Thermodynamics AP Chemistry Ch. 16

First Law of Thermodynamics The energy of the universe is constant. * Law of conservation of energy.* CH4(g) + O2(g)  CO2(g) + H2O(g) + energy a.) How much energy is involved in the change? b.) Does energy flow into or out of the system? c.) What form does the energy assume? d.) but…Why does the reaction proceed in a given direction?

Spontaneous Process: Occurs without outside intervention (may be fast or slow). Ice melting H2O(s)  H2O(l) Iron rusting 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) Ionic solute dissolving NaCl(s)  Na+(aq) + Cl-(aq) Other scenarios to consider spontaneity on page 750

Entropy (S): Thermodynamic function that measures randomness or disorder. Spontaneous process has an increase in entropy.   Entropy describes the number of arrangements available to a system. The tendency to mix is due to the increased volume available to the particles of each component of the mixture. For example, when two liquids are mixed, the molecules of each liquid have more available volume and thus more available positions (positional probability).

Entropy Example 1 For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature. (a) solid CO2 and gaseous CO2   (b) N2 gas at 1.0 atm and N2 gas at 0.010 atm

Entropy Example 2 Predict the sign of the entropy change for each of the following processes: ΔS = Sfinal – Sinitial (a) solid sugar is added to water to form a solution.   (b) iodine vapor condenses on a cold surface to form crystals. Positive (entropy increases) Negative (entropy decreases)

Entropy and the 2nd Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe. ΔSuniv = ΔSsys + ΔSsurr ΔSuniv is +, process is ΔSuniv is -, process is ΔSuniv is 0, process is Energy is conserved in the universe but entropy is not. + change in entropy is spontaneous! 0 at equilibrium state

Effect of Temperature on Spontaneity The vaporization of water: H2O(l)  H2O(g) 1 mol (18 g) = 18 mL liquid = 31 L gas @ 100°C Is this spontaneous? ΔSuniv = ΔSsys + ΔSsurr ΔSsys =  Positive (gas has more disorder than liquid water) Exothermic process: ΔSsurr is positive (+) Endothermic process: ΔSsurr is negative (-) Entropy changes for the surroundings are primarily determined by the flow of energy (heat) endo or exo

Effect of Temperature on Spontaneity Entropy (S) changes in surroundings are primarily determined by heat flow. The significance of exothermicity (magnitude of ΔSsurr)depends on the temperature at which this process occurs.

Summary 1. The sign of ΔSsurr depends on the direction of heat flow. a. Exothermic: ΔSsurr is positive (+) b. Endothermic: ΔSsurr is negative (-) * Remember that this is really entropy (S) described in terms of energy. (Nature seeks lowest possible energy.) 2. The magnitude of ΔSsurr depends upon the temperature a. Exothermic: ΔSsurr = + quantity of heat (J) / temperature (K) b. Endothermic: ΔSsurr = - quantity of heat (J) / temperature (K) ΔSsurr = - ΔH / T

Example Problem In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore: Iron is used to reduce antimony in sulfide ores: Sb2S3(s) + 3 Fe(s)  2 Sb(s) + 3 FeS(s) ΔH = - 125 kJ Carbon is used as the reducing agent for oxide ores: Sb4O6(s) + 6 C(s)  4 Sb(s) + 6 CO(g) ΔH = + 778 kJ Calculate ΔSsurr for each of these reactions at 25°C and 1 atm. ΔSsurr = - ΔH / T 0.419 kJ/K – talk through positive sign of entropy of surroundings -2.61 kJ/K- heat is flowing from the surroundings to the system

Example Problem The element mercury is a silver liquid at room temperature. The normal freezing point of mercury is -38.9°C and its molar enthalpy of fusion is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of liquid mercury freezes at the normal freezing point? -2.44 J/K

Entropy Application Let’s look at the melting of 1 mole of ice in your hand and entropy changes: ΔSsys for phase change (ΔHfusion = 6.01 kJ/mol)   ΔSsurr for hand at body temp (37°C)   ΔStotal = ΔSsys + ΔSsurr (Is this spontaneous? How do you know?) 0.0220 kJ/K -0.0194 kJ/K 0.0026 kJ/K or 2.6 J/K Spontaneous!

Example Problem Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature: a.) H2O(l)  H2O(g) b.) Ag+(aq) + Cl-(aq)  AgCl(s) c.) 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) d.) N2(g) + O2(g)  2 NO(g) Why does water spontaneously freeze below 0°C and melt above 0°C? Fewer molecules mean fewer possible configurations. Generally the more complex the molecule, the higher the standard entropy value. Positive Negative negative d. Positive (appendix 4)

Gibbs Free Energy (G) ΔG = ΔH – TΔS ΔSuniv = - ΔG / T A process carried out at constant temperature will be spontaneous only if ΔG is negative...or, a process (at constant T and P) is spontaneous in the direction in which the free energy decreases (-ΔG means +ΔS).

Let’s look at the melting of ice: H2O(s)  H2O(l) ΔH° = 6.03 x 103 J/mol ΔS° = 22.1 J/K∙mol What is ΔG at -10°C:  What is ΔG at 0°C: What is ΔG at +10°C: When is the melting spontaneous?

Example Problem What is the normal boiling point of liquid bromine? Br2(l)  Br2(g) ΔH° = 31.0 kJ/mol ΔS° = 93.0 J/K∙mol At boiling point change in G is equal to 0 333K

Entropy Changes in Chemical Reactions Remember that entropy changes in the system (reactants and products) are determined by positional probability. Example Problem: Predict the sign of ΔS° for each of the following chemical reactions at constant temperature and pressure. a.) N2(g) + 3 H2 (g)  2 NH3(g)   b.) 4 NH3 (g) + 5 O2 (g)  4 NO(g) + 6 H2O(g) c.) CaCO3 (s)  CaO(s) + CO2 (g) d.) 2 SO2 (g) + O2 (g)  2 SO3 (g)

Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero. Which form has higher entropy of a substance: solid, liquid, or gas? Generally, the more complex the molecule, the higher the standard entropy value [compare H2(g) to H2O(g) to CCl4(g)]  The standard entropy values (S°) of many common substances at 298 K (25°C) and 1 atm are found in Appendix 4 (beginning on page A19). ΔS°reaction = ∑npS°products - ∑ nrS°reactants

2 NiS(s) + 3 O2(g)  2 SO2(g) + 2 NiO(s) Example Problem Calculate ΔS° at 25°C for the reaction 2 NiS(s) + 3 O2(g)  2 SO2(g) + 2 NiO(s) -149J/K

Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g) Example Problem Calculate ΔS° at 25°C for the reaction Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g) 179 J/K

Free Energy and Chemical Reactions For chemical reactions, we are often interested in the standard free energy change (ΔG°), the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. Why is it useful to know ΔG°? One way to calculate ΔG° is with the equation: ΔG° = ΔH° – TΔS° Delta G allows us to compare relative tendency of the reaction to occur.. We need to take into account but kinetics but will account for spontaneity.

Example Problem Consider the reaction 2 SO2(g) + O2(g)  2 SO3(g) carried out at 25°C and 1 atm. Calculate ΔH°, ΔS°, and ΔG° using the following data: -142kJ

Calculate ΔG° for the reaction 2 CO(g) + O2(g)  2 CO2 (g) A second method for calculating ΔG° takes advantage of the fact that free energy (ΔG°) is a state function. Example Problem: Calculate ΔG° for the reaction 2 CO(g) + O2(g)  2 CO2 (g) From the following data: 2 CH4(g) + 3 O2(g)  2 CO(g) + 4 H2O(g) ΔG° = -1088 kJ CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O(g) ΔG° = -801 kJ -514 kJ Like Hess’s Law

Example Problem Cdiamond(s)  Cgraphite(s) Calculate ΔG° for the reaction Cdiamond(s)  Cgraphite(s) From the following data: Cdiamond(s) + O2(g)  CO2 (g) ΔG° = -397 kJ Cgraphite(s) + O2(g)  CO2 (g) ΔG° = -394 kJ -3kJ

ΔG°reaction = ∑np ΔG°f (products) - ∑ nr ΔG°f (reactants) A third method for calculating ΔG° uses standard free energies of formation (ΔG°f): the change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states. The standard free energy of formation values (ΔG°f) of many common substances at 298 K (25°C) and 1 atm are found in Appendix 4 (beginning on page A19).   The standard free energy of formation value of an element in its standard state is zero. ΔG°reaction = ∑np ΔG°f (products) - ∑ nr ΔG°f (reactants)

Example Problem Methanol is a high-octane fuel used in high- performance racing engines. Calculate ΔG° for the reaction 2 CH3OH(g) + 3 O2(g)  2 CO2(g) + 4 H2O(g) Given the free energies of formation: -1378kJ

Example Problem A chemical engineer wants to determine the feasibility of making ethanol by reacting water with ethylene according to the equation C2H4(g) + H2O(l)  C2H5OH(l) Is this reaction spontaneous under standard conditions? -6kJ Yes spontaneous but need to consult kinetics to determine if feasible.

Various Possible Combinations of ΔH and ΔS:

Practice 1.Predict the sign of ΔS° and then calculate ΔS° for each of the following reactions. a. 2 H2S(g) + SO2 (g)  3 Srhombic(s) + 2 H2O(g) b. 2 SO3(g)  2 SO2 (g) + O2 (g) negative; -186 J/K Positive; 187 J/K

Practice From the data in Appendix 4, calculate ΔH°, ΔS°, and ΔG° for each of the following reactions at 25°C. CH4(g) + 2 O2(g)  CO2 (g) + 2 H2O(g) 6 CO2 (g) + 6 H2O(l)  C6H12O6(s) + 6 O2 (g) HCl(g) + NH3(g)  NH4Cl(s) a. H=-803 kJ; S= -4J/K; G= -802 kJ b.H= 2802 kJ; S= -262 J/K; G = 2880kJ d.H=-176kJ; S=-284J/K; G= -91kJ

Practice 3.Given the following data S(s) + ³/₂ O2(g)  SO3(g) ΔG° = -371 kJ 2 SO2 (g) + O2 (g)  2 SO3(g) ΔG° = -142 kJ Calculate ΔG° for the reaction S(s) + O2 (g)  SO2 (g) -300 kJ

Practice For the reaction SF4(g) + F2(g)  SF6(g) The value of ΔG° is -374 kJ. Use this value and the data below to calculate the value of ΔG° for SF4 (g). -731kJ/mol

Free-Response Practice C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l) When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow. (a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C. (b) Calculate the standard heat of formation, ΔH°â, of phenol in kilojoules per mole at 25°C. (c) Calculate the value of the standard free-energy change, ΔG°, for the combustion of phenol at 25°C. (d) If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container when the temperature is changed to 110. °C. (Assume no oxygen remains unreacted and that all products are gaseous.) -3,058 kJ/mol -173 kJ 0.601 atm

How do we calculate ΔG under nonstandard conditions? A system at constant temperature and pressure will proceed spontaneously in the direction that lowers its free energy. This is why reactions proceed until they reach equilibrium. What is the difference between ΔG, ΔG°, and ΔG°f? ΔG = ΔG° + RT ln(Q) = ΔG° + 2.303RT log(Q)

Example Problem CO(g) + 2 H2(g)  CH3OH(l) Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol. What is the meaning of ΔG for a chemical reaction?

Free Energy and Equilibrium (Keq) Equilibrium point (from a thermodynamic point of view): Occurs at the lowest value of free energy available to the reaction system. ΔG = ΔG° + RT ln(K) ; but since ΔG is 0 at equilibrium and Q = K: ΔG° = - RT ln(K) = - 2.303RT log(K) Case 1: When ΔG° = 0   Case 2: When ΔG° < 0 Case 3: When ΔG° > 0

Example Problem Consider the ammonia synthesis reaction N2(g) + 3 H2(g)  2 NH3(g) Where ΔG° = -33.3 kJ per mole of N2 consumed at 25°C. For each of the following mixtures of reactants and products at 25°C, predict the direction in which the system will shift to reach equilibrium. a.) PNH3 = 1.00 atm PN2 = 1.47 atm PH2 = 1.00 x 10-2 atm b.) PNH3 = 1.00 atm PN2 = 1.00 atm PH2 = 1.00 atm

Example Problem The overall reaction for the corrosion of iron by oxygen is 4 Fe(s) + 3 O2(g) ↔ 2 Fe2O3(s) Using the following data, calculate the equilibrium constant for the reaction at 25°C.

Temperature dependence of K

Free Energy and Work The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy: Wmax = ΔG

Individual Practice Beginning on page 783 #23, 25, 27, 29, 31, 33, 35, 47,51, 61