Thermodynamics Chapter 16 • You Are Responsible for Thermo Concepts from CHM Thermodynamics 151

Home , Entropy of vaporization, Spontaneous process, Standard state

Thermodynamics Chapter 16 • You are responsible for Thermo concepts from CHM Thermodynamics 151. You may want to review Chapter 8, specifically GCC CHM152 sections 2, 5, 6, 7, 9, and 10 (except “work”). • Thermodynamics: Study of energy changes in chemical reactions and physical processes. • Thermodynamics allows us to determine IF a reaction will occur. • It also tells us the direction and extent of a reaction. • It does not tell us how fast a reaction occurs (kinetics tell us this!)

o Thermochemistry Review Predicting sign for H rxn

• 1st Law of Thermodynamics: Energy is Predict whether the following processes are endothermic (need heat) or exothermic (release conserved. Energy can be converted from one form heat): to another but it cannot be created or destroyed. Decomposition If a system gives off heat, the surroundings must Acid-Base Neutralization absorb it (and vice versa). Combustion Heat flow (enthalpy, H) is defined with reference to the system Melting • System absorbs heat, H > 0, endothermic Freezing • System gives off heat, H < 0, exothermic Boiling

o o Calculating H rxn Heat of formation rxn & H rxn

• The o symbol refers to the standard state, 1.00 atm • Write heat of formation reaction for H2O(l). pressure, 25.0oC, 1.00 M for solutions • Product = 1 mol H2O(l) o H f = Standard Molar Enthalpy of Formation • Reactants = elements in most stable form

• The enthalpy change for forming 1 mole of a  H2(g) + ½ O2(g)  H2O(l) compound from its elements in their standard states. o o o Calculate H for the following reaction.  Appendix B lists values for H f, G f and S • rxn  Ho = 0 for an element in its most stable form f 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)  Know most stable forms for the elements! o o o H f [NO(g)] = 90.2 kJ/mol; H f [H2O(g)] = -241.8 kJ/mol  H f units: kJ/mol o o H f [NH3(g)] = -46.1 kJ/mol; H f [O2(g)] = ? o o o • H rxn =  n H f (products) -  n H f (reactants) • n = moles (coefficient in balanced reaction) • Answer: Hrxn = -905.6 kJ

1 16.1 Spontaneous Processes Spontaneous Chemical Reactions

• A spontaneous process proceeds on its own • Spontaneous processes without any external influence. The reverse  hot object cools of a spontaneous process is always  Ice melts at T > 0 C nonspontaneous –it does not occur naturally.  iron rusts  Cs reacts with H2O Consider the expansion of a gas into a vacuum. This happens spontaneously. The reverse • Note that spontaneous does not mean fast! process does not!  Rusting of iron is spontaneous but occurs slowly!

• What makes a process spontaneous?  Spontaneity is determined by both the enthalpy and entropy of a reaction.

Enthalpy, Entropy and Spontaneity Entropy (S) Nature tends toward state of lower energy. • Entropy is a measure of disorder or  Many spontaneous processes are exothermic (e.g. combustion reactions have H < 0), but some randomness. spontaneous processes are endothermic (e.g. ice melting has H > 0). • Units of Entropy (S): J/Kmol Nature tends to become more disordered

or random.  Entropy of a system typically increases, S > 0 (+S = • Entropy of element in most stable form is increase in disorder) not zero!  Example: an office naturally becomes more messy over time.

Which has the greater entropy? Enthalpy & Entropy: H & S

• Solid, liquid or gas? (ice, water, or steam) S (s) < S (l) < S (g), Section 10.4

Melting: H ?, S ?

Boiling:

H ?, S ?

2 Which has the greater entropy? Which has the greater entropy?

• Low temperature or high temperature? • Solute and solvent or solution? • Consider gases at different temperatures • Solution has more arrangements, more motion • High T: more motion, more possible • Section 11.2 arrangements • Figure 16.7

Which has the greater entropy? Group Qz #17

• Reactants or products formed from them? Which has greater entropy?

N2(g) + O2(g)  2NO(g)

• The same number of molecules are present & all are gases; therefore no difference in entropy. • Worked example 16.1-3; Problem 16.2, 16.3

Entropy Summary Predict Entropy change

• Temperature changes Increase: S > 0 (more energy, more • Predict whether entropy increases (S > positions, more possible arrangements) 0) or decreases (S < 0) for the following • Phase changes processes. Explain why! 2H (g) + O (g)  2H O(l) Boiling, melting: S > 0 • 2 2 2 + - • Creating more moles • KBr (s)  K (aq) + Br (aq) • 4 Al(s) + 3 O (g)  2Al O (s) S > 0 2 2 3 • I (s)  I (g) • Salts dissolving 2 2 • PCl (s)  PCl (l) + Cl (g) S > 0 (usually) 5 3 2

3 Entropy and Temperature Entropy and Temperature

• 3rd Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. • This allows us to calculate entropy (S) and changes in entropy (S); unlike enthalpy where we can only measure change! • Use Standard Molar Entropies (Table 16.1) - entropy of 1 mole of a pure substance at 1 atm pressure and 25oC (J/mol•K) o  Calculate like H rxn (products - reactants)

Standard Molar Entropies Entropy Changes

o • Notice which phases have lowest values? Highest? • Calculate S rxn for:

• 2Na(s) + Cl2(g)  2NaCl(s)

• Na(s) = 51.05 J/mol•K, Cl2(g) = 223.0 J/mol•K, NaCl(s) = 72.38 J/mol•K

• Worked example 16.3, Problem 16.5

Entropy, 2nd Law of Thermo. Entropy

nd • 2 Law of Thermodynamics: in any • Stotal = Ssystem + Ssurroundings spontaneous process, the total entropy of a system and its surroundings always • Stotal > 0, the reaction is spontaneous increases. • Stotal < 0, the reaction is nonspontaneous  The system is the chemical reaction itself. • Stotal = 0, the reaction is at equilibrium

• Stot is positive for a spontaneous process; S is 0 at equilibrium • All reactions proceed spontaneously in a tot direction that increases the entropy of the  Stot = Ssystem + Ssurroundings system plus the surroundings.

4 nd 2 Law of Thermodynamics Calculating Ssurr

• Exothermic: heat leaves system, enters • We know how to calculate Ssystem (previous surroundings, surroundings have more energy which gives them more disorder calculations); now calculate Ssurr • Figure 16.9  Ssurr > 0 when exothermic: H < 0 • Endothermic: heat enters system, leaves surroundings, surroundings have less energy which gives them more order

 Ssurr < 0 when endothermic: H > 0  Ssurr  -Hrxn and Ssurr  1/T

• Therefore: Ssurr = -Hrxn/T

• STot = Ssys -Hrxn/T

Free Energy Change in Gibbs Free Energy

• Section 8.14: Gibbs Free Energy Change (G) • Gibbs free energy change or free energy change = G  Definition: maximum amount of energy available to do work on the surroundings • E or H: some energy is used to do work  Takes into account enthalpy and entropy to predict within the system (rearrange particles, new spontaneity of a reaction. bonds, ...) and the rest is available to do work on the surroundings (G) • G = H - TS (G in units of kJ/mol) • G < 0  Spontaneous process  H is enthalpy (kJ/mol), T is temperature (K), S is change in entropy (J/mol·K) • G > 0  Nonspontaneous process • In any spontaneous process (constant T and P), • G = 0  Process is at equilibrium the free energy of the system always decreases!

Free Energy Free Energy

• Notice that the TS term is temperature dependent. Temperature plays a part in predicting spontaneity. Endothermic processes are spontaneous at higher temps (TS > H) Exothermic processes are spontaneous at lower temps (TS < H)

5 Free Energy Entropy of Phase Changes

• Phase changes occur at equilibrium (G = 0) • Using ∆G = ∆H – T∆S, we can predict the sign of  H2O(s)  H2O(l) ∆G from the sign of ∆H and ∆S.  H = 6.01 kJ/mol; S = 22.0 J/K•mol  Solid and liquid are in equilibrium • If ∆H is negative and ∆S is positive, ∆G will always be  G = H - TS = 0, solving for temperature tells us the negative. melting point! • If both ∆H and ∆S are negative, ∆G will be – at low T. • T = H/S = 273 K ; this is the melting point of water • If ∆H is positive and ∆S is negative, ∆G will always be positive. • Melting Ice: • If both ∆H and ∆S are positive, ∆G will be – at high T. • 263K (-10oC): G = +0.22 kJ/mol (nonspontaneous T< 0C) • 273K: G = 0.00 kJ/mol (equilibrium; T = 0 C) • 283K: G = -0.22 kJ/mol (spontaneous, T > 0 C)

Calculating Entropy of Vaporization Group Quiz 18 • Phase changes occur at eq: S = H /T • The boiling point of water is 100 oC and the • Calculate the boiling point of ethanol if enthalpy change for the conversion of entropy of vaporization is 111.9 J/K and

water to steam is Hvap = 40.67 kJ/mol. Hvap = 39.3 kJ What is the entropy change for • Watch units! vaporization, Svap, in J/(K·mol)? • Svap = Hvap / T kJ 1000J 40.67  • Svap = mol 1 kJ = 109 J 373K K mol

Find T for spontaneous rxn Standard Free Energy Changes

• Calculate G given H= -227 kJ, • The free energy of a substance depends on S = -309 J/K, T = 1450 K. temperature, pressure, and physical states (like enthalpy and entropy). We must look at standard-state conditions: • Is this process spontaneous at this  Solids, liquids, gases: pure form at 1 atm temperature? If not, calculate the  Solutions at 1 M concentration, gases at 1 atm o temperature (in oC) at which this reaction  Room temperature: 25 C (298 K) becomes spontaneous. • Standard free energy change (Go) is free energy change with reactants and products in their standard states.

6 o Standard Free Energy Changes G f

• Can get Go from Go = Ho - TSo  Use Go to predict spontaneity in the standard state • Which one of these reactions corresponds • Can also get values of Go from free energies o o to G f of H2O(g)? of formation: G f (formation from the elements) o 2H2(g) + O2(g)  2H2O(g) • G f = 0 for an element in its stable form o o o H (g) + ½O (g)  H O(g)  G rxn =  n G fproducts -  n G freactants 2 2 2 • What are the characteristics of a formation H2(g) + ½ O2(s)  H2O(g) reaction? o o o • G = n G f products - n G f reactants  1 mole of product formed  Reactants are elements in standard state. • Values of Go , Ho , So are listed for in App B f f

o o Gf in kJ/mol G calc CaO(s) -604.20 Relating Go to G and K Ca(OH)2(s) -896.76 H2O(g) -228.59 • G is not standard. It changes as a chemical H2O(l) -237.18 reaction proceeds, as concentrations and/or • Calculate Go for these reactions and temperatures change. predict whether they will be spontaneous. • Go IS standard and does NOT change during a reaction. • 2H2(g) + O2(g)  2H2O(g) • G = Go + RT lnQ • CaO(s) + H2O(l)  Ca(OH)2(s) • Q (from Ch. 13 - not necessarily at equil.)

• RT lnQ is correction for non-standard conditions; R = 8.314 J/mol·K

 Example 16.7, Problem 16.12

Go and Equilibrium Free Energy Curve for +G

Figure 16.10 G < 0 (-G), product-favored G > 0, Reactant favored

Eq mix

7 Go and K Go and Equilibrium

G = Go + RT ln Q • At equilibrium, G = 0 and Q = K

Thus, Go = - RT ln K • Go > 0 when K << 1 (lies toward reactants) • Go < 0 when K >>1 (lies toward products) • Go = 0 when K = 1 (roughly equal amounts of reactants and products) • See Table 16.4

Calculating K Group Quiz #19

Go is -24.7 kJ/mol for the formation of Calculate K (the equilibrium constant) for the methanol. following unbalanced reaction at 25.0oC. Is this reaction reactant- or product-favored? C(s) + ½ O2(g) + 2H2(g)  CH3OH(g) Calculate the equilibrium constant, K, at 25  C for this reaction. H2(g) + Br2(l)  HBr(g)

o G f (HBr) = -53.4 kJ/mol

Calculations Practice Calculations Practice

At 25 C, K for acetic acid is 1.810-5. • ∆G˚ for the reaction H2(g) + I2(g)  2 HI(g) a is 2.60 kJ/mol at 25°C. a) Predict the sign of Go for + - CH3COOH(aq) + H2O(l)  H3O (aq) + CH3COO (aq). • In one experiment, the initial pressures are b) Calculate Go at 25 C. P = 4.3 atm, P = 0.34 atm, and P = H2 I2 HI c) Calculate G at 25 C for the acetic acid equilibrium 0.23 atm. + - reaction, when [H3O ] = 0.020 M, [CH3COO ] = 0.010 • Calculate G and predict the direction that o M and [CH3COOH] = 0.10 M. (Use G from part b.) this reaction will proceed.