Chem Lecture Notes 6 Fall 2013 Second law

Chem 340 - Lecture Notes 6 Fall 2013 Second law In the first law, we determined energies, enthalpies heat and work for any process from an initial to final state. We could know if the system did work or gave out heat (exothermic) or the reverse. However, we could not predict if it would actually happen. In nature, and in all our experiences, we know there are processes that always go in one way on their own, spontaneously. Heat always flows from a hotter to colder system, gas always expands from a higher to lower pressure region, rocks roll downhill, not up, Na metal always reacts with water to form NaOH and H 2, salt dissolves in water. These processes happen spontaneously, without outside intervention, unless we counteract them (e.g. with work). The reverse processes do not happen, a heated metal rod does not spontaneously get hot and one end and cold at another, the air in the room never spontaneously moves all to one corner, sugar does not crystalize out of your coke unless you evaporate the water or cool the solution, i.e. do some work on it. These processes are irreversible, spontaneously go in one direction and not in reverse. To reverse them you need to take energy from the surroundings have them work/heat on system. By contrast, reversible processes go in infinitesimal steps and can reverse (at equilibrium). We learned that work done by the system is maximum for reversible processes, so this concept relates to getting the most work out of a system - efficiency. Thermodynamics needs a parameter, measure, ideally a state function, to tell us if a process will occur spontaneously, or in what direction a process/reaction will proceed. That is entropy, and is the heart of the second law of thermodynamics. There are various ways to introduce the concept of entropy into thermodynamics, none of them is possibly ideal. The idea of reversible and irreversible processes opens up a path to a function. Lets assume there is entropy = S and we want it to be a state function, so only depends on initial and final state: S = ds = S f S i, and over a cycle: o ds = 0 S cannot be U or H since those state functions do not indicate direction of a process i.e. system does not just lose energy, e.g. expand ideal gas to vac., U=0, spont. Similarly ds cannot be ᵭq or ᵭw since those variables are not state functions. Second law statements tend to be abstract (and sort of negative) but they always hook up heat and temperature into a combined change: Clausius: heat cannot flow form a colder to a hotter body without work added or It is impossible to devise a continuously cycling engine that produces no effect other than the transfer of heat from a colder to a hotter body. i.e. refrigerator needs engine and power (compressor) 1

Kelvin: one cannot use a cyclic process to transform heat into work without also transferring heat from a hotter to a colder reservoir or It is impossible to devise a continuously cycling engine that produces no effect other than the extraction of heat from a reservoir at one temperature and the performance of an equal amount of mechanical work. (no perpetual motion machines example, ship cannot draw energy from ocean, without losing heat to something colder, text--ball at rest cannot bounce up) Carathéodory statement: In the neighborhood of every equilibrium state of a closed system there are states that cannot be reached from the first state along any adiabatic path. oh well! These statements combine heat and temperature (colder, hotter) and seem to depend on machines. Thermodynamics originally analyzed at heat engines like a steam engine, and evaluated the work derived from the engine by providing energy in form of heat. These models lead to entropy, S. Also they tend to disperse energy (into both heat and work) like text example bouncing ball dispersing energy into floor. timing the contacts drives rod down and up turning the crankshaft So to get at elusive entropy, lets turn to a model engine, the Carnot cycle: Heat engine works by expanding gas diathermal contact cylinder with T hot, and contracting by contact to T cold Model this with a cycle of reversible steps, 1 st a b, isothermal expansion (in contact T hot ) follow by b c, adiabatic expansion, then reverse, lower T cold isothermal compression, c d, follow by adiabatic compression, d a. Since each step reversible and has controlled parameters can be modeled with the ( H, U,q,w) we know. 2

Have an empirical look at the plot, what it means First 2 steps, expansion, system does work, equivalent to area under curve a b c Second 2 steps, surroundings do work on system, equivalent to area under c d a Clearly the cycle does work on surroundings, i.e. design of engine, w tot = area inside Design need mix isothermal and adiabatic, since both process occur on nonintersecting curves, to get a cycle cannot use just one, must combine them Cyclic, so 4 steps restore the system to its initial state. isothermal, T h, U=0 adiabatic, q=0, U=w isothermal, T c, U=0 Since a cycle: ΔU = 0 w cy = w 1 + w 2 +w 3 +w 4 q cy = q h + q c = q ab +q cd ΔU = 0 = w cy + q cy thus - w cy = q h + q c adiabatic, q=0, U=w Work done by system is greater than done on system, w tot <0 (since syst does work) Means q cy >0, or q ab =q h > q cd =q c because heat flow: q h into syst, q c out from syst So to get work, need more heat form surroundings than give back If w cy is negative (work has flowed from the system across the boundary and appeared in the surroundings), any work produced by the system has been produced at the expense of the thermal energy of the surroundings. That is, heat (q h + q c ) has flowed from the surroundings to the system at the same time. (Kelvin 2 nd law) 3

From steps 2 and 4: for w cy note C V dt terms cancel, opp. sign integrals ln(v C /V B ) = -ln(v A /V D ) since both = (C V /R)ln(T c /T h ) rearrange: (V B /V A ) = (V C /V D ) w cy = R( T h T c ) ln(v B /V A ) depends on the temperatures of the two reservoirs and on the ratio (V B / V A ) (i.e. compression ratio) for an ideal gas as the working substance in a reversible cycle. Second law says heat engine must always lose heat to surroundings Process (a) normal, but (b) cannot happen ( perpetual motion of second kind) If engine (b), then T h =T c and PV curve is a line, no area, w cy =0 no machine has 100% efficiency, but efficiency improves for T h >>T c but only ~1 for T h ~, T c ~0 Example: calculate max work done by a reversible heat engine operating between 500 and 200K if 1kJ is absorbed at 500K Efficiency: e = 1 T 200 /T 500 = 1-2/5 = 0.6 Work: (step of expansion) w =. q ab = 0.6 x 1000J = 600 J Go back to efficiency equation: 4

So have a state function, ds = dq rev /T, or S = dq rev /T and it describes entropy technically, dq rev = ᵭq rev inexact differential, but if divide by T, exact (state fct) Because of definition, must calculate entropy for a reversible path, even if process is irreversible, need reversible path to get entropy between initial and final states Categories: 1) reversible adiabatic, q = 0, so S = dq rev /T = 0 2) cyclic process: S = o dq rev /T = 0 state function 3) isothermal expansion, U=0, q rev = -w rev =nrt ln(v f /V i ) from w= - PdV S = dq rev /T = q rev /T = nr ln(v f /V i ) S > 0 for an isothermal expansion, ideal: no energy change but entropy change Note any process V i V f, keep at const. T, has same S 4) reversibly change T, const. V U = q rev, or dq rev = C V dt S = dq rev /T = nc Vm dt/t = nc Vm ln(t f /T i ) 5) const P see same thing, dq rev = C P dt S = dq rev /T = nc Pm dt/t = nc Pm ln(t f /T i ) Note: above assumes C V or C P constant over T range If change both V and T, V i T i V f T f, then break into 2 seg, const T, const V same for P i T i P f T f but P f < P i - expansion, need (-) sign for work (const T) part Example: See extra sheets- S irrev find rev. path 5

Example: Showed on Friday for irreversible path (adiabatic expansion of gas T=500 K,P=100 atm P=1 atm against P ext =1 atm) that first needed to determine T f and V f, which you could do from U=w, since q=0 for adiabatic) Then: create a reversible path from V i T i V f T f and use this path to calculate S=q rev /T For adiabatic, const V -T change, then const T- V change S=q rev /T=R ln(24.8/0.411) + 3/2 R ln (500/302) = (34 + 6.3)JK -1 = 40 JK -1 Now if a reversible expansion know from previous chapter: T f /T i = (V i /V f ) -1 = (V i /V f ) 2/3 S=q rev /T= R ln(v i /V f ) + 3/2 R ln (T f /T i ) = -R ln(v i /V f ) + 3/2 R ln (V i /V f ) 2/3 = 0 So S rev = 0 and q rev = 0 but S irev > 0 Phase change can happen at a const T since system needs heat (or loss of heat) to change phase, stays at T b or T m for boiling or melting (note, vapor press different) S = dq rev /T = q rev /T = H vap /T Same for fusion: S = dq rev /T = q rev /T = H fus /T Example for phase change: supercooled water, irreversible freezing at T < T m, for S calc need reversible cycle, heat to 0 C, freeze, cool to low T, separate page Generalize, if know and i.e. expansion coefficient and compressibility, for substance, then can determine S without equation of state for change: V i T i V f T f S = C V /T dt + dv = C V ln(t f /T i ) + V Alternatively have change: P i T i P f T f S = C P /T dt - V dp For liquids and solids, C P and constant (assume small S = C P ln(t f /T i ) - V P Example for CO expansion, choose reversible path, separate sheet Similarly, for liquid Hg pressure increase, use formula above, separate sheet Entropy was shown to be a state function, and relates to efficiency, and calculable for chemically interesting process (by finding rev path) Back to our goal determine the direction of a process Example: bring 2 metal rods together, one at T 1, and other at T 2, assume T 1 > T 2, what will happen? You know that the hotter one (1) will spontaneously cool and the cooler one (2) will warm up. Does entropy indicate this? Need a reversible path: slowly (rev) cool rod 1 by coupling to reservoir, const P dq P = C P dt or T change caused by q: T = 1/C P dq P = q P /C P (assume C P const) since done under const P, path defined, so not restrict how fast heat go, q P = H = q rev if system isolated heat only from 1 into 2, i.e. q = q 1 + q 2 = 0, entropy change is then S = q 1 /T 1 + q 2 /T 2 = q P (1/T 1-1/T 2 ) uses: q P = q 1 = - q 2 6

--So look at normal process, heat flow from hot cold, T will get smaller, q P = q 1 =(-) T 1 > T 2 so (1/T 1-1/T 2 )<0 and q P < 0 makes: ds > 0 --but if opposite, unnatural, if heat flow from cold to hot, then q P > 0 and ds < 0 Thus S (+) indicates natural direction of process, if isolated system Example: Next consider a volume containing an ideal gas that suddenly collapses to half its previous volume, constant T with no forces, spontaneous partition into gas and vacuum, which we all recognize as unnatural. Design a reversible analog to compute S, i.e. imagine dripping water on piston to slowly increase pressure and compress gas to half its volume. Since ideal, this is isothermal (recall U not depend on V), so V i T i ½V i T i and U = 0, so q = - w S = dq rev /T = q rev /T = -w rev /T = nr ln(½v i /V i ) = -nr ln 2 < 0 (recall: ln½ = - ln 2) If reverse the process, gas spontaneous expands to fill the volume S = dq rev /T = nr ln2 > 0 Consistent: i.e. spontaneous, irreversible process is the one with S > 0 General: For any irreversible process in an isolated system, there is a unique direction of spontaneous change: S > 0 for spontaneous change, S < 0 for the opposite or nonspontaneous change. For these S = 0 for a reversible process only works if no energy flow: no heat or work on the system, isolated! Contrast: U is neither created or destroyed, but S can be increased for an isolated system, S > 0, but not destroyed imbalance! For S < 0, need work Aside: Interpretation S (+) when system becomes less ordered, more random e.g. expansion of gas, more volume, more places to be, less order molecules same with heating rods, heat (vibrations of motions of atoms) more dispersed Statistical view: S = k ln where = # ways system can be in that state S = k ln ( / recall k = R/N A - and concept more states more probable System then tends to a state of higher probability S (+) Can think of number of states as / V 2 /V 1 S = k ln V 2 /V 1 ) N = Nk ln V 2 /V 1 ) = nr ln V 2 /V 1 ) were N = nn A 7

The Clausius inequality recall: du = dq + dw = dq rev + dw rev rearrange: dq rev - dq = dw - dw rev now the system can do maximum work for reversible process: dw dw rev but definition is work by system is (-), or -dw -dw rev or dw - dw rev 0 Substitute: dq rev - dq = dw - dw rev 0 or dq rev dq Now recall entropy: ds = dq rev /T ds dq/t the entropy is greater or equal to than heat change divided by temperature Example (text) consider transfer of heat from hot to cold body (same example again) ds dq h /T h + dq c /T c but dq h = - dq c so ds - dq c /T h + dq c /T c or ds dq c (1/T c - 1/T h ) since dq c > 0 and T c < T h then ds > 0 finally if system is isolated from surroundings, then ds 0, isolated system cannot reduce its entropy second law Clausius extension for cyclic integral, dq/t around a path Imagine step 1 2 is reversible and then back, 2 1 is irreversible o dq/t = 12 dq rev /T + 21 dq irev /T = - 21 dq rev /T + 21 dq irev /T But dq rev > dq irev so o dq/t 0 i.e. neg. (recall S = 0 = o dq rev /T so S o dq/t ) Now consider the surroundings, this is a large temperature bath, so const T, heat in has negligible effect, stays in equilibrium no matter what happens to system Const V change: q sur = U sur or const P : q sur = H sur H and U state fct, path indep ds sur = dq sur /T sur or macroscopic: S sur = q sur /T sur (acts rev, since state fct) this is the real heat of process, path indep, whereas for system need rev path Example: compare process for reversible and irreversible compression of a gas, evaluate S tot = S sys + S sur and compare (see separate sheet) In example see S tot = S sys + S sur = 0 for reversible path, equilibrium no direction but S tot = S sys + S sur > 0 for irreversible path, for spontaneous process. Looking at this more globally, consider system and surrounding, S tot 0 or entropy of the universe is always increasing Entropy of mixing: U = 0, q = -w, T= P t = V t =0, know spont. (ideal gas), P i, V i 0 rev. path: S a,b =n a,b Rln(V t /V a,b ) S t = S a + S b = -n t R(x a lnx a +x b lnx b ) >0 (lnx a <0) In some way this defines time. If watch a process like mixing 2 gases, know S tot > 0, and we know we are watching it in a real time direction, but if it were filmed and played backward, it would separate, but this tells us that S tot < 0, so not real time We will use this Clausius inequality relationship to define Gibbs and Helmholtz free energies and those will tell us directions of process, using system properties alone Absolute entropy and the Third law U and H were defined as changes, du = dq + dw, or dh = du + d(pv) and an absolute value was not established, but relative values to a standard state were used: H o f Entropy is different, can define an absolute, but need to know what is zero 3 rd law 8

Third law of thermodynamics: the entropy of a pure, perfectly crystalline substance (element of compound) is zero at 0 K. Critical issue: idea of absolute temperature (from Kelvin and gas thermometer) and idea of a perfect crystal. In perfect crystal, all the molecules are identical, so interchanging them give the same state, perfect crystal has only one state, so: S = k ln = k ln 1 = 0 How to calculate entropy relates to heat and temperature, so need heat capacity, but for all the phases from perfect crystal to other solid phases, to liquid and to gas, allow for heats of phase changes (fusion, vaporization, and sol 1 sol 2, etc.) along the way Plot of C p vs, T for O 2 shows growth from 0 K, initially like T 3, but change at each phase change To get entropy, divide by T, integrate those curves, and then add in H/T for phase changes, as below: - entropy increases monotonically, not lost, - heat capacity grows or drops with phase, each phase change: H ph (+), S(+), -vapor the big one: H vap /T and then relatively const, all C p > 0 gives trend: S vap >S liq >S sol - molar entropy increases with molecule size, more vibrations (degree of freedom) - gases translate relatively freely, rotate, vibrate (3N-6 modes), liquids in hindered motion (trans., rot. collide), solids only vibrate (like big coupled springs) - strongly bound solids higher energy vibrations, less accessible, lower entropy 9

Example - copies Engel calculation of the entropy of O 2 for all phases at T = 298K C p values measured from 12K, lower ones fit to T 3 law, each phase change has H See largest changes for vaporization (75.6 J/K), then heating (C p T) gas and liquid Standard states Although we have absolute values for entropy, U and H are defined relative to standard states, so can also do that for entropy. Can use heat capacity to define entropy for T=298 K and P = 1 bar, but for gas phase, entropy highly dependent on pressure At const T, ideal gas: S m = R ln(v f /V i ) = - R ln(p f /P i ) if let P i = P o = 1 bar S m (P) = S m o R ln(p/p o ) Graph shows S m vs. P for gas, see max, S m, P 0 Meaning is that V, P 0, so infinite states Solids and liquids, P dependence small, can often neglect Entropy in Chemical reactions Since entropy is a state function we can do the same approach as for H and U for reactions, Hess s law works, just sum up reactants and subtract from sum of products ΔS o rxn= S o prod - S o react since absolute values do not need Recall, for H we used H o f which were referenced to elements in standard states, so reference was same for reactants and products (mass law atoms same) But S o use absolute values for reactant and product, reference to perfect crystal, 0 K Example: synthesis of glucose, C 6 H 12 O 6 (s), from carbon dioxide and water, T = 298 K: 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) Determine if spontaneous by testing if S tot = S sys + S sur > 0 S sys = S o rxn = S o (glu) + 6 S o (O 2 ) - 6 S o (CO 2 ) - 6 S o (H 2 O) = (209 + 6x205 6x214 6x70)JK -1 mol -1 = - 262 JK -1 (for 1 mole rxn) See S sys (-) but need to consider S sur use heat transfer form reaction to surrounding H sys = H o f (glu) + 6 H o f (O 2 ) - 6 H o f (CO 2 ) - 6 H o f (H 2 O) = [-1273 + 6x0 6x(-294) 6x(-286)]kJmol -1 = 2801 kj (for 1 mole rxn) S sur = H sur /T = H sys /T = -2802 kj/298 K = -9398 JK -1 S tot = S sys + S sur = (-262-9398) JK -1 = 9.66 kjk -1 < 0 So reaction as written is not spontaneous at 298 K, logic is to form solid glucose from liquid and gas phase molecules, high disorder to more order, must do work on system Also need to consider T dependence of S Kirchoff s law again - must account for phase changes in the entropy, integrate between them 10

Refrigerator, Heat Pump, Reverse Carnot cycles: Goal of Refrigerator is to pump heat form cold body to a hot one, since second law says this cannot happen without work. Efficiency then defined as coefficient of performance c = r = q c / w Energy deposited in hot body is q h = q c + w rearrange: w = q h - q c 1/c = ( q h - q c )/ q c = q h / q c - 1 From previous efficiency use : q h /q c = -T h /T c 1/c = 1-T h /T c or c = r = T c /(T h -T c ) if T c decrease relative to T h, c = r decreases Example: refrigerator operates freezer at T f = 255 K, refrigertor at T r = 277 K and room about T h = 295 K Let T f = T c, maximum efficiency is r = 6.5, which implies 1 J work produces 6.5 J cooling, actual r ~ 1.5 The difference is loss of efficiency fo irreversible cycle Alternate, heat pump, bring heat out of ground into house (illustrated below), now c = hp = q h / w Same picture here, if inside is T h = 295 K (~72 F) and outside is T c = 273 K (freezing), then c = hp = 13.4 but this again is ideal efficiency (max) real is hp ~ 2-3 But if you use the ground, which deep enough is always at T c ~ 285 K, it helps, and even with efficiency hp ~ 2-3, get 2-3x advantage over direct electric heat Entropy considerations, Process is not spontaneous, however, can convert work to heat at 100%, so hp calculated for minimum work input to make S favor a spontaneous process for withdrawing q c and depositing q h +w Beyond this consider solar, wind, and fuel cells as alternate energy sources. Electrical energy can be directly converted to work without heat engine loss 11