Chapter 16 Creative Commons License

Thermodynamics  Images and tables in this file have been used from the following GCC CHM152 sources:  OpenStax: Creative Commons Attribution License 4.0.  ChemWiki (CC BY-NC-SA 3.0): Unless otherwise noted, the StatWiki is licensed under a Creative Commons Attribution- Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at [email protected].  Principles of General (CC BY-NC-SA 3.0): http://2012books.lardbucket.org/pdfs/principles-of-general- chemistry-v1.0.pdf

Chemistry: OpenStax 2

Thermodynamics Thermochemistry Review • You are responsible for Thermo concepts from CHM 1st Law of Thermodynamics: Energy is conserved. 151. You may want to review Chapter 5, specifically Energy can be converted from one form to another sections 1-3. but it cannot be created or destroyed. • Thermodynamics: Study of energy changes in If a system gives off , the surroundings must chemical reactions and physical processes. absorb it (and vice versa). • Thermodynamics allows us to determine IF a reaction Heat flow (, H) is defined with will occur. reference to the system • It also tells us the direction and extent of a reaction. • System absorbs heat, H > 0, endothermic • It does not tell us how fast a reaction occurs (kinetics • System gives off heat, H < 0, exothermic tell us this!)

o o Predicting sign for H rxn Calculating H rxn Predict whether the following are endothermic The o symbol refers to the standard state, 1.00 atm (need heat) or exothermic (release heat): pressure, 25.0oC, 1.00 M for Decomposition + Ho = Standard Molar Enthalpy of Formation Acid-Base Neutralization - f The enthalpy change when 1 mole of a compound is Combustion - formed from its elements in their standard states. Melting o o o +  Appendix G lists values for H f, G f and S  Ho = 0 for an element in its most stable form Freezing -  f  Know most stable forms for the elements! Boiling + o  H f units: kJ/mol Remember breaking bonds requires energy = endo Ho =  n Ho (products) -  n Ho (reactants) but making new bonds releases energy = exo. rxn f f • n = moles (coefficient in balanced reaction)

1 1 o Heat of formation rxn & H rxn Spontaneous Processes Write heat of formation reaction for H2O(l). A spontaneous process proceeds on its own • Product = 1 mol H2O(l) without any external influence. The reverse • Reactants = elements in most stable form of a spontaneous process is always nonspontaneous –it does not occur naturally.  H2(g) + ½ O2(g)  H2O(l) Consider the expansion of a into a Calculate Hrxn for the following reaction. vacuum. This happens spontaneously. The 4NH (g) + 5O (g)  4NO(g) + 6H O(g) reverse process does not! 3 2 2 o o H f [NO(g)] = 90.2 kJ/mol; H f [H2O(g)] = -241.8 kJ/mol o o H f [NH3(g)] = -46.1 kJ/mol; H f [O2(g)] = ?

Answer: Hrxn = -905.6 kJ

Spontaneous Chemical Reactions Enthalpy, and Spontaneity

Spontaneous processes Nature tends toward state of lower energy.  hot object cools  Many spontaneous processes are exothermic (e.g.  Ice melts at T > 0 C combustion reactions have H < 0), but some  iron rusts spontaneous processes are endothermic (e.g. ice melting has H > 0).  sodium reacts with H2O • Note that spontaneous does not mean fast! Nature tends to become more disordered  Rusting of iron is spontaneous but occurs slowly! or random.  Entropy of a system typically increases, S > 0 (+S = • What makes a process spontaneous? increase in disorder)  Spontaneity is determined by both the enthalpy and  Example: an office naturally becomes more messy entropy of a reaction. over time.

Entropy (S) Entropy of Molecules

Entropy is a measure of disorder or randomness. • Units of Entropy: J K  mol

• S: standard molar entropy of an element in its most stable form is not zero! The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry -v1.0/s19-chemical-equilibrium.html, CC-BY-NC-SA 3.0 license 12

2 2 Which has the greater entropy? Entropy and temperature Solid, or gas? (ice, water, or steam) • Consider at different temperatures • High T: More motion & higher kinetic

energy so there are more possible arrangements.

S (s) < S (l) < S (g)

Entropy and solubility Entropy Summary Dissolving a solute usually increases entropy since a • Temperature changes has more possible arrangements and is more dispersed but some ionic compounds have a negative entropy change upon Increase: S > 0 (more energy, more dissolution. positions, more possible arrangements) Ionic compounds: entropy can decrease or increase • Phase changes NaCl (s): Ssoln = +43.4 J/K AlCl3 (s): Ssoln = -253.2 J/K Boiling, melting: S > 0 • Creating more moles S > 0 • Salts dissolving: S +/- but usually • Also entropy usually as complexity of molecule increases.

Predict Entropy change Answers • 2H (g) + O (g)  2H O(l): Predict whether entropy increases (S > 0) 2 2 2  S < 0; fewer particles, phase change or decreases (S < 0) for the following • KBr(s)  K+(aq) + Br-(aq): processes. Explain why!  S > 0; more particles/phase change

• 2H2(g) + O2(g)  2H2O(l) • 4 Al(s) + 3 O2(g)  2Al2O3(s): • KBr (s)  K+(aq) + Br-(aq)  S < 0; fewer particles/phase change • I2(s)  I2(g): • 4 Al(s) + 3 O2(g)  2Al2O3(s)  S > 0; phase change • I (s)  I (g) 2 2 • PCl (s)  PCl (l) + Cl (g): PCl (s)  PCl (l) + Cl (g) 5 3 2 • 5 3 2  S > 0; more particles/phase changes

3 3 o Standard Molar Entropy Standard Entropy Values (S ) Which phase has the lowest entropy? Highest? rd 3 Law of Thermodynamics: The entropy of a Substance So (J mol-1 K-1) Substance So (J mol-1 K-1) perfectly ordered crystalline substance at 0 K is C (s, diamond) 2.38 H (g) 114.6 zero. C (s, graphite) 5.740 H2 (g) 130.57 • This allows us to calculate entropy (S) and Al (s) 28.3 CH4 (g) 186.3 changes in entropy (S); unlike enthalpy where Al2O3 (s) 50.92 HCl (g) 186.8 BaSO (s) 132.2 H O (g) 188.8 we can only measure changes! 4 2 CO (g) 197.7 Standard Molar (Table 16.2) - entropy of H2O (l) 70.0 CO2 (g) 213.8 1 mole of a pure substance at 1 atm pressure and Hg (l) 75.9 C H (g) 219.3 25oC (J/mol•K) 2 4 H2O2 (l) 109.6 C2H6 (g) 229.5 o S rxn = nS(products) - nS(reactants) Br2 (l) 152.23 CH3OH (g) 239.9 CCl (l) 214.4 CCl (g) 309.7 4 4 Note: The standard entropy for a pure element in its most stable o form is NOT zero (like for H f). 20

Entropy Change for a Reaction 2nd Law of Thermodynamics

• 2nd Law of Thermodynamics: in any Calculate So for 2Na(s) + Cl (g)  2NaCl(s) rxn 2 spontaneous process, the total entropy of a S values: Na(s) = 51.05 J/mol•K, Cl2(g) = system and its surroundings always 223.0 J/mol•K, NaCl(s) = 72.38 J/mol•K increases. (system = reaction)

o • Stotal = Ssystem + Ssurroundings • S rxn = -180.3 J/K • Stotal > 0, the reaction is spontaneous • S < 0, the reaction is nonspontaneous • Entropy change is < 0. There is more order due total • Stotal = 0, the reaction is at equilibrium to the phase change. • All reactions proceed spontaneously in a direction that increases the entropy of the system plus the surroundings.

2nd Law of Thermodynamics Calculating Ssurr • Ssys is the entropy change for the reaction that we just calculated. • Exothermic: heat flows from system to surroundings, surroundings have more energy = • Ssurr is related to the enthalpy of the more disorder. (Hotter particles have more motion) system; if Heat flows into the  Ssurr > 0 when exothermic: H < 0 surroundings then their entropy increases • Endothermic: heat flows into system from due to increased motion. surroundings, surroundings have less energy which gives them more order.

 Ssurr < 0 when endothermic: H > 0

• Therefore: S = -H /T Entropy surr sys Entropy Heat Heat • STot = Ssys -Hsys/T

4 4 (G) or free energy Predicting sign for Free Energy Change • G < 0  Spontaneous process G is the maximum amount of energy available to do work • G > 0  Nonspontaneous process on the surroundings. • G = 0  Process is at equilibrium  Takes into account enthalpy and entropy to predict spontaneity of a reaction. G = H - TS • G = H - TS (G units: kJ/mol)  T must be in Kelvin! Watch units!

 The actual amount of work (wmax) obtained is always less than the maximum available because of energy lost in carrying out a process (given off as heat, light, sound, etc. energy). • In any spontaneous process (constant T and P), the free energy of the system always decreases!

Free Energy sign is T dependent when Calculating Entropy of enthalpy and entropy have same sign! • Phase changes occur at eq: S = H /T • Notice that the TS term is temperature • The of water is 100 oC and the dependent. Temperature plays a part in enthalpy change for the conversion of water to predicting spontaneity. steam is Hvap = 40.67 kJ/mol. What is the entropy change for vaporization, Svap, in Endothermic processes are spontaneous at J/(K·mol)? higher temps (TS > H) • S = H / T Exothermic processes are spontaneous at vap vap kJ 1000J lower temps (TS < H) 40.67  J • Svap = mol 1 kJ = 109 K mol 373K

Find T for spontaneous rxn Find T spontaneous rxn

• Calculate G given H= -227 kJ, S = -309 • G = +221 kJ; not spontaneous J/K, T = 1450 K.

• G = 0; H = TS

Is this process spontaneous at this • T = H / S = -227 kJ / -0.309 kJ/K temperature? If not, calculate the temperature • T = 735 K = 462 oC (in oC) at which this reaction becomes • Spontaneous when T < 462 oC spontaneous. (Hint: set G = 0 and solve for • (entropy & enthalpy are both -, so it is T!) spontaneous at T below equilibrium T.)

5 5 Standard Free Energy Changes Standard Free Energy Changes

G depends on T, P, [ ] and physical states (like 1) Can get Go from Go = Ho - TSo H and S).  Use Go to predict spontaneity in the standard state Standard-state conditions: 2) Go can also be calculated from standard free o  Solids, , and gases in pure form energies of formation: (G f) o o o  1 M Solutions , gases at 1 atm pressure G rxn =  n G f(products) -  n G f(reactants) o o  Room temperature: 25 C (298 K) G f = standard free energy of formation: the Standard free energy change (Go) is the free free energy change when 1 mole of a energy change when reactants in their standard compound is formed from its elements in their standard states. states are converted to products in their standard o states.  G f = 0 for an element in its stable form o o o  Values of G f, H f, S are listed in Appendix

G o in kJ/mol o Go calculation f Reaction for G f CaO(s) -604.20 Ca(OH)2(s) -896.76 • Which one of these reactions corresponds H2O(g) -228.59 o H O(l) -237.18 to G f of H2O(g)? 2 o o o 2H2(g) + O2(g)  2H2O(g) G = n G f (products) - n G f (reactants) H (g) + ½O (g)  H O(g) 2 2 2 Calculate Go for these reactions and H (g) + ½ O (s)  H O(g) 2 2 2 predict whether they will be spontaneous.

• 2H2(g) + O2(g)  2H2O(g)

• CaO(s) + H2O(l)  Ca(OH)2(s) o  G (H2O)= -457.18 kJ, spontaneous o  G (Ca(OH)2)= -55.38 kJ, spontaneous

Relating G to Go and Q Go and Equilibrium G < 0 (-G), product-favored G is the actual free energy – this is the free energy change under nonstandard conditions. It changes as a reaction proceeds, as concentrations and/or pressures change. Go IS standard and does NOT change during a reaction. G = Go + RT lnQ • Q = Reaction Quotient (plug the given [ ]’s and P’s into Q) • R = 8.314 J/mol·K, T in Kelvin

6 6 Free Energy Curve for +G Free Energy and Equilibrium G > 0, Reactant favored G = ΔGo + RT ln Q At equilibrium, ΔG = 0 and Q = K: ΔGo = - RT ln K

Go K Comment Go < 0 K > 1 Equilibrium mixture contains mostly products. Go > 0 K < 1 Equilibrium mixture contains mostly reactants. Go = 0 K = 1 Equilibrium mixture contains appreciable amount of reactants and products at

equilibrium 38

Calculating K from G Calculation G from K

o -5 G is -24.7 kJ/mol for the formation of At 25 C, Ka for acetic acid is 1.810 . methanol. a) Predict the sign of Go for CH COOH(aq) + H O(l)  H O+(aq) + CH COO-(aq). C(s) + ½ O2(g) + 2H2(g)  CH3OH(g) 3 2 3 3 Calculate the equilibrium constant, K, at 25  C b) Calculate Go at 25 C. for this reaction. ln K = -Go/RT = • K < 1 so G is + -(-24.7 kJ/mol /(0.008314 kJ/mol·K  298 K) • Go = -(8.314 J/molK)(298 K)(ln 1.810-5) 9.969 4 ln K = 9.969; K = e = 2.14 x 10 • Go = 2.7104 J/mol (or 27 kJ/mol)

Find G for weak acid solution Calculate Nonstandard Free Energy

∆G˚ for the reaction H2(g) + I2(g)  2 HI(g) is Calculate G at 25 C for the acetic acid 2.60 kJ/mol at 25°C. + equilibrium reaction, when [H3O ] = 0.020 M, In one experiment, the initial pressures are [CH COO-] = 0.010 M and [CH COOH] = P = 4.3 atm, P = 0.34 atm and P = 0.23 atm. 3 3 H2 I2 HI 0.10 M. (Use Go from part b.) • Calculate G and predict the direction that this reaction will proceed. Q = (0.020)(0.010)/(0.10) = 2.010-3 Q = 0.0362 G = 27 kJ/mol + (0.008314 kJ/Kmol)(298 K)(ln 0.0020) G = 2.60 kJ/mol+(8.314x10-3kJ/Kmol)(298 K)(ln 0.0362) G = -5.6 kJ/mol G = 12 kJ/mol (nonspontaneous in forward direction) -5 Spontaneous so Eq shifts right Note: Q > Ka = 1.8x10 so reaction reverses to reach eq.

7 7 Calculations Practice

• Calculate G for S(s) + O2(g)  SO2(g) when o PO2 = 0.140 atm and PSO2 = 1.24 atm at 25 C.

• Need to calculate Go and Q. o o o  G f (S(s)) = 0 kJ/mol; G f (O2(g)) = 0 kJ/mol; G f (SO2(g)) = -300.2 kJ/mol • Q = 1.24 atm / 0.140 atm = 8.8571 • Go = -300.2 kJ • G = -295 kJ

8 8