2) = 33. 0 atm, p(H 2) = 99. 0 atm, p(NH 3) = 2. 0 atm, G < 0. 200 o. C and 750 atm gives near 100% yield. But there are safety concerns of equipment running at high pressure. In practice, commercial process is carried out around 200 atm and 500 o. C to yield 10 -20% to minimize on the cost and safety concerns of the plant. CHEM 3310 21

Synthesis of Ammonia: N 2(g) + 3 H 2(g) ⇌ 2 NH 3(g) CHEM 3310 22

Change in Gibbs Free Energy, G Nonspontaneous changes can be made to occur if they are coupled with spontaneous changes. Spontaneous Example: Iron rusts 4 Fe (s) + 3 O 2 (g) 2 Fe 2 O 3 (s) G = -1484. 4 k. J The reverse of this reaction is NOT spontaneous. Commercially, many tons of Fe 2 O 3 are used to produce iron. How do they do that? 2 Fe 2 O 3 (s) 4 Fe (s) + 3 O 2 (g) G = +1484. 4 k. J 6 CO (g) + 3 O 2 (g) 6 CO 2 (g) G = -1542. 7 k. J 2 Fe 2 O 3 (s) + 6 CO (g) 4 Fe (s) + 6 CO 2 (g) G = -58. 3 k. J Spontaneous! Iron ore (primarily rust) is reduced to iron metal in a blast furnace by a reaction with carbon monoxide, or carbon. CHEM 3310 23

Change in Gibbs Free Energy, G Gibbs energy allows us to make predictions based on thermodynamic properties of the reactants and products themselves, eliminating the need to carry out the experiment. But while thermodynamics always correctly predicts whether a given process is spontaneous, it is unable to tell whether it will take place at an observable rate. CHEM 3310 24

Go = - n. FEo Change in Gibbs Free Energy, G G is the free energy that is available to do useful work such as electrochemical work. G = - n. FE and G° = - n. FE° An electrochemical cell is able do work if its cell potential, E, is positive because G < 0. where n is the number of moles of electrons that flow in the cell, F is Faraday’s constant, 96485 coulombs / mole of electrons, Eo is the standard voltage measured of the electrochemical cell. Example: Eo=1. 10 V Zinc and Copper voltaic cell CHEM 3310 25

Change in Gibbs Free Energy, G Go = - n. FEo How much useful work can be obtained from a Zn and Cu battery operating at standard conditions when 0. 15 g Zn react? Therefore, 490 J of useful work can be obtained CHEM 3310 26

Change in Gibbs Free Energy, G Go = - n. FEo Example: Hydrogen fuel cell Use solar energy to break apart water to generate H 2 and O 2 as fuel for the fuel cell. Eo=1. 23 V Anode: 2 H 2 (g) 4 H+ (g) + 4 e- Cathode: O 2 (g) + 4 H+ (g) + 4 e- 2 H 2 O (l) CHEM 3310 2 H 2 (g) + O 2 (g) 2 H 2 O (l) E = 1. 23 V 27

Summary of Thermodynamic, Equilibrium and Electrochemical Relationships 28

Summary: • The spontaneity of a process that is carried out under constant T and P can be predicted by G G = H - T S • G < 0 is a spontaneous process. • G = 0 is when the process is in equilibrium. • G > 0 is a nonspontaneous process. G = Go + RT ln Q Go = - RT ln Keq Go = - n. FEo CHEM 3310 29