Chapter 16 - Spontaneity, Entropy, and Free Energy 1 st Law of Thermodynamics - energy can be neither created nor destroyed. Although total energy is constant, the various forms of energy can be interchanged in physical and chemical processes. Thermodynamics vs. Kinetics Spontaneous Process - Chemical Reactions like other things in nature occur due to 2 basic reasons: 1) 2) REALLY: Entropy a thermodynamic function that describes the number of arrangements (positions/energy levels) that are available to a system existing in a given state. Key concept: the more ways a particular state can be achieved Positional probability Which has the highest positional entropy? 1. lawn before raking/lawn after raking 2. salt before dissolving/ salt after dissolving 3. solid/ liquid/ gas 4. mixture of 3 kinds molec./ mixture of 2 kinds molec. 5. N 2 gas at 1 atm/n 2 gas at 1.0 x 10-2 atm 6. 1 mol CO 2 (at STP)/ 1 mol CO 2 (at room conditions) 7. C 6 H 12 O 6 / C 12 H 22 O 11 1
2nd Law of Thermodynamics - in any process there is always in the entropy of. if Suniv is (+) if Suniv is (-) if Suniv is zero Ssys Sign determined by Magnitude determined by Ssurr Sign determined by Magnitude determined by Suniv = Ssys + Ssurr Exothermic Rxns: H ΔS surr Endothermic Rxns: ΔH S surr Predict the sign for Ssys and Ssurr for each of the following: C 2 H 5 OH (l) C 2 H 5 OH (g) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) + 46.1 kj CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) H = -891 kj Ssys Ssurr The Effect of Temperature on Spontaneity: Entropy changes in the surroundings are determined primarily by the flow of energy into or out of the system as heat. CONSIDER: H 2 O (l) --> H 2 O (g) endo / exo S surr +/- S sys +/- S univ +/-? At high temps At low temps 2
The key idea as to why temperature controls the situation is that the entropy s in the surroundings are primarily determined by heat flow. the significance of exothermicity as a driving force depends on the temperature at which the process occurs. The transfer of a given quantity of energy as heat to or from the surroundings will have a greater impact at low/high temperatures. CONCLUSION: 1) The tendency toward min. energy is dominant @ temps. 2) The tendency toward max. randomness is dominant @ temps. OR *As temperature increases, the factor becomes more important. ---------------------------------------------------------------------------------------------------------- Free Energy: Free energy is another function used to predict if. G = H - T S (T is temp in K) Reactions at constant pressure and temperature, go in such a direction as to the free energy of the system. If G is negative, the reaction is If G is positive, the reaction is If G is zero, the system is. 3 rd Law of Thermodynamics - (Standard entropy values (S ) can be found in Appendix 4.) Entropy (like enthalpy) is a Entropy is also an property, which mean it depends on Thus, ΔS =. 3
Can calculate using at what temps? H - standard change in enthalpy H f S - standard change in entropy S G - standard change in free energy G f EXAMPLE: Calculating G Method 1) Use H and S to calculate G for the following reaction at 25 C. C 3 H 8 (g) + 5 O 2 (g) ---> 3 CO 2 (g) + 4 H 2 O (l) Change in enthalpy is H = n products H f - n reactants H f Does this move to lower or higher energy? Change in entropy is S = n products S - n reactant S Change in free energy is G = H - T S Method 2) Using G f calculate G G = n products G f - n reactants G f Free energy changes are properties of state (like enthalpy and entropy) Is the reaction spontaneous? What has more entropy reactants or products? What has more stored energy reactants or products? What has more bond energy reactants or products? 4
Temperature Dependence of G revisted: Some observations about signs of G, H, & S : H S G Results - + spontaneous at Temps + - spontaneous at Temps + + spontaneous at Temps - - spontaneous at Temps G can be calculated at different temperatures using G = H - T S : Calculate G at 450. K for the reaction at the top of this page. Summary of calculating G At 25 C: At other temperatures: Temp. effects: H vs. H same/about the same/totally different S vs. S same/about the same/totally different G vs. G same/about the same/totally different 5
G and phase changes: At equilibrium G = 0 or 0 = H -T S Thus at equilibrium T= Can be used to determine boiling point. Find the b.p. of ether, given H vap = 26.0 kj/mol and S vap = 84.4 J / mol K ----------------------------------------------------------------------------------------------------- Method 3) Using G rxn Calculate G: Free energy changes are properties of state (like enthalpy and entropy) Thus additivity of G s can be handled like H s for reactions. If 2 Fe(s) + 3 / 2 O 2 (g) ---> Fe 2 O 3 (g) ΔG = -742.2 kj CO(g) + 1 / 2 O 2 (g) ---> CO 2 (g) ΔG = -257.1 kj Fe 2 O 3 (g) ---> 2 Fe(s) + 3 / 2 O 2 (g) G = 3 CO(g) + 3 / 2 O 2 (g) ---> 3 CO 2 (g) G = Fe 2 O 3 (g) + 3 CO (g) ----> 2 Fe(s) + 3 CO 2 (g) G = The Meaning of ΔG for a Chemical Reaction A A B C B Problem Set 16.2 Notes Systems will achieve the lowest possible free energy by going to equilibrium, not by going to completion. G G G Fraction of Reactants used Fraction of Reactants used Fraction of Reactants used 6
ANOTHER USEFUL RELATIONSHIP: G = G + RT ln(q) R is gas law constant in joules (8.31 J / mol K ) T is temperature in K G is free energy change in joules Q is the reaction quotient (from Ch. 13!) ΔG is pressure/concentration dependent: S lg. vol S low pressure S sm vol. S high pressure G - G - Example: Consider the ammonia synthesis reaction: N 2 (g) + 3 H 2 (g) 2 NH 2 (g); ΔG = -33.3 kj Calculate ΔG for each of the following: a. P NH3 = 1.00 atm, P N2 = 1.47 atm, P H2 = 1.00 x 10-2 atm b. P NH3 = 1.00 atm, P N2 = 1.00 atm, P H2 = 1.00 atm c. P NH3 = 2.00 atm, P N2 = 2.00 atm, P H2 = 2.00 atm 1) Calculate Q 2 Predict the direction in which the system will shift to reach equilibrium for each of the above. 3) Calculate ΔG Ex. Given the following reaction: CO (g) + 2 H 2 (g) CH 3 OH (l) ΔG = -38 kj A. Predict which way the equilibrium would shift in order to reach equilibrium if the pressures of both gases (CO and H 2 ) were both 1 atm. B. Calculate G for the reaction assuming P CO = 5.0 atm, P H2 = 3.0 atm. C. Given your answer to B, is the reaction more/less/equally spontaneous with these new pressures? D. Explain how increasing the partial pressures of both reactants affect the spontaneity of the reaction. (Key: affecting spontaneity is NOT the same as which direction it will shift.) E. Explain how increasing the temperature affects the spontaneity of the reaction. 7
----------------------------------------------------------------------------------------------------- @ equil G =, Q =, so. G = G + RT ln(q) G = - R T ln K NOTE: If G is (-) G products G reactants K 1 If G is (+) G products G reactants K 1 If G is zero G products G reactants K 1 Ex: Given G = +18.0 kj for HF(aq) <===> H + (aq) + F - (aq) Find value of Ka @ 25 C. Predict the magnitude (small/large) and sign of G for the following: Magnitude Sign CH 3 COOH H + + CH 3 COO - H + + CH 3 COO - CH 3 COOH Mg(OH) 2 Mg 2+ + 2 OH - H + + + NH 3 NH 4 The Temperature Dependence of K G = - RT ln (K) = H - T S or Free Energy and Work w max = G ln (K) = - H 1 + S R T R G is from a process @ const. Temp. and pressure. 8