P a g e 1 Ch 18 Free Energy and Thermodynamics: Homework: Read Ch 18, Work out sample/practice exercises in the sections as you read, Ch 18: 27, 31, 33, 41, 43, 47, 51, 55, 61, 63, 67, 71, 77, 87 Check for MasteringChemistry due dates. Turn in your Research topic by the due date Chemical Thermodynamics: Chemical Thermodynamics studies energy relationships between Enthalpy, Entropy, Equilibrium, and Gibbs free energy. Thermodynamics can predict the spontaneous direction of a reaction under specified conditions, but not the rate of the reaction. Thermodynamics vs. Kinetics: Which one deals with state functions, independent of path?
P a g e 2 Spontaneous Processes: Spontaneous process: A spontaneous process is one that proceeds in a specific direction on its own without any outside assistance under specified conditions (T, P). The products are considered thermodynamically more stable. The reverse reaction is nonspontaneous. Some reactions are always spontaneous, others are never spontaneous and many reactions can be either, depending on conditions. Most spontaneous reactions favor exothermic processes (- H) and increased entropy (+ S). Often one factor may override the other. Equilibrium reactions can shift to be spontaneous in the forward or reverse direction with just slight changes in the environment/conditions. A spontaneous process always moves a reaction mixture toward equilibrium. Reversible process will restore all components to the original conditions and states with no net change to the system or surroundings, an example that is reversible is the melting or freezing of ice at 1 atm and 0 C. Reversible processes only occur for situations at equilibrium, S univ =. Irreversible process will not restore to original, an example is to mix equal portions of hot and cold liquid together that end with an in between final temperature. Any spontaneous process is irreversible, S univ >. HISTORY: Carnot Engine: In 1824 a French Engineer, Sadi Carnot published an analysis of factors that determine efficiency of a steam engine. An ideal engine that converts all the energy content of fuel to useable work is impossible because heat is always lost to the surroundings. This study led to engineering more efficient engines and was the first study that led to the discipline thermodynamics. Rudolph Clausius: Forty years later the ratio of q p /T (entropy) was found to have a special significance and was comparable to energy. Ideal Engine: An ideal engine that has maximum efficiency operates under ideal conditions in which all processes are reversible. The Energy Tax: Every energy transition results in a loss of energy, heating up the surroundings. An ideal engine is not possible. Heat Tax: Fewer steps are generally more efficient. The heating of your home with natural gas directly (1 step) is much more efficient than heating with electricity, which may take 5 or more steps.
P a g e 3 Entropy, S: Second Law: Entropy Principle: Every spontaneous change increases the total entropy change of the universe. This does not mean local decreases in entropy cannot take place. For the entropy of a system to decrease, the entropy of the surroundings must increase more. S univ > The universe tends toward greater randomness. for reversible process Suniv = 0 Third Law: Zero Entropy Established: The entropy of a pure, perfect, crystalline substance (no disorder) is zero at the temperature of absolute zero, 0 Kelvin.
P a g e 4 Entropy: Entropy is a state function that measures randomness, ways of being. This includes all possible motions: Vibration, rotation, translation. S= q p /T = H/T S = klnw (entropy = Boltzmann s constant, 1.38 x 10-23 J/K time the natural log of the ways of being, number of microstates) S= S final S inital (S, entropy values can be found in the Appendix) S= klnw final klnw inital = kln(w final /W initial ) Entropy has units of J/mol K and can be found in Appendix C. Be aware that the units for entropy are in J and not kj! Predicting the sign for S: (+ more ways, favorable) 1) Phase changes: s l g H 2 O (l) H 2 O (g) (+ S)
P a g e 5 2) Temp: warming + S O 2 (g) 25 C O 2 (g) 80 C 3) Volume: expanding + S 10 ml gas 1.0 liter of same gas 4) Mixing: dissolving... + S NaCl (s) + H 2 O (l) NaCl (aq) 5) Rearrange to create more ways of being + S H 2 (g) + Cl 2 (g) 2 HCl (g) 6) # parts: increase + S 10 g rock 10 g sand 7) More moles gas + S N 2 O 4 (g) 2 NO 2 (g) Example 1: a) What is the entropy of a system with only 1 microstate? S = klnw b) What sign is S for the following: Temperature decreases Volume increases gas liquid increase the moles of gas Dissolving sugar in hot tea N 2 (g) + 3H 2 (g) 2 NH 3 (g) Ag + (aq) + Cl -1 (aq) AgCl (s) c) Circle the one with the higher entropy in each of the following pairs. 1 mole O2 (g) ; 25 C or 1 mole O2 (g) ; 50 C 1 mole N 2 O4 gas or 2 mole NO2 gas NaCl (s) or NaCl (aq) 1 mol N 2 gas in 22.4 L or 1 mol N 2 gas in 2.2 L 1 mol H 2 O (l) or 1 mol H 2 O (s)
P a g e 6 Solving Srxn: Just like enthalpy, entropy can be solved using the Appendix values. S rxn = ns products - ns reactants Example 2: S for CO 2 (g) = 213.8 J/mol K a) Predict the sign of the standard entropy change for the reaction CH 4 (g) + 2 O 2 (g) CO 2 (g)+ 2 H 2 O (l) b) Calculate S using the appendix values. Example 3: a) Predict the sign of the entropy change for the vaporization of water b) Calculate S using the appendix values.
P a g e 7 Trouton s Law: Most liquids follow Trouton s Law- the molar heat of entropy of vaporization is approximately 88 J/mol K; S vap = 88+/- 5J/mol K Example 4: Most liquids follow Trouton s rule. Normal boiling points and enthalpies of vaporization of several organic liquids are in the table. Substance Normal Boiling Pt Hvap (kj/mol) Acetone, (CH 3 ) 2 CO 56.1 29.1 Dimethyl ether, -24.8 21.5 (CH Ethanol, ) O C 2 H 5 OH 78.4 38.6 Octane, C 8 H 18 125.6 34.4 Pyridine, C 5 H 5 N 115.3 35.1 a) Estimate Svap for each liquid using Svap= Hvap/T bp Acetone, (CH 3 ) 2 CO Dimethyl ether, (CH 3 ) 2 O Ethanol, C 2 H 5 OH Octane, C 8 H 18 Pyridine, C 5 H 5 N b) Which do not follow Trouton s Law? c) Explain the exceptions using intermolecular forces? d) Should water follow Trouton s Law? Check the accuracy of your conclusion using the Appendix. e) Chlorobenzene boils at 131.5 C, estimate the Hvap for C 6 H 5 Cl.
P a g e 8 Example 5: Estimate the normal boiling point in degrees Celsius of CCl4. Use the appendix values. Gibbs Free Energy: Gibbs Free Energy is the maximum amount of useful energy obtainable in the form of work. It uses both enthalpy and entropy to predict spontaneity. G f, Gibbs Free Energy can be found in the Appendix G = H - T S (remember to convert S into kj for consistency) G rxn= n G f (products) - n G f(reactants) G under nonstandard conditions is a reliable indicator of spontaneity. G > 0; a positive free energy is nonspontaneous G= 0; a zero free energy is at equilibrium G< 0; a negative free energy is spontaneous
P a g e 9 Example 6: Predict the sign and calculate the Gibbs free energy change under standard conditions (25 C) for the reaction Use the appendix values. CH 4 (g) + 2 O 2 (g) CO 2 (g)+ 2 H 2 O (l) a) From this equation G rxn= n G f (products) - n G f (reactants) b) From this equation G rxn= H rxn - T S rxn Gibbs Free Energy and Temperature: H f and S data do not significantly change with varying temperatures. H (kj/mol) S (kj/mol) G = H - T S Spontaneity conditions _ + _ Spontaneous at all Temps -196 +126 _ 2H 2 O 2 (l) 2 H 2 O (l) + O 2 (g) + _ + Nonspontaneous at all Temps +285-137 + 3 O 2 (g) 2 O 3 (g) + or - Spontaneous at low Temps Nonspontaneous at high Temps -233-424 Equil T =? 2H 2 S (g) + SO 2 (g) 2 H 2 O (l) + 3 S (s) + + + or - Spontaneous at high Temps Nonspontaneous at low Temps +176 +285 Equil T =? NH 4 Cl (s) NH 3 (g) + HCl (g) The equilibrium temperature can be found for the bottom two conditions by setting G to zero; H = T S; = H / S (use kj, T is in Kelvin)
P a g e 10 Example 7: Solve for the equilibrium temperatures in the table above. Specify the spontaneous temperature range/conditions at 1 atm, (above or below equilibrium temperature). Example 8: An important reaction in the production of sulfuric acid is the oxidation of sulfur dioxide gas into sulfur trioxide gas. a) Write the chemical reaction. b) Use appendix: Solve for G rxn c) Is the reaction spontaneous at standard conditions? d) Solve H f and S e) Will this reaction be spontaneous at 900 C? f) Solve for the equilibrium temperature and determine the spontaneity conditions.
P a g e 11 Gibbs Free Energy and Equilibrium Constant, Keq: G = G + RTlnQ eq and at equilibrium G = -RTlnK eq For a spontaneous reactions G< 0 is negative and K >>1 is very large.
P a g e 12 Example 9: Consider the reaction 2 NO2 (g) N2O4 (g) a) Calculate G at 25 C using Appendix values. b) Calculate G at 25 C if the partial pressures of NO2 and N2O4 are 0.40 atm and 1.60 atm, respectively. Example 10: Consider the decomposition of barium carbonate, BaCO3(s) BaO(s)+ CO 2 (g) a) Calculate the equilibrium pressure of CO 2 using Appendix values at 25 C. b) Calculate the equilibrium pressure of CO 2 at 1100 K.
P a g e 13 Example 11: The Ka for HNO2 = 4.5 x 10-4. a) Write the Ka equation for HNO2. b) Calculate G at 25 C using Ka. c) What is G at equilibrium. d) What is G when [NO 2-1 ] = 6.0 x 10-4 M, [HNO 2 ] = 0.20 M, and ph = 1.30 Calculate Coupling Reactions: Consider the direct decomposition of rust to iron 2 Fe 2 O 3 (s) 4 Fe (s) + 3 O 2 (g) : G = +1487 kj This large positive value for free energy tells us the reaction is nonspontaneous. However this does not mean you cannot change the rust back to pure iron; you just need to do some useful work to reduce it by coupling with a reaction with a large negative free energy. 2 CO (g) + O 2 (g) 2 CO 2 (g) : G = -514 kj The common term in both reactions is O 2 (g). Notice we have 3 as product in the rust decomposition and 1 O 2 (g) in the CO (g) combustion. 2 Fe 2 O 3 (s) 4 Fe (s) + 3 O 2 (g) : G = +1487 kj 3 x [2 CO (g) + O 2 (g) 2 CO 2 (g) : G = -514 kj] Overall: 2 Fe 2 O 3 (s) + 6 CO (g) 4 Fe (s) + 6 CO 2 (g) ; G = -56 kj This reaction occurs in a blast furnace. In Biochemistry ATP provides the coupling energy necessary to accomplish many reactions in a cell. -2 ATP + H 2 O (l) ADP + HPO 4 (aq) ; G = -30.5 kj Glucose + Fructose sucrose + H 2 O (l) ; G = +29.3 kj -2 Glucose + Fructose + ATP sucrose + ADP + HPO 4 (aq); G = -1.2 kj
P a g e 14 Example 12: Given the two step reaction to form methane gas from CO and H2... Step 1 CO (g) + 2 H2 (g) CH3OH (l) Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g) And using only the information on the table solve all parts in this problem. compound H f S f G f CO(g) ------ ------ -137.2 kj/mol H2 (g) you know this! ------ you know this! CH3OH (l) -238.7 kj/mol 126.8 J/mol K -166.3 kj/mol CH4 (g) -74.87 kj/mol 186.1 J/mol K ------ O2 (g) you know this! 205.0 J/mol K you know this! a) What is H f and G f for H2 (g) and O 2 (g)? b) Calculate G rxn in kj/mol for step 1 (25 C) Step 1 CO (g) + 2 H2 (g) CH3OH (l) c) Is step 1 spontaneous under standard conditions? d) Given that H is negative and you have the ability to predict the sign of S, answer under what conditions is step 1 spontaneous? (always, at low temperatures, at high temperatures, never) e) Calculate H rxn in kj/mol for step 2 Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g) f) Calculate S rxn in J/mol*K for step 2 Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g)
P a g e 15 g) Calculate G rxn in kj/mol for step 2 (25 C) h) Is step 2 spontaneous under standard conditions? i) At what temperature will step 2 have G = 0? j) Specifically, what temperature range is step 2 spontaneous? k) Write the overall coupled reaction of step 1 and step 2 when together and Calculate the coupled overall G under standard condition, 25 C. Coupled reaction Step 1 CO (g) + 2 H2 (g) CH3OH (l) G Step 2 CH3OH (l) CH4 (g) + 1/2 O2 (g) G G l) Is the coupled reaction spontaneous under standard conditions, 25 C? m) You are designing a production facility. Considering the information above, would you carry out the reactions in separate steps under different temperature conditions or coupled simultaneously together in the same vessel under the same conditions.