Lecture 7: The Second and Third Laws of Thermodynamics. Reading: Zumdahl 10.4,10.5, 10.6 Outline

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1 Lecture 7: The Second and Third Laws of Thermodynamics Reading: Zumdahl 10.4,10.5, 10.6 Outline Definition of the Second Law Determining ΔS for a reaction Definition of the Third Law Problems: Z , Z , Z , Z , Z

2 4 3 1 V 1 V 2 Volume Any quantity that is zero when summed over all changes through a cycle (bringing the system back where it started) is a state function. 2 Entropy from Heat Engine V 2 V 2 q1 = nrthigh ln and q3 = nrtlow ln V1 V1 V qtotal = q1+ q3 = nrδt V1 0 = q 1 T high + q 3 T low q Δ S = = T 2 ln 0 final initial Thermodynamic definition of entropy: The heat transferred under ersible conditions divided by the temperature. What if the transfer is not ersible? Then the amount of heat transferred is not the same but the entropy is the same. dq T 2

3 Calculating Entropy: Different Processes Calculus Refresher x2 dx x 2 = ln x and = ln x 1 dx x x x 1 ΔS = final initial dq T ΔV q = nrt V ΔT = 0 ΔS = final initial dq T = nrln V final V initial q = qv =Δ E = ncvδt ΔV = 0 ΔS = final initial dq T = nc v ln T final T initial ΔP = 0 q = qp =Δ H = ncpδt ΔS = final initial dq T = nc p ln T final T initial 3

4 Calculating Entropy (Z10.8) Example: What is ΔS when heating one mole of an ideal monatomic gas isochorically from 298 K to 350 K? Δ V = 0 q = q =ΔE final dq Tfinal 3 Δ S = = ncvln where CV = R T Tinitial 2 initial V 3 350K Δ S = (1 mol) R ln 2 J = 2 298K K Z10.8 Which is larger ΔS at constant pressure or ΔS at constant volume? Provide a conceptual rationale. Z10.27 shows several examples of using these expressions. 4

5 Connecting with Boltzmann s Theory From Thermodynamics for isothermal expansion: q V nrt final Vfinal Δ S = = ln = nrln T T Vinitial Vinitial Exactly the same as derived based on possible arrangements! (at constant energy, i.e. isothermal). ΔS = Nk ln V final = nrln V final V initial V initial Compute the entropy change for the gas expansion from one bulb to two bulbs (Joule s experiment). Reconcile Δ S > 0 with q=0 in Joule s experiment. (Z10.26) Note q = w = TΔS (for Isothermal Change) 5

6 The Second Law The Second Law: In any spontaneous (or possible) process, there is always an increase in the entropy of the universe. Source: ersible work is always greater than any other kind of work. w w Δ E = q + w = q+ w 0 Isolated System 0 From our definitions of system and surroundings: w w = TΔS q q TΔS ΔS 0 Δ Suniverse =Δ Ssurrounding +ΔSsystem 6

7 Use of The Second Law Three possibilities: IfΔS univ > 0..process is spontaneous If ΔS univ < 0..process is spontaneous in opposite direction. If ΔS univ = 0.equilibrium Here s the catch: We need to know ΔS for both the system and surroundings to predict if a reaction will be spontaneous! 7

8 The Second Law (cont.) Consider a reaction driven by heat flow from the surroundings at constant P. Exothermic Process: ΔS surr = heat/t Endothermic Process: ΔS surr = -heat/t Heat transferred = q P,surr = - q P,system = -ΔH sys ΔS surr = ΔH sys T 8

9 Entropy of Surroundings Example For the following reaction at 298 K: Sb 4 O 6 (s) + 6C(s) 4Sb(s) + 6CO 2 (g) ΔH = 778 kj What is ΔS surr? ΔS surr = -ΔH/T = -778 kj/298k = -2.6 kj/k Being endothermic (on its own) is not spontaneous. (See Z10.9). Is the reaction spontaneous? We need ΔS system and add them up. If the sum is positive, then yes. 9

10 The Third Law Recall, in determining enthalpies we had standard state values to use, as reference points. Does the same thing exist for entropy? The third law: The entropy of a perfect crystal at 0K is zero. The third law provides the reference state for use in calculating absolute entropies. 10

11 What is a Perfect Crystal? Perfect crystal at 0 K S=0 Crystal deforms at T > 0 K S>0 11

12 Standard Entropies With reference to this state, standard entropies have been tabulated (Appendix 4). Recall, entropy is a state function; therefore, the entropy change for a chemical reaction can be calculated as follows: ΔS o reaction = S o S o prod. react. 12

13 Compute Entropy for a Reaction Balance the following reaction and determine ΔS rxn. (Z10.41) Fe(s) + H 2 O(g) Fe 2 O 3 (s) + H 2 (g) 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (s) + 3H 2 (g) ΔS rxn = ( S (Fe 2 O 3 (s)) + 3S (H 2 (g) ) -( 2S (Fe(s) )+ 3S (H 2 O(g)) ) Use the absolute molar entropies of formation in Appendix A to compute reaction entropy: ΔS rxn = J/K 13

14 Spontaneity Example Is the following reaction spontaneous at 298 K? (Is ΔS univ > 0?) 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (s) + 3H 2 (g) ΔS rxn = ΔS system = J/K ΔS surr = -ΔH sys /T = -ΔH rxn /T ΔH rxn = ΔH f (Fe 2 O 3 (s)) + 3ΔH f (H 2 (g)) -2ΔH f (Fe (s)) - 3 ΔH f (H 2 O(g)) 14

15 Example (cont.) ΔH rxn = -100 kj ΔS surr = -ΔH sys /T = 348 J/K ΔS univ = ΔS sys + ΔS surr = J/K J/K = J/K ΔS univ > 0 ; therefore, reaction is spontaneous. What is driving the reaction? It is entropically forbidden; But Enthalpically favorable. At lower temperatures it will be more favorable 15

16 Entropy and Phase Changes Phase Change: Reaction in which a substance goes from one phase of state to another. Example: H 2 O(l) H K Phase changes are equilibrium processes such that: ΔS univ = 0 16

17 ΔS and Phase Changes H 2 O(l) H K Now, ΔS rxn = S (H 2 O(g)) - S (H 2 O(l)) = J/K J/K = J/K And, ΔS surr = -ΔH sys /T = kj/373 K = J/K Therefore, ΔS univ = ΔS sys + ΔS surr = 0 17

18 Phase Transition Temperature Determine the temperature at which liquid bromine boils: Br 2 (l) Br 2 (g) Now, ΔS rxn = S (Br 2 (g)) - S (Br 2 (l)) (for the system) = J/K J/K = 93.2 J/K At equilibrium ΔS universe =0 18

19 Phase Transition Example Now, ΔS sys = 93.2 J/K = -ΔS surr = ΔΗ sys /T Assume ΔH sys and ΔH sys are independent of temperature. Therefore, calculate ΔH sys and solve for T! Now, ΔH rxn = ΔH f (Br 2 (g)) - ΔH f (Br 2 (l)) = kj - 0 = kj (standard state) 0 Therefore: 93.2 J/K = kj/t T boiling = K Z10.48 The higher temperature form is is entropically favored, and the lower temperature form is enthalpically favored (and is more ordered). 19