Unit 5: Spontaneity of Reaction You need to bring your textbooks everyday of this unit. THE LAWS OF THERMODYNAMICS 1 st Law of Thermodynamics Energy is conserved ΔE = q + w 2 nd Law of Thermodynamics A system NOT at equilibrium will move toward it. 3 rd Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero 1
Processes that are spontaneous in one direction are NOT spontaneous in the other direction. *Spontaneous does NOT mean RAPID. Be careful not to mix up thermodynamics with kinetics. Thermodynamics tells us about direction and extent of movement, but we need KINETICS to tell us about rate. Together, kinetics and thermodynamics describe a reaction. Standard Heat of Formation: ΔHf compounds formed from elements in their standard states. Exothermic heats of formation are ( ), endothermic are (+). Heat of formation of an element in its standard state = 0. Enthalpy and Spontaneity A spontaneous change involves going to a lower energy state. Endothermic and Exothermic changes can be spontaneous. Spontaneous changes often occur for combustion reactions, neutralization reactions, melting, dissolving. 2
Here are some specific examples: Combustion: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) ΔHf = 890.4 kj Neutralization: H + (aq) + OH (aq) H 2 O(l) ΔHf = 55.9 kj Melting: H 2 O(s) H 2 O(l) ΔHf = +6.01 kj Dissolving: NH 4 NO 3 (s) NH 4 +(aq) + NO 3 (aq) ΔHf = +25 kj An exothermic enthalpy change favors spontaneity, but another thermodynamic property needs to be considered, entropy Entropy: Symbol is S Units of S = J/mol K (vs. kj/mol for ΔH) It is the measure of order/disorder, the higher the degree of disorder, the greater the entropy. Second Law of Thermodynamics restated: When a spontaneous change occurs, the entropy of the universe as a whole increases. The driving force of any spontaneous change is an increase in the entropy of the universe. There is a drive toward the states that have the highest probability of existing. 3
Systems will move toward the state with the most possible arrangements or the highest probability of existing. Third Law of Thermodynamics: the entropy of a pure crystalline substance at absolute zero is zero. ie: S (at 0 K) = 0 Entropy changes: ΔS = S final S initial (Δ is always final initial) ΔS > 0 (ΔS = +) disorder is increasing, system becoming less organized, more random. ΔS < 0 (ΔS = ) disorder is decreasing, system getting more organized. ΔS > 0 is thermodynamically favorable disorder can be reduced by energy input. 4
Entropy Changes for Physical Phase Changes For a given substance: S(s) < S(l) < S(g) For freezing (l to s) molecules are more ordered so ΔS is negative. Entropy Changes for Chemical Reactions 1. NH 3 (g) + HCl(g) NH 4 Cl(s), ΔS is negative gas to solid more order 2. 2O 3 (g) 3O 2 (g), ΔS positive more moles of gas after reaction 3. H 2 (g) 2H(g), ΔS is positive more moles of gas after reaction 4. Ag + (aq) + Cl (aq) AgCl(s), ΔS negative solid more order than aqueous solutions ions 5
S increases as volume increases particle mobility increases (s l g) temperature increases particles dissolve moles of gas increases Example 17.1 Which of the following involve an increase in the entropy? a. melting a solid b. sublimation c. freezing d. mixing e. separation f. boiling 6
Example 17.2 Predict the sign of ΔS for each of the following changes a. Na (s) + ½ Cl 2 (g) NaCl (s) b. N 2 (g) + 3H 2 (g) 2NH 3 (g) c. NaCl (s) Na 1+ (aq) + Cl 1 (aq) d. NaCl (s) NaCl (l) e. CaCO 3 (s) CaO (s) + CO 2 (g) f. 2SO 2 (g) + O 2 (g) 2SO 3 (g) Practice Problems 1-10 in textbook. 7
Calculating ΔS rxn ( = standard thermodynamic conditions, 1 atm, 25 C, 1 M solution) ΔS rxn = Σ S products Σ S reaction S = absolute entropy of a substance at 25 C (can be calculated based on the fact that S = 0 at 0 K) *S for an element in its standard state is NOT zero. (No element has a standard state defined where T = 0 K.) Example 17.3 Predict the sign of ΔS then calculate ΔS for each of the following reactions. a. 2SO 3 (g) 2SO 2 (g) + O 2 (g) b. Fe 2 O 3 (s) + 3H 2 (g) 2Fe (s) + 3H 2 O (g) c. H 2 (g) + ½ O 2 (g) H 2 O (l) 8
Example 17.3 Predict the sign of ΔS then calculate ΔS for each of the following reactions. (continued) d. N 2 (g) + 3H 2 (g) 2NH 3 (g) e. HCl (g) H 1+ (aq) + Cl 1 (aq) 9
Example 17.4 For the reaction: C 2 H 2 (g) + 4F 2 (g) 2CF 4 (g) + H 2 (g) ΔS is equal to 358 J/K. Use this value and data from Thermodynamic Data Sheet to calculate S for CF 4 (g) Practice Problems 12-16 evens in textbook. 10
The Two Factors Affecting Spontaneity 1. Tendency to go to a lower enthalpy. 2. Tendency to go to a higher entropy. Gibb's Free Energy: ΔG = ΔH TΔS ΔG = energy free to do useful work ΔH = change in heat content (enthalpy) TΔS = energy associated with changing order/disorder ΔG depends on the nature of products, reactants, and the conditions (P, T, and concentration) not on the path for which the the rx. is carried out. Two Ways to Calculate ΔG o rxn 1. Use ΔG rxn = ΔH TΔS where temperature is in Kelvin. (K = o C + 273.16) Calculate ΔH rxn = Σ ΔH prod Σ ΔH react and Calculate ΔS rxn = Σ S prod Σ S react Substitute and calculate ΔG rxn = ΔH TΔS 2. Use ΔG rxn = Σ ΔG f prod Σ ΔG f react ΔG f for an element in its standard state is 0 kj/mol. 11
Relationship between ΔG and spontaneity: ΔG = ( ), spontaneous ΔG = 0, equilibrium ΔG = (+), non spontaneous, spontaneous in the reverse rx. What Does Spontaneous or Non spontaneous Mean? 1. Physical change: if spontaneous, these go to completion, 100% change. There is no tendency for an ice cube to stop melting at 10 C. H 2 O(s) H 2 O(l) at 10 C 2. Chemical reactions: Spontaneous means a reaction that gives predominantly products. The system may reach equilibrium, the right side will be favored. 12
Non spontaneous does not mean "no reaction." Here, spontaneous, non spontaneous does not mean "go or no go." Non spontaneous means the reaction mixture is mostly reactants Non spontaneous reactions can be made to occur by use of an outside energy source. An example, electrolysis of water and other electrolytic reactions. Qualitative Relationship Between ΔG and Position at Equilibrium 1. When ΔG is ( ) The more ( ) the ΔG, the higher the % reaction. Large ΔG rxns go more toward completion than low ΔG reactions. Reactants are always higher than products, but the bottom of the curve skews left or right between 50 and 100%. 2. When ΔG is (+) The more (+) the ΔG, the lower the % reaction. Reactants are always lower than products, but the bottom of the curve skews left or right between 0 and 50%. The more ( )ΔG rxn, the higher the % completion of the reaction at 25 C. The more (+)ΔG rxn, the lower the % completion of the reaction at 25 C. Reactions that are non spontaneous in one direction will be spontaneous in the other. Also, remember that conditions will often not be standard. 13
Example 17.5 Consider the reaction 2SO 2 (g) + O 2 (g) 2SO 3 (g) carried out at 25 C and 1 atm. Calculate ΔH, ΔS, and ΔG. Example 17.6 Methanol, CH 3 OH (l), is a high octane fuel used in high performance racing engines. Calculate ΔG f for it s combustion reaction. Note water vapor is a product. 14
Example 17.7 A chemical engineer wants to determine the feasibility of making ethanol, C 2 H 5 OH by reacting water ethylene C 2 H 4, according to the equation: C 2 H 4 (g) + H 2 O (l) C 2 H 5 OH (l) Is this reaction spontaneous under standard conditions? Example 17.8 Given the following data S (s) + 3/2 O 2 (g) SO 3 (g) 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔG = 371 kj ΔG = 142 kj Calculate ΔG for the reaction: S (s) + O 2 (g) SO 2 (g) 15
Ex 17.5 pg. 456 Calculate ΔG at 230 C for the reduction of one mole of Fe 2 O 3 with hydrogen; the products are iron metal and water vapor Homework #3 pgs. 468-469 18a, 18b, 20a, 20b, 22a, 24 all, 26 all, 28 all, 55 all, and 56 all 16
Example 17.9 #29 for our textbook: The alcohol in most liqueurs is ethanol, C 2 H 2 OH. It is produced by the fermentation of the glucose in fruit or grain. C 6 H 12 O 6 (aq) 2C 2 H 5 OH (l) + 2CO 2 (g) ΔH = 82.4 kj ΔG = 219.8 kj at 25 C a. Calculate ΔS for this reaction. Is the sign reasonable b. Calculate S for C 6 H 12 O 6 (aq) c. Calculate ΔH f for C 6 H 12 O 6 (aq) Example 17.9 #29 for our textbook: The alcohol in most liqueurs is ethanol, C 2 H 2 OH. It is produced by the fermentation of the glucose in fruit or grain. c. Calculate ΔH f for C 6 H 12 O 6 (aq) C 6 H 12 O 6 (aq) 2C 2 H 5 OH (l) + 2CO 2 (g) ΔH = 82.4 kj ΔG = 219.8 kj at 25 C 17
The Effect of Temperature on Spontaneity ΔH and ΔS are virtually temperature independent. So, ΔG = ΔH T ΔS can be used to calculate ΔG at any temp. Temperature is considered in T ΔS. 1. ΔG = ΔH T ΔS ΔH and + ΔS 2. ΔG = ΔH T ΔS + ΔH and ΔS 3. ΔG = ΔH T ΔS + ΔH and + ΔS 4. ΔG = ΔH T ΔS ΔH and ΔS Situations 2 and 3 where the signs for ΔH and ΔS are ALIKE are temperature dependent. Situations 1 and 4, where the signs are OPPOSITE are not temperature dependent. 18
Calculating the Temperature at Which a Reaction Becomes Spontaneous/Non spontaneous ΔG = ΔH T ΔS Example 17.11 #33a from textbook Discuss the effects of temperature change on the spontaneity of the following reactions at 1 atm. 2PbO (s) + 2SO 2 (g) 2PbS (s) + 3O 2 (g) ΔH = +830.8 kj ΔS = +168 J/K 19
35a from textbook At what temperature does ΔG become zero for the above reaction. Explain the significance of your answer Example 17.11 #33b from textbook (continued) Discuss the effects of temperature change on the spontaneity of the following reactions at 1 atm. 2As (s) + 3F 2 (g) 2AsF 3 (l) ΔH = 1643 kj ΔS = 0.316 kj/k 20
35b from textbook At what temperature does ΔG become zero for the above reaction. Explain the significance of your answer 33c from textbook Discuss the effects of temperature change on the spontaneity of the following reactions at 1 atm. CO (g) C (s) + ½ O 2 (g) ΔH = +110.5 kj ΔS = 89.4 J/K 21
35c from textbook At what temperature does ΔG become zero for the above reaction. Explain the significance of your answer Example 17.12 #43 from textbook Red phosphorus is formed by heating white phosphorus. Calculate the temperature at which the two forms are at equilibrium. White P: ΔH f = 0.00 kj/mol Red P: ΔH f = 17.6 kj/mol S = 41.1 J/mol K S = 22.8 J/mol K 22
Homework #4 pgs. 469-470 34, 36, 38, 42, 44, 46, 48 ΔG = Σproducts Σreactants ΔG = ΔH TΔS ΔG = ΔG + RTlnQ 23
Relationship Between ΔG, Equilibrium Expressions, and Reaction Quotients ΔG = standard free energy change ΔG = ΔG + RTlnQ (standard conditions: P= 1 atm, [ ] = 1M, T = 25 C or 298K) ΔG = free energy under any conditions. T = temp in Kelvins Q = Reaction quotient, from the equilibrium constant equation. gases represented by pressure aqueous solutions represented by molar concentration, (solids, liquids do not get used, consider them to be the value of 1) 24
ΔG = ΔG + RTlnQ Reaction Quotient aa(g) + bb (g) cc(g) + dd (g) Q = (P C c) X (P D d)or Q = [C] c X [D] d (P A a) X (P B b) [A] a X [B] b R = 8.314 X 10 3 kj/mol K (8.314 J/mol K) If conditions are NOT standard, ΔG must be calculated, NOT ΔG. ΔG changes as the reaction proceeds, with changes in Q. Driving forces (H and S) change until equilibrium is achieved. Example 17.12 Consider the above reaction: Zn (s) + 2H 1+ (aq) Zn 2+ (aq) + H 2 (g) at 25 C. Calculate a. ΔG 25
Example 17.12 Consider the above reaction: Zn (s) + 2H 1+ (aq) Zn 2+ (aq) + H 2 (g) at 25 C. Calculate b. ΔG when P H2 = 750 mm Hg, [Zn 2+ ] = 0.10 M, [H 1+ ] = 1.0 X 10 4 M Example 17.12 Consider the above reaction: Zn (s) + 2H 1+ (aq) Zn 2+ (aq) + H 2 (g) at 25 C. Calculate c. [H 1+ ] when ΔG = 100.0 kj and all other species involved in the reaction are at standard conditions. 26
The Free Energy Change and the Equilibrium Constant At equilibrium, ΔG = 0, and Q = K. ΔG is a true measure of spontaneity under any conditions. At equilibrium, ΔG = 0 = ΔG + RTlnK, therefore ΔG = RTlnK Note: K is equilibrium constant (K gas, K w, K a, K b, K sp, ) A frequently confusing issue results from the fact that ΔG implies 25 C (298K), however ΔG may be calculated for other temperatures. The position of equilibrium is related to ΔG. If K > 1, ΔG is ( ), mostly product and spontaneous If K = 1, ΔG = 0, system at equilibrium If K < 1, ΔG = (+), mostly reactants and non spontaneous. 27
Example 17.13 Using ΔG f tables in Appendix 1, calculate the solubility product constant, K sp, for PbCl 2 at 25 C Operations Involving Logarithms Multiplication: ln (xy) = ln x + ln y Division: ln (x/y) = ln x ln y Raising to a Power: ln (x n ) = n ln x Extracting a Root: ln(x 1/n ) = 1 ln x n Taking a Reciprocal: ln (1/x) = ln x 28
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