where R = universal gas constant R = PV/nT R = atm L mol R = atm dm 3 mol 1 K 1 R = J mol 1 K 1 (SI unit)

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1 Ideal Gas Law PV = nrt where R = universal gas constant R = PV/nT R = atm L mol 1 K 1 R = atm dm 3 mol 1 K 1 R = J mol 1 K 1 (SI unit) Standard molar volume = 22.4 L mol 1 at 0 C and 1 atm Real gases approach ideal gas behavior at low P & high T

2 Ideal Gas Equation of State

3 General Principle!! Energy is distributed ib t d among accessible configurations in a random process. The ergodic hypothesis Consider fixed total energy with multiple particles and various possible energies for the particles. Determine the distribution that occupies the largest portion of the available Phase Space. That is the observed distribution.

4 Energy Randomness is the basis* of an exponential distribution of occupied energy levels n(e) A exp[-e/<e>] Average Energy <E> ~ k BT n(e) A exp[-e/k B T] This energy distribution is known as the Boltzmann Distribution. * Shown later when we study statistical mechanics.

5 Maxwell Speed Distribution Law 1dN Ndu 32 m 2 2 mu B B 4 ue 2 kt 2k T 1dN is the fraction of molecules per unit speed interval Ndu

6 Distinguish between System & Surroundings

7 Internal Energy Internal Energy (U) is the sum of all potential and kinetic energy for all particles in a system U is a state function Depends only on current state, not on path U = U final -U initial

8 Internal Energy, Heat, and Work If heat (q) is absorbed by the system, and work (w) is done on the system, the increase in internal energy (U) is given by: U = q (heat absorbed by the system) + w (work done on the system)

9 Reversible and Irreversible Work Reversible or Irreversible Processes Reversible: carried out through a sequence of equilibrium states Irreversible: anything else!

10 Internal Energy (2) U is a state function It depends only on state, not on path to get there U = U final - U initialiti This means mathematically that du is an f exact differential: U du For now, consider a system of constant composition. U can then be regarded as a function of V, T and P. Because there is an equation of state relating them, any two are sufficient to characterize U. So we could have U(P,V), U(P,T) or U(V,T). i

11 Enthalpy Defined Enthalpy, H U + PV At Constant P, H = U + PV U = q + w q= q P = U - w, w = -PV q P = U + PV= H At constant V, q = U = H

12 Comparing H and U at constant P H = U + PV 1. Reactions that do not involve gases V 0 and H U 2. Reactions in which n gas = 0 V V 0 and H H UU 3 Reactions in which n 0 3. Reactions in which n gas 0 V 0 and H U

13 Heat Capacity at Constant Volume or Pressure C V = dq V /dt = (U/T) V Partial derivative of internal energy with respect to T at constant V C P = dq P /dt = (H/T) P Partial derivative of enthalpy with respect to T at constant P Ideal Gas: C P = C V + nr

14 Endothermic & Exothermic Processes H = H final -H initial H H Positive Positive amount of heat absorbed by the system Endothermic Process H Negative Negative amount of heat absorbed (i.e. heat released by the system) Exothermic Process

15 Thermochemical h Equations CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) (a combustion reaction) c H = 890 kj Reaction must be balanced Phases must be specified H is an extensive property Sign of H changes when reaction is reversed

16 Standard State The Standard State of an element is defined to be the form in which it is most stable at 25 C and 1 bar pressure Some Standard States of elements: Hg (l) O 2 (g) Cl 2 (g) Ag (s) C (graphite) The standard enthalpy of formation ( f H ) of an element in its standard state is defined to be zero.

17 Enthalpies of Formation The standard enthalpy of formation ( f H ) of a compound is the enthalpy change for the formation of one mole of compound from the elements in their standard state. Designated by superscript o, Ø, or o: H,.. For example, CO 2 : C (graphite) + O 2 (g) CO 2 (g) rxn H = kj/mol Table 2.7 rxn f H CO 2 (g) = kj/mol

18 Enthalpies of Reaction The enthalpy of reaction can be calculated from the enthalpies of formation of the reactants and products H O rxn = H O f (Products) - f H O (Reactants)

19 Example: Find rxn H (using Standard Enthalpies of Formation) CH 4 (g) + 2 O 2 (g) CO 2 (g) +2HO 2 (l) f H (from Tables 2.4&5, text): CH 4 (g) kj/mol O 2 (g) 0 CO 2 (g) H 2 O (l) rxn H = (285.8) 0 (-74.6) kj/mol Therefore, rxn H = kj/mol

20 Statement of law Second Law Spontaneity (system vs. surroundings) Entropy defined Calculation of S as f(t,p,phase) Standard entropy Applied to cyclic processes Carnot engine valid for all reversible engines Details of Carnot cycle processes Gibbs and Helmholtz Free Energies Enables a system only perspective Changes with P,V,T, quantities Applications to chemical processes

21 Entropy change in a general process with an ideal gas V 1, T 1 V 2 2, T 2 Make a two component reversible path: Isothermal (expan or compr) from V 1, T 1 to V 2, T 1 Isochoric heating or cooling to V 2, T 2 P V T S S S nr ln C ln V V T 1 1 P 1 P 2 Path 1 Path 2 V 1 V V2

22 Steps of Carnot Cycle q=0 w done on system (+) q h absorbed (+) from hot reservoir T h w done by system ( ) q=0 w done by system ( ) q discarded ( ) into cold q c discarded ( ) into cold reservoir T c w done on system (+)

23 Summary of 4 Steps in Carnot Cycle Step U q rev w rev S AB 0 RT h lnv B /V A J RT h lnv B /V A 1678 J RlnV B /V A 4.19 J/K BC C v (T c -T h ) 1247 J 0 C v (T c -T h ) 1247 J 0 CD 0 RT c lnv D /V C 1259 J RT c lnv D /V C J RlnV D /V C 4.19 J/K DA C v (T h -T c ) J 0 C v (T h -T c ) J 0 Net 0 R(T h -T c )lnv B /V A +419 J R(T h -T c )lnv B /V A 419 J 0

24 Net work = work done by system work done on system In example above, net work = = 1678 J J J 1247 J = 419 J What is the efficiency?

25 Efficiency of Carnot Engine Efficiency,, of Engine: = Work Performed Heat Absorbed = w/q/ h = T h -T c h c T h

26 H 2 O (l) -5 C S in Phase Changes: Example Freezing (-5 C) H 2 O (s) -5 C Step 1 heating Step 3 cooling J H 0 C H O 0 C 2 O (l) 2 (s) Step 2 Phase transition (freezing) But fusion is tabulated! 273 J S1 Cpln mol K J 6008 J S2 mol K mol K 268 J S ln mol K Putting it all together (Freezing at -5 C), J Soverall S1 S2 S mol K The entropy change is negative!!! Does this process happen spontaneously? YES!! Why?? Spontaneous S total 0 So we need to consider the surroundings!!

27 Gibbs Energy, G Consider the condition where T sys = T surr The entropy change in the surroundings in any process is ds surr dq T surr surr dq T So we can write the second law solely in terms of the system. ΔS univ = ΔS surr + ΔS sys 0 for a spontaneous process ds univ = - dq sys + ds sys T sys sys = 0 for equilibrium dq-tds = 0 at equilibrium, or dq-tds = 0 with all properties those of the system! Note: In this form, we have dq TdS 0 for a spontaneous process.

28 Thermodynamic Equilibrium at constant pressure dq-tds = 0 at equilibrium For a process taking place at constant P, dq= dh and dh TdS 0 This relationship suggests a new state function, G, now known as the Gibbs function, G: G H TS At equilibrium with constant T and P, we have dg =0, and for a spontaneous process at constant P and T, dg <0 The Gibbs energy goes toward a minimum in a spontaneous process.

29 Thermodynamic Equilibrium at constant volume dq-tds = 0 at equilibrium For a process taking place at constant V, dq= du, and du TdS 0 0 This relationship suggests another new state function, A, now known as the Helmholtz function: A U TS At equilibrium with constant T and V, we have da =0, and for a spontaneous process at constant V and T, da < 0 The Helmholtz energy also goes toward a minimum in a spontaneous process at constant V,T.

30 Thermodynamic Equilibrium

31 The Standard Gibbs Energy Change, G S univ = S surr + S sys S univ = H sys + S sys T S univ = H sys TS sys G sys = H sys TS sys Spontaneous reaction: S univ >0 G sys < 0

32 G = H - TS < 0 for Spontaneous Processes Reaction is spontaneous at all temperatures H < 0 and S > 0 Reaction is nonspontaneous at all temperatures H > 0 and S < 0 Reaction is spontaneous only at higher temperatures H > 0 and S > 0 Reaction is spontaneous only at lower temperatures Reaction is spontaneous only at lower temperatures H < 0 and S < 0

33 Gibbs Energy of Formation, G f º I. Since Gibbs energy is not on an absolute scale, we need to assume a reference state t of the elements. Convention is to choose the element reference (standard) state as the element in its most stable form at K and 1 Bar pressure. G f 0 0 II. Then for the elements in this form. G f III. For a non-elemental substance, its is the Gibbs energy for formation of 1 mole of the substance at standard T and P, made from the elements in their standard states. IV. We can then calculate G for a reaction taking place at standard conditions: G n G n G rxn i i f i j j f j products reactants V. We use this construct to determine reaction spontaneity. 0

34 Chemical Potential (4) G G G G dg dt dp dn dn A T P P T n n A B P, n, n T, n, n T, P, n T, P, n A B A B B A Calculate the change in the Gibbs energy as a result of the progress of the reaction. At constant T and P, dn d and dn d, so A B dg dn dn d d G and, A A B B B A T P G B A R G R B A A A B B B reaction goes forward reaction goes backward mixture at equilibrium 0 1 B composition A

35 Chemical Potential (6) aa g bb g b P B P P 0 P B A G b a brt ln 0 art ln 0 G RT ln R B A R a P P Standard P A Gibbs Energy 0 Change P o G0 G RT ln KQ R R At equilibrium, i G R = 0, Q = K P, and we have o G RT ln K R P P K, the equilibrium constant P for (in pressure) reaction Q This special value of Q that comes at equilibrium is called K P,the equilibrium constant. A very useful connection between G and the equilibrium constant

36 Determining the Components of the Rate Equation aa + bb yy + zz Rate k[a] [B] The coefficients and components of the rate equation Must be found by experiment Cannot be deduced d d from stoichiometry Do not necessarily tell us the physical mechanism Also, the reaction orders can be fractional, negative, zero, or even not-definable

37 Rate Laws We have seen how to obtain the differential form of rate laws based upon experimental observation. As they involve derivatives, we must integrate the rate equations to obtain the time dependence of concentrations. We will do this for a few cases, all involving these empirical rate laws. Here the rate law need not bear any relationship to the stoichiometry of the reaction. Our next step will be to understand the origin of the empirical laws. This leads to the concept of elementary reactions where the reaction is direct and occurs (or nor) in a single encounter. We will see how the more complex rate laws arise from multiple elementary processes. Understanding the nature of the elementary reactions and the role of elementary reactions in complex processes will occupy the remainder of the semester.

38 First Order Reactions, n=1 Differential form: Integrated form: A X = -k[a] 1 d [A] dtd k[a] d[a] d[a] k[a] or dt d [A] kdt Integrating, g, [ A] t d [A] [A] k dt or ln [A] [A] A 0 which says that A A e [ ] kt kt

39 Worse for higher order reactions Table 9.3 This is not what we had earlier. We got = -k Oh! WHY??

40 [A] X X X X X X X X X X ConcepTest X X X X 1 X X X X X X X X X X X X X X XXXXXXX X X X X X X What is the X order X X X of Xthis reaction? X X X X X X X X X X X A. X X Zero X XXXXXXX order X X X X X X X X X X B. X X First X X order X X X X X X X X X X C. X X Second X X X order X X X X X XXXXXXX X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X XXXXXX X X X X X X X X X X X

41 Different Types of Composite Reactions Simultaneous: A Y and A Z Consecutive: A X Y Opposing: A B Z Consecutive with opposition: A X Y In order for the system to be at complete equilibrium, the time derivative of all species concentrations must be zero!

42 Kinetics for a Simple Consecutive Reaction k 1 2 A X and X Z k with 1 A A e kt 0 d X k X k A (2) 2 1 dt becomes d X kt 1 k X k A e 2 1 dt 0 d X kt 1 k X k A e a "standard form" o.d.e. with solution 2 1 dt 0 k 2 kt 1 k2t X e e A k k and Z A A X A ke k 2 1 kt ke k kt

43 Quasi Steady State Approximation k 1 2 A X and X Z Consider the differential rate equations for [A], [X] and [Z]. d A k A (1) Let [A] 1 dt t=0 = [A] 0 and [X] t=0 = [Z] t=0 = 0 d X k X k A 0 (2) 2 1 dt 1 A A e kt 0 k k 1 1 kt 1 or X A A e ( 2.5) Make QSS approx k k t kt 1 d Z 1 k X (3) 0 2 dt 0 k1t 1 e A k Z k A e dt 0 For (2) and (2.5) to be compatible, we must have (k 1 /k 2 ) << 1

44 Compare Exact and Steady State Solutions for k 2 = 20 k 1 k 1 t

45 ConcepTest 1 Reaction: 2O 3 3O 2 k 1 Mechanism: O 3 O + O 2 k 1k2 (fast) O + O 3 2 O 2 (slow) What is the rate eequation for O 3 loss? A. Rate = k [O 3 ] 2 B. Rate = k [O 3 ] [O 2 ] C. Rate = k [O 3] 2 [O 2] 1 B. Rate = k [O 3 ] 1 [O 2 ]

46 Solution Reaction: 2 O 3 3 O 2 k 1 k 1 k 2 Mechanism: O 3 O + O 2 (fast) O+O O 3 2O 2 (slow) What is the rate equation for O 3 loss? The slow step is generally the rate determining step. Rxn 1 equilibrium : k 1O 3 k O O 1 2 k O 1 3 O k O 1 2 with a slow second step, the equilibrium is maintained d O Rate dt 2 3 k O O k O O k O k k O O 1 2 2

47 Lindemann Hinshelwood Mechanism A+B AB* Complex with excess energy Given enough time, it will fall apart AB* A+B Unless a collision i removes the excess energy AB*+ M AB + M*

48 Lindemann Hinshelwood Mechanism Consider only one species A. This is easy to generalize. d[a*] 2 A + A A* + A ka[a] dt d[a*] ' A* + A A + A ka[a*][a] dt d[a*] A* P kb[a*] dt The rate law for formation of A* is (with s.s. approximation) d[a*] dt 2 ' a a b k A k A A* k A* 0

49 Lindemann Hinshelwood Mechanism Given With the result A* k b k a k A ' a 2 A 2 d[p] kk[ ] A* a b A kb dt k k A d [P] kk[ A] a b k A where k dt eff eff ' k k A b a ' 1 ka 1 k k k k A eff a b a This is not a first order reaction. However, if the rate for stabilization of A* is much greater than the unimolecular decay, A*A, k a [A*][A] ] >> k b b[ [A*] or just k a [A] >> k b, then k eff is cons t, a pseudo 1 st order Rxn Such a reaction goes from 2 nd order to 1 st order as the pressure increases. This is a very common gas phase mechanism, as collisions supply (take away) energy. b ' a

50 Lindemann Hinshelwood Mechanism O + NO NO 2 * NO 2 * + M NO 2 + M* Normally, bath gas dominates

51 Final Exam Exam will be comprehensive ; planned length ~ 1.5 hours ~ 7-8 problems (weights given) budget your time Closed book Don t memorize formulas/constants This exam will have some numeric problems, similar to the homework in complexity. If a problem seems lengthy, do another problem & come back later Understanding homework and old exams will be quite useful EXAM held in JILA Auditorium, Dec 11, 4:30 PM

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