Chem 1721 Brief Notes: Chapter 18

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1 Chem 1721 Brief Notes: Chapter 18 Chapter 18: Thermodynamics and Equilibrium A quick refresher on Thermochemistry (chapter 6) understanding of energy changes that accompany chemical and physical change enthalpy, ΔH = q P endothermic vs. exothermic change heat absorbed heat released H f > H i H f < H i ΔH is + ΔH is Hess s Law of Heat Summation as a way to calculate an unknown ΔH rxn standard state and standard reaction enthalpies; ΔH standard enthalpy of formation, ΔH f and using ΔH f to calculate ΔH rxn ΔH rxn = Σ n ΔH f (products) Σ n ΔH f (reactants) Extend our discussion of thermodynamics to consider: What factors determine where a system reaches equilibrium? Will K be large or small? In what direction will a reaction proceed? Spontaneous processes proceed on their own without any external influence the reverse of a spontaneous process is nonspontaneous (requires external influence) a spontaneous process always moves a system toward equilibrium Enthalpy, Entropy and Spontaneity What do spontaneous processes have in common? Are they all exothermic? H 2 O (s) H 2 O (l); ΔH = +6 kj NaCl (s) Na + (aq) + Cl (aq); ΔH = kj spontaneous processes move a system toward greater disorder; tendency toward randomness entropy, S is the thermodynamic quantity that related to randomness when disorder increases S f > S i ; ΔS is + when disorder decreases S f < S i ; ΔS is factors to consider when evaluating S f and S i (and therefore predicting sign of ΔS) 1. physical state; S solid < S liquid < S gas 2. number of particles; more particles greater disorder 3. volume (for gas samples); when volume occupied by a gas sample increases, so does the positional entropy 4. temperature; kinetic and thermal energy increase with T therefore greater disorder at higher T s predict the sign of ΔS for the following CO 2 (s) CO 2 (g) NaCl (s) Na + (aq) + Cl (aq) 1 mol N 2 (g) in 5.0 L 1.0 mol N 2 (g) in 25.0 L 1 mol N 2 (g) at 100 C 1.0 mol N 2 (g) at 200 C CaSO 4 (s) CaO (s) + SO 3 (g) N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

2 statistical interpretation of entropy there are more possible ways to have a disordered system than an ordered one disordered states are more probable consider 5 quarters in a box; how many possible head/tail configurations? all T s 3 T s and 2 H s 1 T and 4 H s 4 T s and 1 H 2 T s and 3 H s all H s each of these is called a macrostate; here there are 6 macrostates but each individual quarter can be H or T; there are 2 5 possible specific arrangements each specific arrangement is called a microstate; here there are 32 microstates microstate macrostate #microstates microstate macrostate # microstates per macrostate per macrostate TTTTT 5 T 1 TTHHH 2T:3H 10 THTHH HTTTT 4T:1H 5 THHTH THTTT THHHT TTHTT HTTHH TTTHT HTHTH TTTTH HTHHT HHTTH HHTTT 3T:2H 10 HHTHT HTHTT HHHTT HTTHT HTTTH HHHHT 1T:4H 5 THHTT HHHTH THTHT HHTHH THTTH HTHHH TTHHT THHHH TTHTH TTTHH HHHHH 5H 1 the most probably macrostate corresponds to the maximum entropy each microstate has the same probability; 1/32 (3.125%) 3T:2H and 2T:3H tie as the most probable macrostate; 10/32 (31.25%) probability of coins arranging in a 3:2 macrostate is 20/32 (62.5%) the number of microstates in a given macrostate: # = n! (H!)(T!) Ludwig Boltzman: S = k B lnw where k B = Boltzman s constant = x J/K W = # microstates per macrostate larger W more microstates more probable more disordered population greater entropy ΔS = S f S i = k B ln(w f /W i ) calculate the molar entropy change for an ideal gas that expands to occupy 2x its initial volume

3 2 nd Law of Thermodynamics the entropy of the Universe is increasing. Universe = System + Surrounding if ΔS univ > 0 the process is spontaneous if ΔS univ < 0 the process is nonspontaneous if ΔS univ = 0 the system is at equilibrium all spontaneous processes proceed in the direction that will increase the entropy of the Universe Calculations of: ΔS sys, ΔS surr (and dependence on temperature), and ΔS univ calculation of ΔS sys use tabulated standard molar entropy data, S (Appendix C) units kj/mol 3 rd Law of Thermodynamics: the entropy of a perfectly crystalline solid at 0 K is 0. note S 0 for elements ΔS = Σn S (products) Σn S (reactant) calculate ΔS for the following reaction: N 2 O 4 (g) 2 NO 2 (g) S = J/K mol for N 2 O 4 and J/K mol for NO 2 ΔS = [(2mol NO 2 )(240.0 J/K mol)] [(1mol N 2 O 4 )(304.2 J/K mol)] = J/K calculation of ΔS surr for a reaction at constant pressure, ΔS surr is directly proportional and opposite in sign to ΔH, and inversely proportional to temperature ΔS surr = ΔH/T consider: ΔS surr ΔH for an exothermic reaction: ΔH is, heat is released by system, energy flow is system surroundings energy and thermal motion in the surroundings increases therefore disorder and entropy increase for an endothermic reaction: ΔH is +, heat is absorbed by system, energy flow is surrounding system energy and thermal motion in the surroundings decreases therefore disorder and entropy decrease consider: ΔS surr 1/T subtle interpretation of adding or removing energy in the surroundings in the form of heat think of the surroundings as a constant temperature environment what is the relative change in disorder when energy is added or removed here? at high T surroundings are already highly chaotic and disordered; addition of more energy increases disorder, but the ΔS is relatively small at low T surroundings are fairly ordered; addition of energy causes a rather significant increase in disorder so ΔS is relatively large Calcualte ΔS univ for: 4 Fe (s) + 3 O 2 (g) 2 Fe 2 O 3 (s). Is this reaction spontaneous or nonspontaneous at 25 C? some data: S (units J/K mol) ΔH f (units kj/mol) Fe (s) O 2 (g) Fe 2 O 3 (s) 87.4 J ΔS sys = Σn S (products) Σn S (reactant) ΔS surr = ΔH /T; recall: ΔH rxn = Σ n ΔH f (products) Σ n ΔH f (reactants) ΔS univ = ΔS sys + ΔS surr

4 Gibbs Free Energy, G G = H TS ΔG = ΔH TΔS relationship to spontaneity of a process? for a spontaneous process ΔS univ is + it would be easier to have a thermodynamic quantity for the system to serve as an indicator of spontaneity if: ΔG = ΔH TΔS divide both sides by T ΔG /T = ( ΔH /T) + ΔS ; ΔS = ΔS sys ΔH /T = ΔS surr ΔG /T = ΔS univ when ΔS univ is +, ΔG is so: ΔG is for spontaneous processes ΔG is + for nonspontaneous processes when ΔG = 0 the system is at equilibrium when ΔG < 0 the reaction is spontaneous ΔH and ΔS are both important factors in determining spontaneity What role does temperature play? ΔH ΔS ΔG spontaneous process? + spontaneous at all T s OR + spontaneous at low T s when ΔH > TΔS + + nonspontaneous at all T s + + OR + spontaneous at high T s when TΔS > ΔH for cases where spontaneity depends on temperature there is a crossover T; T when ΔG = 0 ΔG = 0 = ΔH TΔS T = ΔH /ΔS consider the following reaction: Fe 2 O 3 (s) + 3 H 2 (g) 2 Fe (s) + 3 H 2 O (g) ΔH = 98.8 kj ΔS = J/K Is this reaction spontaneous at 25 C? If not, at what temperature does it become spontaneous? ΔH and ΔS are both +, so the reaction will be spontaneous at high T s. when T = 25 C: ΔG = 98.8 kj (298 K)( kj/k) = kj the reaction is not spontaneous at 25 C temperature where ΔG = 0? T = (98.8 kj)/( kj/k) = 698 K the reaction will be spontaneous at temps above 698 K (425 C) Calculating ΔG for reactions: ΔG is the standard free energy change for a reaction; the change in free energy for reactants in their standard states forming products in their standard states at 1 atm P, a defined T (usually 25 C), and solution concentration of 1 M. 3 ways to calculate ΔG : 1. use ΔG = ΔH TΔS (see previous example) recall: ΔH = ΣnΔH f (products) ΣnΔH f (reactants) [see chapter 6.8, probs 77 85, 112, 114, 118] ΔS = ΣnΔS (products) ΣnΔS (reactants)

5 2. use Hess s Law [see chapter 6.7, probs 71 75, 110] 3. use ΔG f values ΔG f is the free energy change corresponding to the formation of 1 mol of a substance in its standard state from its elements in their standard states. units kj/mol tabulated in Appendix C ΔG f = 0 for elements in their standard state and most stable form ΔG f is a measure of thermodynamic stability of a substance with respect to its elements; the larger and more negative a ΔG f the more thermodynamically stable a substance with respect to decomposition to its elements; a compound with a + ΔG f should spontaneously decompose into its constituent elements (thermo vs. kinetics) for a reaction: ΔG = ΣnΔG f (products) ΣnΔG f (reactants) Calculate ΔG for 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (l) using ΔG f data (given below). ΔG f (kj/mol) = Free Energy (ΔG), Reaction Composition (Q), and Equilibrium (K) For a reaction at non-standard conditions... ΔG = ΔG + RTlnQ where R = J K 1 mol 1 Q = reaction quotient ΔG and Q are related both predict the direction of a reaction as it approaches equilibrium both are dependent on reaction composition both are related to the equilibrium position of a system Q vs K? Q > K, reaction proceeds in reverse direction toward equilibrium Q < K, reaction proceeds in forward direction toward equilibrium Q = K, reaction is at equilibrium ΔG + or? ΔG is, the reaction is spontaneous in the forward direction ΔG is +, the reaction is spontaneous in the reverse direction ΔG = 0, the system is at equilibrium Calculate ΔG for: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) at 25 C when P N2 = 1.0 atm, P H2 = 3.0 atm, and P NH3 = atm. ΔG = 33.0 kj/mol. Q = (P NH3 ) 2 (P N2 )(P H2 ) 3 Q = 1.5 x 10 5 ΔG = 33.0 kj/mol + (8.314 x 10 3 kj/k mol)(298k)ln(1.5 x 10 5 ) = 60.5 kj/mol

6 think about how ΔG changes as a reaction proceeds... 2 extremes: if [reactants] are v. large and [products] are v. small Q << 1 RTlnQ < 0 (or ) ΔG < 0 (or ) reaction is spontaneous in the forward direction if [products] are v. large and [reactants] are v. small Q >> 1 RTlnQ > 0 (or +) ΔG > 0 (or +) reaction is spontaneous in the reverse direction somewhere in the middle the system reaches equilibrium energy minimum equilibrium composition corresponds to the lowest free energy (and maximum entropy of the universe) ΔG = 0 when moving toward equilibrium ΔG is ; when moving away from equilibrium ΔG is + ΔG and K if, at equilibrium ΔG = 0 and Q = K... then... ΔG = ΔG + RTlnQ becomes ΔG = RTlnK temperature dependence of K for a given reaction at constant pressure: ΔG = ΔH TΔS AND set these two equations equal to one another and rearrange... ΔG = RTlnK lnk = ΔHo R 1 T + ΔSo R look at this equation in the form of y = mx + b; a plot of ln K vs. 1/T should yield a straight line slope = ΔH /R and y intercept of ΔS /R Calculate K at 25 C for: CO (g) + 2 H 2 (g) CH 3 OH (g). ΔG f (in kj/mol): ΔG = [(1mol CH 3 OH)( kj/mol)] [(1 mol CO)( kj/mol)] = 24.7 kj lnk = ΔG /( RT) = 9.97 K = 2.14 x 10 4