Spontaneity, Entropy, and Free Energy Chapter 12 Spontaneity, Entropy, and Free Energy

Thermodynamics The study of energy changes in a chemical reaction that determine the direction in which the reaction is spontaneous. Focuses on the initial & final states of the reaction. Does not require knowledge of the pathway of the reaction.

Kinetics The study of reaction rates (Ch 11) Dependent upon things such as activation energy, temperature, concentration, and catalysts. Focuses on the pathway of the reaction.

Thermodynamics/Kinetics We must understand both thermodynamics & kinetics in order to fully describe a reaction. Potential Energy diagram-activation energy = kinetics; enthalpy = thermo

Spontaneous Processes & Entropy Can we use thermodynamics to predict IF a reaction will occur? What is entropy? How can we choose the greatest positional entropy?

Spontaneous Processes & Entropy Spontaneous process – one that occurs without outside intervention (no matter if it occurs quickly or slowly) A ball rolls __________ a hill but never spontaneously rolls back up a hill. If exposed to air and moisture, steel _______ spontaneously. A gas fills its container uniformly. It never spontaneously collects at one end of the container. Heat flow always occurs from a _____ object to a ______ onel Wood burns spontaneously in an exothermic reaction to form __________ and ___________. At temperatures below 0°C, water spontaneously _____________, and at temperatures above 0°, ice spontaneously ____________.

Spontaneous Processes & Entropy Thermodynamics explains why some processes are spontaneous while others are not. The spontaneity of a reaction is unequivocally determined based on the combination of _________________ & _____________________

Energy Energy – the capacity to do __________ or produce ______________ Energy – the capacity to do work or produce heat Ex. Accelerating an object (motion) Lifting things up (position) Produce electrical power (electrical) Raise the temperature of a substance (thermal) Produce sound or light (waves of energy) Chemical reactions (chemical) Phase changes (heat/thermal)

Energy is the driving force for _________. ____________ ______________ Energy is the driving force for change. Chemical Change Physical Change (chemical reactions) (phase changes)

Energy Potential Kinetic *Chemical* – E stored in atoms and bonds of molecules Mechanical – E stored in objects by application of force Nuclear- E stored in the nucleus of atoms Gravitational – E of position Electrical – movement of electrical charges Radiant – EME that travels as transverse waves Thermal – vibration & movement of atoms and molecules internally *Motion* – movement of objects from one place to another Sound – movement of E through substances as longitudinal waves

Energy KEobject = ½ mv2 Kinetic Energy (KE) – energy of an object due to its _____________. Depends on the object’s _____________ and _______________. KEobject = ½ mv2 Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

Energy Calculate the kinetic energy (in Joules) of a 1.0 x 10-5 g object with a velocity of 2.0 x 105 cm/s. (Recall that 1 Joule = kg · m2) s2 Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

Energy Potential Energy (V) – energy due to __________________ or ____________________ Examples in chemistry: “Chemical Potential Energy” – E stored in the __________ of chemicals “Coulombic Potential Energy” – E due to the _________________ and ___________ between particles In this chapter we will study _______ potential E Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

Energy Transfer Law of Conservation of Energy : energy can be _______________ from one substance to another, but it cannot be _____________ nor ___________________. The energy of the universe is __________________. Energy is conserved. “1st law of thermodynamics”… Law of Conservation of Energy: energy can be converted from one form to another but can be neither created nor destroyed.

Energy Transfer Euniverse = Esystem + Esurroundings Universe = ______________________ System = _______________________ Surroundings = ___________________ Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

Energy Transfer Two ways to transfer E: 1) heat (q) – a transfer of E that causes a ____________________ 2) work (w) – a transfer of E that causes something with mass to ___________________ ___________________ Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) + E(heat) Heat Transfer “The Woosh Bottle” C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) + E(heat) System = reactants & products Surroundings = everything else Is E flowing in or out of the system? Exo-/Endothermic? +/- value for q? What does this say about the amount of energy required to break the bonds in the reactants compared to the amount of E released in forming the bonds of the reactants? What does this say about the difference in potential energies between the products and the reactants?

Heat Transfer System = reactants & products “Water Glue” Ba(OH)2•8H2O (s) + 2NH4SCN (s) + E (heat) --> Ba(SCN)2 (s)+ 2NH3(g) + 10H2O(l) System = reactants & products Surroundings = everything else Is E flowing in or out of the system? Exo-/Endothermic? +/- value for q? What does this say about the amount of energy required to break the bonds in the reactants compared to the amount of E released in forming the bonds of the reactants? What does this say about the difference in potential energies between the products and the reactants?

Internal Energy Internal Energy (E) – the sum of the _____________ and ______________ energies of all the particles in a system - can be changed by a transfer/flow of _________, _________, or ________. Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

ΔE – change in the system’s internal energy (J or kJ) ΔE = q + w ΔE – change in the system’s internal energy (J or kJ) q = heat (J or kJ) w = work ( J or kJ) Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

Internal Energy The sign (-/+) for each variable in the equation reflects the system’s point of view. Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

Internal Energy Internal energy (E) is a _________ function, meaning its value is not dependent upon the pathway. Heat (q) and work (w) are not _______ functions. Potential Energy (V) – energy due to position or composition Ex. Coulombic Potential E (attractive & repulsive forces between particles) Energy due to bonds breaking and forming in a reaction.

Note: V = Vfinal - Vinitial Work 1. Calculate the work (with proper sign) associated with the contraction of a gas from 75 L to 30 L at a constant external pressure of 6.0 atm in L atm b. Joules (1 L atm = 101.3 J) Note: V = Vfinal - Vinitial

Internal Energy (E) Recall: ΔE = q + w -ΔE = means the system loses/releases energy +ΔE = means the system gains/absorbs energy   Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J.

Energy, Heat, & Work A piston compresses a gas from a volume of 8.3 L to 2.8 L with a constant pressure of 1.9 atm. In the process, there is a heat gain by the gas of 350 J. Calculate the change in internal energy of the system.

Change during a phase change!!!! Heating Curve Energy changes also occur in physical processes. L → G (vaporization) Gas boiling point Liquid S → L (melt) **NOTE – Temp DOES NOT Change during a phase change!!!! melting point Solid

Cooling Curve

Specific Heat Specific Heat (C): amount of energy (in Joules) needed to increase the temperature of 1 gram of a substance by 1oC. See Reference Tables q = mCΔT (+q = endothermic, -q = exothermic) q = heat (joules or calories) m = mass (grams) C = specific heat → ΔT = change in Temperature = Tf - Ti (oC) J goC

Remember!!! Temperature does not change during a phase change! (Energy goes into pulling molecules farther apart.) Therefore, we cannot use this equation for phase changes: q = mCΔT

Latent Heats Heat of Vaporization (Hv): amount of energy absorbed when one gram of a liquid is changed into a vapor (or, the amount of energy released when 1 gram of a vapor is changed into a liquid). q = mHv Hv = heat of vaporization (J/g) Hv of water = 2260 J/g

Heat of Fusion Heat of Fusion (Hf): amount of energy needed to convert 1 gram of a solid into a liquid. (Or the heat released when 1 gram of liquid → solid.) q = mHf Hf = heat of fusion (J/g) Hf of water = 334 J/g

Applying Q when heating a substance across different phases: 1) q = mcΔT 2) q = mHf 3) q = mcΔT 4) q = mHv 5) q = mcΔT

Example Problems – Level One 1. How much heat is added if 100 grams of liquid water increases in temperature from 30oC to 70oC? 2. How much heat is absorbed if 200 g of ice increases in temperature from -15oC to -5oC? q = mCΔT q = (100 g)(4.18 J/goC)(70o – 30o) q = 16,720 J q = (200 g)(2.05 J/goC)(-5o – -15oC) q = 4100 J

Continued… 5. How much heat is released when 50 grams of water vapor is changed into liquid water at 100oC? q = mHv q = (50 g)(2260 J/g) q = 113,000 J → -113,000 J

Example Problems – Level Two 6. How much heat is absorbed if 30 grams of ice at -10oC is converted into liquid water? 2 steps: (1) heat the ice and (2) melt ice → liquid (1) q = mCΔT q = (30g)(2.05 J/goC)(0o – -10o) q = 615J (2) q = mHf q = (30g)(334J/g) q = 10,020J Add both steps together! Total heat = 615J + 10,020J Total heat = 10,635 J

Calorimetry Calorimetry: the precise measurement of the heat flow out of a system for chemical and physical processes.

Calorimetry Practice If 40.5 J of heat is added to a 15.4 g sample of silver, how much will the temperature increase by?The specific heat of silver is 0.235 J/g o C. 2. What is the specific heat of silicon if the temperature of a 4.11 g sample of silicon is increased by 3.8oC when 11.1 J of heat is added? 3. A chemistry student dissolves 4.51 grams of sodium hydroxide in 100.0 mL of water at 19.5°C (in a calorimeter cup). As the sodium hydroxide dissolves, the temperature of the surrounding water increases to 31.7°C. Determine the heat of solution of the sodium hydroxide in J/g

Enthalpy ΔHreaction = Hproducts - Hreactants Enthalpy (ΔH): the difference in energy between the products and reactants in a chemical change ΔHreaction = Hproducts - Hreactants ΔHreaction is negative for exothermic reactions. Why? ΔHreaction is positive for endothermic reactions. Why? Nature favors lower energy → exothermic

Hess’s Law the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or several steps.

Hess’s Law if two or more chemical equations can be added together to produce an overall equation, the sum of the enthalpy equals the enthalpy change of the overall equation. This is called the Heat of Summation, ∆H

Analogy for Hess's Law There is an old Chinese proverb which says: There are many ways to the top of a mountain, but the view from the top is always the same.

Hess’s Law Read through the whole question Plan a Strategy Evaluate the given equations. Rearrange and manipulate the equations so that they will produce the overall equation. Add the enthalpy terms.

H2O(g) + C(s) → CO(g) + H2(g) Example 1 Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2(g) + ½ O2(g) → H2O(g) ∆H = -241.8kJ

H2O(g) + C(s) → CO(g) + H2(g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2O(g) → H2(g) + ½ O2(g) ∆H = +241.8kJ _____________________________________ C(s) + H2O(g) → H2(g) + CO(g) ∆H=+131.3kJ

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/mol C(s) + O2(g) → CO2(g) ∆H= -393.5kJ/mol H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol Read through the whole question Plan a Strategy Evaluate the given equations. Rearrange and manipulate the equations so that they will produce the overall equation. Add the enthalpy terms. REWRITE THE CHANGES.

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/mol C(s) + O2(g) → CO2(g) ∆H= -393.5kJ/mol H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol C(s) + O2(g) → CO2(g) ∆H= -393.5kJ/mol H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4(C(s) + O2(g) → CO2(g)) ∆H= 4(-393.5kJ/mol) H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) distribute the 4 ∆H= 4(-393.5kJ/mol) H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) distribute the 4 ∆H= 4(-393.5kJ/mol) 5(H2(g) + ½O2(g) → H2O(g)) distribute the 5 ∆H= 5(-241.8kJ/mol)

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol) 5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol)

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol) 5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol) _____________________________________________________ ∆H = -125.6kJ/mol

Spontaneous Processes & Entropy Entropy (S) A condition of molecular randomness or disorder. A thermodynamic function that describes the number of arrangements that are available to a system existing in a given state. The more ways a particular state can be achieved, the greater the likelihood of finding that state. Nature spontaneously proceeds toward the states that have the highest probabilities of existing. A measure of positional probabilities.

Spontaneous Processes & Entropy Microstate Each configuration that gives a particular arrangement . Problem 1: Consider 4 gas particles in the following flask: How many different microstates are there? Which state is most likely to be found? Are the gas particles more likely to spread out or be confined to one side of the flask? What arrangement would have the highest entropy? The lowest entropy?

Spontaneous Processes & Entropy Microstates Problem 2: Which substance in the following pairs is likely to have the higher positional entropy per mole at a given temperature? A) Solid or gaseous phosphorus B) CH4(g) or C3H8 (g)

Spontaneous Processes & Entropy Problem 3: Predict the sign of the entropy change for each of the following processes. A) Solid sugar is added to water to form a solution. B) Iodine vapor condenses on a cold surface to form crystals.

Entropy & the 2nd Law 1st Law of Thermodynamics – the energy of the universe is constant (Law of Conservation) 2nd Law of Thermodynamics – In any spontaneous process, there is always an increase in the entropy of the universe. ΔSuniv = ΔSsys + ΔSsurr

Entropy & the 2nd Law ΔSuniv = ΔSsys + ΔSsurr If ΔSuniv is (+), then the entropy of the universe is increasing, and the process is spontaneous in the direction written. If ΔSuniv is (-), then the entropy of the universe is decreasing, and the process is spontaneous in the opposite direction. If ΔSuniv is zero, then the process has no tendency to occur (equilibrium is reached).

Entropy & the 2nd Law ΔSuniv = ΔSsys + ΔSsurr Problem #4 : Under what circumstances can the entropy of the system decrease for a spontaneous process?

Temperature & Spontaneity What is the importance of exothermic reactions to entropy? How do we calculate the change in entropy of the surroundings from the heat of reaction of the solution?

Temperature & Spontaneity Problem 5: Consider the following process: H2O (l)  H2O (g) A) If we define the water to be the system, would the entropy of the system increase or decrease? B) What is the sign of ΔSsys? C) Is the process endothermic or exothermic? D) Based on your answer to part c and the heat flow that would occur, would the entropy of the surroundings increase or decrease? E) What is the sign of ΔSsurr? F) How must ΔSsys and ΔSsurr compare to one another in magnitude if this is a spontaneous process?

Temperature & Spontaneity Problem 5: Consider the following process: H2O (g)  H2O (l) A) If we define the water to be the sysem, would the entropy of the system increase or decrease? B) What is the sign of ΔSsys? C) Is the process endothermic or exothermic? D) Based on your answer to part c and the heat flow that would occur, would the entropy of the surroundings increase or decrease? E) What is the sign of ΔSsurr? F) How must ΔSsys and ΔSsurr compare to one another in magnitude if this is a spontaneous process?

Temperature & Spontaneity The sign of ΔSsurr depends on the _________________________________. The magnitude of ΔSsurr depends on the _________________________. Water spontaneously vaporizes at T > 100°C, and spontaneously condenses at T < 100°C.

Temperature & Spontaneity ΔSsurr = -ΔH T (in Kelvin) Signs of Entropy Change ΔSsys ΔSsurr ΔSuniv Spontaneous? + - ?

Temperature & Spontaneity Problem #7: ΔSsurr and the Heat of Reaction Calculate ΔSsurr for each of the following reactions at 25°C and 1 atm. A) C3H8(g) + 5O2(g)  3CO2(g) + 4H2O (g) ΔH = -2045 kJ B) (NH4)2Cr2O7(s)  N2(g) + Cr2O3(s) + 4H2O ΔH = -315 kJ C) H2O (l)  H2O (g) ΔH = +44 kJ

Temperature & Spontaneity Problem 8: Reaction Spontaneity Determine if the values for entropy in each of the following will produce a spontaneous process. Also, which of the following processes is endothermic (from the perspective of the system)? A) ΔSsys = 30 J/K ΔSsurr = 50 J/K B) ΔSsys = -27 J/K ΔSsurr = 40 J/K C) ΔSsys = 60 J/K ΔSsurr = -85 J/K

Free Energy Free Energy (G)- describes unequivocally whether a reaction will be spontaneous (it reflects ΔSuniv).

(when no subscript appears, assume it is referring to the system) Free Energy Two important relationships: 1) ΔG = ΔH – TΔS (when no subscript appears, assume it is referring to the system) -an explicit way to calculate free energy -shows that there are circumstances under which temperature will determine whether a reaction is spontaneous (consider the temperature dependence of melting ice)

Free Energy Two important relationships: 2) ΔSuniv = -ΔG T -When a process is spontaneous, ΔSuniv is ________. - When a process is spontaneous, ΔG is ________.

Free Energy Problem #9: Given the values for ΔH, ΔS, and T, determine whether each of the following sets of data represent spontaneous or nonspontaneous processes. ΔH (kJ) ΔS (J/K) T (K) a. 40 300 130 b. 150 c. -300 d. -40 e.

Free Energy Problem #10: You know that the boiling point of water is 373K. How does this compare to the minimum temperature for the boiling of water determined using thermodynamics data: H2O(l)  H2O(g) where ΔH = 44kJ and ΔS = 119 J/K

Entropy Changes in Chemical Rxns How can we predict the sign of entropy changes for a given reaction? How can we calculate ΔS from thermodynamic data tables?

Entropy Changes in Chemical Rxns For chemical rxns, the system is defined as the reaction reactants & products, while the surroundings is defined as everything else that comes into contact with the system. ΔSsurr is dependent upon ________________ from the system ΔSsys is dependent upon the _______________ __________________ for each of the reactants.

Entropy Changes in Chemical Rxns For a chemical reaction involving only the gas phase, entropy is related to the total number of moles on either side of the equation. A decrease in the number of moles corresponds to a decrease in entropy, while an increase in the number of moles corresponds to an increase in entropy.

Entropy Changes in Chemical Rxns Problem #11: For the reaction 2H2(g) + O2(g)  2H2O(g) what is the sign for entropy? ΔS = ___89 kJ (at 25°)

Entropy Changes in Chemical Rxns For a chemical reaction involving different phases, the production of a gas will (in general) increase the entropy much more than an increase in the number of moles of a liquid or solid.

Entropy Changes in Chemical Rxns Problem #12: For the reaction… 2HNO3(aq) + Na2CO3(s)  2NaNO3(aq) + H2O(l) + CO2(g) what is the sign for entropy? ΔS = ___88 kJ (at 25°C)

Entropy Changes in Chemical Rxns Problem #13: Predict the sign of Δso for each of the following reactions: A) (NH4)2Cr2O7(s)  Cr2O3(s) + 4H2O(l) + N2(g) B) Mg(OH)2(s)  MgO(s) + H2O(g) C) PCl5(g)  PCl3(g) + Cl2(g)

Entropy Changes in Chemical Rxns Just like ΔH0, ΔS0 for a reaction can be determined if the entropy of each reactant and each product is known: Δ S0rxn = ΣnpS0products - ΣnrS0reactants See Appendix 4

Entropy Changes in Chemical Rxns Problem #14: Calculate ΔS0 for each of the following reactions using data from Appendix 4 in the textbook. A) N2O4(g)  2NO2(g) B) Fe2O3(s) + 2Al(s)  2Fe(s) + Al2O3(s) C) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

Free Energy and Chemical Rxns Standard Free Energy (ΔG0) – the free energy change that occurs if reactants in their standard states (1 atm, 25°C) are converted to products in their standard states A standard set of conditions to compare properties of reactions Cannot be measured directly

Free Energy and Chemical Rxns How can we calculate the standard free energy of formation and use it to predict spontaneity of chemical reactions?

Free Energy and Chemical Rxns Three ways of calculating ΔG0: 1. ΔG0 = ΔH0 -TΔS0 2. By manipulating known equations, as in Hess’s law problems for ΔH0 3. ΔGf0 = ΣnpGf0products – ΣnrGf0reactants

Free Energy and Chemical Rxns Problem #15: Standard free energy from Entropy & Enthalpy Using data for ΔH0 and ΔS0, calculate ΔG0 for the following reactions at 25°C and 1 atm. A) Cr2O3(s) + 2Al(s)  Al2O3(s) + 2Cr(s) B) C3H8(g) + 5O2(g)  3CO2(g) + 4H2O (g)

Free Energy and Chemical Rxns Problem #16: Standard free energy from Hess’s Law (Combining Eqns) Given the following data, calculate ΔG0 for the “goal equation”: Eqn 1: S(s) + 3/2O2 (g)  SO3(g) ΔG0 = -371 kJ Eqn 2: 2SO2(g) + O2(g)  2SO3(g) ΔG0 = -142 kJ Goal Eqn: S(s) + O2(g)  SO2 (g)

Free Energy and Chemical Rxns Problem #17: Standard free energy from Energies of Formations (“Products - Reactants”) Calculate ΔG0 for the following reaction using ΔG0 data from Appendix 4. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O (g) Compare with 15b.

Free Energy and Pressure How do we calculate ΔG0 at pressures other than 1 atm (nonstandard conditions)?

Free Energy and Pressure For an ideal gas, enthalpy is not pressure- dependent, however, entropy is affected by pressure. Slow pressure > Shigh pressure More positions are possible at low pressures (think of volume changes).

Free Energy and Pressure ΔG = ΔG0 +RTln(P) ΔG = ΔG0 +RTln(Q) G0 = free E under std pressure (1 atm) G = free E under nonstd pressure or Q conditions of a substance or reaction P = pressure (atm) R = 8.3145 J/mol K T = temperature in Kelvin Q = reaction quotient (the mass action expression relating to initial quantities)

Free Energy and Pressure Problem 18: a) Calculate ΔG at 700K for the following reaction: C(s, graphite) + H2O(g) ↔ CO(g) + H2 (g) Initial pressures are PH2O = 0.85 atm, PCO = 1.0 x 10-4 atm, PH2 = 2.0 x 10-4 b) Is the reaction spontaneous at standard conditions? c) What about for the conditions given?

Free Energy and Equilibrium Equilibrium occurs at the lowest free energy available to the system… Gforward rxn = Greverse rxn (ΔG = 0 at equilibrium) Can we interpret the direction of a reaction? How do we interconvert between K and ΔG0?

Free Energy and Equilibrium Greactants > Gproducts Goes right until Greact= Gprod ΔG > 0 (+) Greactants < Gproducts Goes left until Greact= Gprod ΔGo = -RT ln K

Free Energy and Equilibrium Problem #19: Determine the direction of the reaction where T = 700K and ΔG0 = 92 kJ if the following initial pressure of each gas is PH2O = 0.67 atm, PCO = 0.23atm, PH2 = 0.51 C(s, graphite) + H2O(g) ↔ CO(g) + H2 (g)

Free Energy and Equilibrium Problem #20: Given the values for ΔG0 that were calculated for Cr2O3(s) + 2Al(s)  Al2O3(s) + 2Cr(s) in problem 15a, calculate K for the reaction at 25°C.