Outline of the Course

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Ira McDowell
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1 Outline of the Course 1) Review and Definitions 2) Molecules and their Energies 3) 1 st Law of Thermodynamics 4) 2 nd Law of Thermodynamics 5) Gibbs Free Energy 6) Phase Diagrams and REAL Phenomena 7) Non-Electrolyte Solutions & Simple Mixtures 8) Chemical Equilibrium 9) Kinetics

2 2 nd Law of Thermodynamics Chapter 5 most of it!

3 2 nd Law of Thermodynamics GOALS (1) Understand difference between spontaneous and non-spontaneous (1) Define Entropy (s) in terms of statistical probabilities (2) Define Entropy in terms of thermodynamics (3) Define 2 nd Law of Thermodynamics (4) Calculate Entropy changes for systems undergoing physical change (5) Calculate Entropy changes for phase transitions (6) Calculate Entropy changes for chemical reactions

4 2 nd Law of Thermodynamics Why do we need a 2 nd Law. 1st Law of Thermodynamics. Law of Conservation of Energy..Energy can neither be created nor destroyed..the total amount of Energy in the Universe remains constant. But.1 st Law doesn t predict whether or not a process can occur and it doesn t predict the direction of a process.. so we need the 2 nd Law this Law relies on something called Entropy (S) changes in Entropy allow us to predict the direction of a process

5 Processes only occur spontaneously in one direction. under the same conditions the process cannot occur spontaneously in the opposite direction.. if it could nothing would ever happen.a sugar cube melts in a cup of coffee..an ice cube melts in your hand...a leaf falls to the ground.. you won t see a leaf on the ground rise into the air and makes its own return to the branch on the tree.. water will not spontaneously form into an ice cube in your hand WHY?

6 A process (e.g. change from state 1 to state 2) occurs spontaneously in only one direction. The reverse (i.e. change from state 2 to state 1) cannot occur spontaneously under the same conditions. If this was not true nothing would ever occur.

7 Would it make sense to draw the arrows in the other direction? Figure Physical Chemistry 7 th Edn. Atkins P., de Paula J.

8 Why can t t the reverse of a spontaneous process occur by itself?.. an example to understand this consider a rubber ball held 1 meter from the floor..you drop the ball and it bounces. after every bounce the ball does not rise quite as high eventually it comes to rest on the floor.. each time the ball hits the floor some of its kinetic energy is dissipated among molecules in the floor floor becomes warmer eventually ball comes to rest all of its KE transferred to floor

9 .we just reviewed an example of a spontaneous process How come the reverse process does not occur spontaneously? can a ball at rest on the floor suddenly start bouncing into the air? NO! for this process to occur we need the ball to absorb sufficient energy from the floor the thermal energy of the floor is random molecular motion the amount of energy needed to raise the ball would require all molecules within the floor to line up and vibrate in phase with one another

10 for the ball to acquire sufficient energy all molecules within the floor would need to align and vibrate in phase and all atoms in these molecules must be moving upward for proper energy transfer..this might be possible for a million molecules but the magnitude of energy transfer we need would require alignment of 6.0 x molecules No Way given the random nature of molecular motion! 6.0 x molecules This explains why no one has ever witnessed the spontaneous rising of a ball from the floor.

11 Spontaneous Not Spontaneous

12 spontaneous non-spontaneous Why don t all of the molecules in this container collect in the same corner???.they could it s just extremely unlikely. Phys. Chem. With Applications in Biology P.Atkins

13 consider again a ball at rest acquiring enough energy to enable it to spontaneously bounce into the air....this would require approx. 6 x molecules to move in a particular direction simultaneously...there is no reason in the absence of an external force that all of these molecules would align in the same direction.in fact we could say that such an event is statistically very unlikely.. let s try to understand entropy from a statistical perspective.

14 The Statistical Definition of Entropy for a cylinder containing He atoms the probability of finding any one atom in V 2 (the entire volume of the cylinder) is 1 the probability of finding an He atom in ½ the volume (i.e. V 1 ) is only ½ the probability of finding two He atoms in V 1 is (½)(½) or ¼ the probability = W = (½) N where N is no. of atoms.so as N increases the probability of finding all atoms in V 1 decreases

15 if N is on the order of 6 x probability, W = (½) 6 x 1023 thus W 0 Initial State Final State Initial State = all atoms compressed in V 1.then allow gas to expand on its own..eventually we will end up with all atoms evenly distributed over entire volume V 2 this situation is MOST PROBABLE STATE DIRECTION of SPONTANEOUS CHANGE is from all atoms in V 1 to atoms distributed evenly over entire volume V 2 spontaneous change occurs from state of low probability of occurring to state with maximum probability..

16 we now know how to predict the direction of spontaneous change in terms of probabilities of initial and final states Entropy may then be said to be proportional to probability S = k B W k B is proportionality constant.this equation is actually INCORRECT Actually S = k B lnw i.e. as W inc. to W 2.S inc. to 2S remember lnw 2 = 2 ln W this is known as the Boltzmann Equation where k B is the Boltzmann constant = 1.38 x J/K Initial State Final State S 1 = k B ln W 1 S 2 = k B ln W 2 Entropy is a state function so path independent

18 Entropy is a state function.so depends ONLY on probability of state occurring NOT on way in which state was created S = S 2 S 1 = k B ln (W 2 /W 1 ) for the 1 2 process Question: What is the probability of a gas expanding from V 1 to V 2? (isothermal expansion) Remember probabilities of each state are.. W 1 = (CV 1 ) N and W 2 = (CV 2 ) N [ in example above: C=(1/V 2 ) W 1 =(V 1 /V 2 ) N = (V 2 /2V 2 ) N = (1/2) N W 2 =(V 2 /V 2 ) N = (1) N ] S = S 2 S 1 = k B ln [ (CV 2 ) N / (CV 1 ) N ] = k B ln (V 2 /V 1 ) N S = (N/N A ) R ln (V 2 /V 1 ) = n R ln (V 2 /V 1 ) where: R = k B N A *** this eqn. only holds true for an isothermal expansion as Entropy is affected by changes in Temperature

19 Calculate the entropy change when 2.0 moles of an ideal gas are allowed to expand isothermally from an initial volume of 1.5 L to 2.4 L. Estimate the probability that the gas will contract spontaneously from the final volume to the initial one. Initial State (V 1 = 1.5 L) Final State (V 2 = 2.4 L) S = n R ln (V 2 /V 1 ) = (2.0 mol)( Jmol -1 K -1 ) ln [(2.4L)/(1.5L)] = 7.8 J K -1 Spontaneous contraction will then be accompanied by a decrease in entropy equivalent to 7.8 J K -1 S = k B ln (W 1 /W 2 ) (since process is now defined as 2 1) -7.8 J K -1 = (1.381 x J K -1 ) ln (W 1 /W 2 ) ln (W 1 /W 2 ) = x W 1 /W 2 = e -5.7 x 10 23

20 we have only defined entropy in terms of statistical probabilities let s define in terms of thermodynamics remember we just derived equation for S for isothermal expansion of an ideal gas S = n R ln (V 2 /V 1 ) Heat absorbed by an ideal gas in a reversible, isothermal expansion: q rev = n R T ln (V 2 /V 1 ) so (q rev /T) = n R ln (V 2 /V 1 ) thus S = (q rev /T) the entropy change for a reversible, isothermal process is equal to the heat absorbed divided by the temperature

21 S = (q rev /T) this equation is true for any reversible process occurring at constant temperature. In general: ds = dq rev /T..entropy is a state function so path independent BUT q is not path independent for an irreversible process the work done by the gas on the surroundings would be less and heat absorbed would be less. so q irr q rev but S = S rev = S irrev so S > q irr /T

22 To this point we have only discussed the SYSTEM must also consider the surroundings. system system is SO SMALL compared to surroundings surroundings is an infinitely large reservoir any heat or work exchanged between system and surroundings can only alter the properties of the surroundings by an infinitesimal amount Remember..infinitessimal changes are characteristic of reversible processes.. so we can say that d(qsurr)rev = d(qsurr)irrev = dqsurr we can forget about path thus S surr = (q sur /T)

23 If no work is done in the process there is no heat exchanged between system and surroundings What does this mean??? U = q + w U = 0 for an isothermal process So q = - w thus for an isothermal expansion the heat absorbed by the gas is equal to the work done by the ideal gas on its surroundings. also, for an isothermal process if w = 0 (no work done) then q = 0

24 Total Change in Entropy for the Universe: S univ = S sys + S surr = (q sys /T) + (q surr /T) consider again the isothermal expansion of an ideal gas where we said that the heat absorbed by the system is q sys = n R T ln (V 2 /V 1 ) so heat lost by the surroundings is q surr = - n R T ln (V 2 /V 1 ) (T is constant since isothermal.t surr = T sys = T) S univ = n R ln (V 2 /V 1 ) + [- n R ln (V 2 /V 1 )] = 0 thus for a reversible process S univ = 0

25 For an irreversible expansion.extreme case.gas is expanding into a vacuum.. S sys = n R ln (V 2 /V 1 ) Yet since no work is done in the process there is no heat exchanged between system and surroundings q surr = 0 and S surr = 0 S univ = n R ln (V 2 /V 1 ) For a REVERSIBLE process S univ = 0 For an IRREVERSIBLE process S univ 0 Overall then S univ 0

26 Sample Problem A quantity of 0.50 mol of an ideal gas at 20 C expands isothermally against a constant pressure of 2.0 atm from 1.0L to 5.0L. Calculate the values of S sys, S surr, S univ. Solution From conditions given we can calculate initial pressure of the gas (12 atm).

27 Solution continued the process is spontaneous.which is expected considering the initial pressure of the gas.

28 The Second Law of Thermodynamics.. the entropy of an isolated system increases in an irreversible process and remains unchanged in a reversible process.other statements on 2 nd Law are in Appendix 5.1 (Chang text) Entropy CAN NEVER DECREASE ****** S surr or S sys can be a negative quantity but their sum must never be less than zero.

29 Entropy Change Due to Phase Transitions S An entropy change is associated with phase transitions.. l g entropy increases when going from a solid to a liquid to a gas because number of possible states for molecule increases.. (i.e. degree of molecular order changes) Phys. Chem. With Applications in Biology P.Atkins

30 Let s take a look at the values for standard molar entropy for some molecules..(appendix B in Chang text) Standard Molar Entropies: S ( at 25 C ) H 2 O (s) : H 2 O (l) : H 2 O (g) : 45 J / K mol 69.9 J / K mol J / K mol H 2 (g) : N 2 (g) : J / K mol J / K mol C as graphite : C as diamond : 5.7 J / K mol 2.4 J / K mol

31 Temperature Dependence of Entropy S = n C P ln (T 2 /T 1 ).we assume here that C P may be treated as a constant S for Phase Transitions S = H trans /T trans Example Calculate S when a sample (of one mole) of H 2 O(l) is heated from 25 C to 100 C, and then following vaporization, the H 2 O(g) is further heated to 160 C. S = C P (H 2 O (l) ) ln (373/298) + H vap /373 + C P (H 2 O (g) ) ln (433/373)

32 Heat of Fusion ( H f ) solid liquid e.g. f S o H 2 O = 22 J/K mol (T = K) T f Heat of Vaporization ( H vap ) Liquid gas e.g. vap S o H 2 O = J/K mol (T = 373 K) T b in general vap S o > f S o for the same substance. solid to liquid transition results in a relatively small increase in molecular disorder arrangement of molecules in gaseous state is completely random liquid to vapor transition..large increase in disorder

33 Sample Problem Calculate S for the following reaction: (1 mol) H 2 O (s) 0 C, 1 atm (1 mol) H 2 O (g) 100 C, 1 atm Given H fus = 6008 J/mol for H 2 O at 0 C H vap = 40,670 J/mol for H 2 O at 100 C = 75.3 J/mol K for liquid water Solution C P

34 Note: S for Phase Transitions S sys = H trans /T trans Here we are calculating the entropy change for the system. Heat is absorbed reversibly in an equilibrium process so the changes in entropy for the surroundings at a phase transition will be given by: S surr = - H trans /T trans [e.g. S surr = - H fus /T f S surr = - H vap /T v ] Therefore, the total change in entropy for the universe will be zero (at a phase transition).

35 Entropy Changes for Chemical Reactions We can also keep track of Entropy changes in a reaction by subtracting ΔS (products) from ΔS (reactants) Δ R S = Σ n ΔS (products) Σ n ΔS (reactants) As an example, the reaction to produce ammonia : N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ΔS sys = Σ n ΔS (products) Σ n ΔS (reactants) = (2 mol)(193 J/mol K ) (1 mol)(192 J/mol K) (3 mol)(131 J/mol K ) = 386 J/K 192 J/K 393 J/K = 199 J/K ** here we see that we start with 4 molecules of gas and produce only 2 molecules of gas

36 Can you see a trend (1) CaCO 3 (s) CaO(s) + CO 2 (g) Δ R S = J/K (2) 2H 2 (g) + O 2 (g) 2H 2 O (l) Δ R S = J/K (3) N 2 (g) + O 2 (g) 2NO(g) Δ R S = J/K

37 Consider the reaction to produce water: 2 H 2 (g) + O 2 (g) 2 H 2 O (l) Δ R S = J/ K.the entropy change is negative both because the net number of molecules of gas decrease and we are producing a liquid from gases One would expect that this reaction would NOT OCCUR. BUT don t forget we have only considered the Entropy of the system.what about Entropy of the SURROUNDINGS?

38 S univ = S sys + S surr 2 H 2 (g) + O 2 (g) 2 H 2 O (l) Δ R S = J/ K S sys = J/ K S surr = H surr /T = ( J)/(298 K) = 1919 J/K Δ f H of water = 286 kj/mol Δ f H for 2 moles of water = -572 kj = J S univ = J/ K J/K = 1593 J/K Overall Entropy is positive..this is SPONTANEOUS!

39 2 nd Law of Thermodynamics Goals (1) Understand difference between spontaneous and nonspontaneous (2) Define Entropy (s) in terms of statistical probabilities (3) Define Entropy in terms of thermodynamics (4) Define 2 nd Law of Thermodynamics (5) Calculate Entropy changes Progress (1) Spontaneous change occurs from state of low probability to state of maximum probability (2) S = k B ln W (3) S = (q rev /T) for Isothermal ; ds = (dq rev /T) in General (4) 2 nd Law: the entropy of an isolated system increases in the course of a spontaneous change: S tot 0.

Chemical thermodynamics the area of chemistry that deals with energy relationships

Chemistry: The Central Science Chapter 19: Chemical Thermodynamics Chemical thermodynamics the area of chemistry that deals with energy relationships 19.1: Spontaneous Processes First law of thermodynamics