Chemistry 223H1 Fall Term 2010

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1 Chemistry 223H1 Fall Term 2010 Instructor: Dr. D.F. McIntosh Office hour: Thursdays 3:00 4:00 pm Rm 223 in Lash Miller Lectures: T : LM 159 3:00 4:30 pm F : LM 159 2:00 3:30 pm Textbook: Physical Chemistry for the Biological Sciences by Raymond Chang. (in the Bookstore) Midterm: Wednesday, November 03, 2010

2 Other Suggested Reading.these additional texts also provide very useful discussions of topics covered in this course.. 1. Physical Chemistry 7 th Edition Atkins P., de Paula J. 2. The Elements of Physical Chemistry with Applications in Biology Atkins P. *** Please note that if not specified the figures within the slides are taken from the Chang Text. *** The majority of the problems are also taken from the Chang Text.

3 Problem Sets & Tutorials Problems Recommended problems will be assigned for each chapter. They will not be graded. Full solutions will be covered in detail at the TUTORIALs. TUTORIALS: Mondays 3:00-4:00 pm (A L) GB 244 Megan Oh (M Z) GB 248 Brian Lam A teaching assistant will go over solutions to problems and answer any specific questions.

4 Course Evaluation 45 % Midterm Wednesday, November 03, % Final Exam To Be Scheduled

5 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

6 SECTION 1.0. Review and Definitions Relevant Sections in Text.Chapter 2 (sections )

7 SECTION 1.0. Review and Definitions GOALS: 1) to review some basic definitions 2) to review the gas laws (i.e. Boyle s law, Charles law, Avogadro s law) and the ideal gas equation 3) to understand difference between an ideal gas and a real gas (i.e. Z, van der Waals equation) Why is this important: we need these basic concepts to understand molecules and their energies.

8 1.1. Review and Definitions A) Define System: Surroundings OPEN: mass and energy exchange CLOSED: only energy exchange ISOLATED: no mass nor energy exchange SYSTEM Surroundings Surroundings CLOSED OPEN ISOLATED

9 B) Define Properties of Matter Extensive Properties properties whose values are directly proportional to amount of material in system e.g. mass, area, volume, energy Intensive Properties.properties whose values are not dependent on amount of material in system e.g. temperature, pressure, density.intensive properties are usually defined as ratio of two Extensive properties Pressure = Force/Area Density = Mass/Volume

10 C) Temperature How do we define temperature?.. hot or cold Need more precise scientific definition.. - temperature is a variable - consider closed systems A & B (no mass exchange) An ISOTHERM P A V A A P A V A P B V B P B V B A B B heat exchange

12 1.2. Review of GAS Laws A) Boyle s Law relationship between Pressure (P) and Volume (V) V 1 / P.PV = constant so P 1 V 1 = P 2 V 2 (constant n and T) B) Charles Law relationship between V and T so V 1 /T 1 = V 2 /T 2 V T

13 ABSOLUTE ZERO plot of V versus T yields straight line (at a given P) extend line to zero V..intercept is C C is known as absolute zero Kelvin Temperature Scale: T(K) = t ( C) Absolute zero = C or 0 K ** Absolute temperatures are always used in gas law and thermodynamic calculations!

14 C) Avogadro s Law..equal volumes of gas at the same T and P contain the same number of molecules V n OK.we have three laws (Boyle s, Charles and Avogadro s let s put them together IDEAL GAS EQUATION D) Ideal Gas Equation P V = n R T * How did we get this value for R? R* = L atm mol -1 K -1 R = J mol -1 K -1

15 Experimentally it is found that 1 mole of an ideal gas occupies L at 1 atm and K (STP) R = the gas constant R = (PV)/(nT) = (1 atm)(22.414l) (1 mol)(273.15k) = L atm mol -1 K -1

16 Sample Problem Nitrogen is heated to 500 K in a vessel of constant volume. If it enters the vessel at P = 100 atm and T = 300 K, what pressure would it exert at the working temperature if it behaves as an ideal gas?

17 3.0. REAL gases versus IDEAL gases Ideal Gas: Gas molecules possess negligible volume No interaction between molecules (attractive or repulsive) No Such Gases EXIST! * Ideal Gas Equation is only useful for gases at high T, low P Why???

Real Gases deviations from the ideal gas law are most important at high pressures and low temperatures..especially when a gas is on the verge of becoming a liquid.

18 Real Gases deviations from the ideal gas law are most important at high pressures and low temperatures..especially when a gas is on the verge of becoming a liquid.. Real gases deviate from ideal gas law due to intermolecular interactions Attractive Intermolecular Interactions.exist when molecules are close together but not in actual contact. Repulsive Intermolecular Interactions.important when molecules are in actual contact..these are short-range interactions. Figure Physical Chemistry 7 th Edn. Atkins P., de Paula J.

19 INTERMOLECULAR INTERACTIONS ATTRACTIVE these interactions are present over distance equal to a few molecular diameters attractive interactions are responsible for condensation of gases into liquids at low T.at low T, molecules don t have enough KE to escape from each other so they stick together. when molecules are separated by a few molecular diameters they attract each other.when they are in direct contact they repel each other REPULSIVE these interactions are reason that solids and liquids have volume and don t completely collapse.

20 Z = Compressibility Factor...plot Z versus P to measure deviation from ideality Z = PV/nRT = PV/RT V is molar volume (V/n) For an ideal gas Z = 1 for any value of P at a given T in limit of P approaching zero Z = 1 for all gases Why do all gases have Z > 1 at high pressures????.at high P real gases have a larger molar volume than ideal gases due to repulsive intermolecular interactions.

22 Van der Waals Equation. attempts to account for (1) finite volume of individual molecules (2) the attractive forces between molecules [ P + (an 2 )/(V 2 ) ] [V nb] = nrt 1st term accounts for attractive intermolecular forces 2nd term accounts for the finite volume of molecules a and b are constants characteristic of the gas under study rough correlation between a and boiling point * VdW equation becomes unreliable at very high T and low P

23 Values for CO 2 determined using ideal gas and VdW equation P observed/atm P ideal/atm P vdwaals/atm

24 Sample Problem To synthesize ammonia using the Haber process, 2000 moles of N 2 are heated in an 800 L vessel to 625 C. Calculate the pressure of the gas if N 2 behaves (a) as a van der Waals gas and (b) as an ideal gas. Note: a = 1.35 atm L 2 mol -2 b = L mol -1

25 Problem continued.. *** Pressure calculated by van der Waals equation is approx. 6 % higher..for industrial processes it s important to have the accurate pressure so to ensure the reaction vessel can handle it. (page 22 in Chang Text)

26 Sample Problem Find the density of F 2 at 20 C at atm. Mol. Wt. = g/mol Solution 1- the unknown is the density () of F 2 2- since the gas is at relatively low pressure we assume it behaves ideally (so use equation PV = nrt)

27 Our Progress to DATE.. Definitions..system, properties and Temperature Gas Laws.and we put them together to get the IDEAL GAS equation Difference between IDEAL and REAL gases Van der Waals equation

28 2.0. Molecules and their Energies Relevant Sections in Text.Chapter 3 (section 3.8)

29 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 8) Chemical Equilibrium 9) Chemical Kinetics and Rates of Processes

30 2.0. Molecules and their Energies Goals (1) to understand how molecules store and use energy (2) to define specific heat capacity and heat capacity (3) to test equipartition of energy theorem using heat capacity measurements Why is this important..need these basic concepts to understand thermodynamics.

31 Molecules and their Energies (Section 3.8) The average kinetic energy for one molecule is (3/2) k B T..where k B is the Boltzmann Constant = x JK -1..since k B N A = R then average kinetic energy for one mole of a gas is KE = ( 3/2 ) RT How is this Energy distributed? to understand lets consider a theory based on Classical Mechanics.

32 The Equipartition of Energy Theorem.. energy of a molecule is equally divided among all types of motion or degrees of freedom energy for each molecule can be divided: E TOT = E TRANSLATION + E ROTATION + E VIBRATION + E ELECTRONIC Let s consider a monatomic gas first.(simplest case ) for a monatomic gas (e.g. He, Ne or Ar) each atom needs three coordinates to completely define its position an atom has three translational degrees of freedom, each of which has an energy of ½ k B T : 3 (½k B T ) = (3/2)(R/N A )T so E TOT = (3/2) RT (no rotation or vibration for monatomic)

33 U = (3/2) RT where U is the molar internal energy (no rotation or vibration for monatomic)

34 Consider molecules..a a little more complicated.. for molecules must consider rotational, vibrational and translational motion for a molecule with N atoms we need a total of 3N coordinates to describe its motion 3 translational components 2 (linear) or 3 (non-linear) rotational components (3N 5) linear or (3N 6) non-linear vibrational components*** *** it is difficult to count the number of vibrational modes so just subtract translational and rotational modes...now we can calculate the total energy for atoms and molecules

37 Mechanical Degrees of Freedom and the Equipartition of Energy Translational: Non Linear and Linear Molecules: 3 Degrees of Freedom ie. movement in the x, y and z directions Rotational: Non Linear Molecules: 3 Degrees of Freedom ie. rotation about the x, y and z axes R x, R y, R z Linear Molecules: 2 Degrees of Freedom ie. rotation about the x and y axes R x, R y

38 Mechanical Degrees of Freedom and the Equipartition of Energy Vibrational: Non Linear Molecules with N atoms: (3N 6) Degrees of Freedom ie. (3N 6) vibrations Q 1, Q 2, Q 3,, Q 3N 6 Linear Molecules with N atoms: (3N 5) Degrees of Freedom ie. (3N 5) vibrations Q 1, Q 2, Q 3,, Q 3N 5

39 Kinetic Theory of Gases (Translational Motions) All Kinetic Energy terms U trans = 3 RT 2 1 = M 2 v = 2 1 M ( 2 v x y v + 2 z v ) Therefore: 1 RT per Degree of Translational Freedom 2

40 Same is True for Rotational Motions All Kinetic Energy terms U rot = 1 2 I1 ω I2 ω I3 ω 3 2 where: I j = Moment of Inertia = m m i = mass of ith atom r i = distance of ith atom from Centre of Mass N i1 2 i r i ω i = angular velocity of rotation (radians per sec) Therefore: 1 RT 2 per Degree of Rotational Freedom

41 However Vibrations Have Both Kinetic And Potential Energies U vib = 1 dx 2 μ 2 dt 1 + k x 2 2 where: μ = Reduced Mass = m1m 2 (m m 1 2 ) (diatomic molecule) k = Force Constant Therefore: (1) RT per Degree of Vibrational Freedom

42 Heat Capacity At Constant Volume C V = U T V Monatomic Gases: No rotations or vibrations only translations C V = U T trans V = T 3 RT 2 V 3 = R 2

43 Linear Molecules with N Atoms 3 translational degrees of freedom 2 rotational degrees of freedom (3N 5) vibrational degrees of freedom U total = 3 RT RT (3N 5)[(1)RT] = N RT C V = V total T U = V RT 2 5 3N T = N R

44 Non-Linear Molecules with N Atoms 3 translational degrees of freedom 3 rotational degrees of freedom (3N 6) vibrational degrees of freedom U total = RT RT 2 + (3N 6)[(1)RT] = [3N 3] RT C V = U T total V T = 3N 3 RT V = N 3 3 R

45 Equipartition of Energy for ONE MOLE of GAS Kind of motion Translational No. of modes Energy per mole Rotational No. of modes Energy per mole Vibrational No. of modes Energy per mole TOTAL ENERGY Monatomic Molecule Polyatomic Molecule Linear Non-linear /2 RT 3/2 RT 3/2 RT RT 3/2 RT 0 3N 5 3N 6 0 (3N-5)RT* (3N-6)RT* 3/2 RT (3N (5/2))RT (3N-3)RT 3)RT * For vibrational energy there are two terms one for PE (½ RT)and one for KE (½ RT)

46 What does this mean??? (1) Molecules can store energy.. If you heat up a gas the energy provided by the increase in temperature should be taken up by each of the 3N components...molecules can TRANSLATE, ROTATE, VIBRATE, or EXCITE to store energy (2) We can now calculate the Total Energy for a monatomic or polyatomic gas.. Example: Total Energy (U) for 1 mole of a linear diatomic gas is then: U = (3/2) RT + RT + RT = (7/2) RT remember you are calculating U using values from previous table which are for ONE MOLE of gas..thus this is molar internal energy

47 But what we really want to know is.. how much energy can molecules hide as HEAT at any temperature? What is the molecule s s HEAT CAPACITY? Heat Capacity is defined as the amount of energy ( U) required to raise the temperature of a substance by 1 K Heat Capacity (C) = ΔU / ΔT At constant volume. C V = [U / T ] v (see p. 58 and 66 in text)

48 Is C = U/ T or is C v = U/ T? ANSWER: C = U/ T page 90 or expressed in partial derivatives: C V = [U / T ] v page 58 & 90 PLEASE NOTE..when you use the table of values (p.58) given to calculate U these values are per mole of gas therefore you are calculating molar internal energy (U)..which in turn will give you C v C v = {[(3/2)RT] / T} v = (3/2)R Please refer p in text for more on this.

49 EXAMPLES: For a monatomic gas. molar Heat Capacity is C v = {[(3/2)RT] / T} v = (3/2)R = JK -1 mol -1 For a diatomic gas. molar Heat Capacity is C v = [ [(7/2)RT] / T ] v = (7/2)R = JK -1 mol -1 can also calculate for polyatomic..

51 To this point 1) we reviewed the Equipartition of Energy Theorem 2) used this theorem to calculate the total Energy (U) for atoms or molecules 3) used calculated value for U to calculate C v 4) compared measured values with our calculated values for C v - only see agreement for monatomics not for molecules Why? to this point we have used classical mechanics to predict values for C v discrepancies for molecules can be explained using quantum mechanics..

52 According to Quantum Mechanics. Electronic, Vibrational and Rotational energies of a molecule are quantized meaning: different molecular energy levels are associated with each type of motion Spacing between energy levels varies depending on type of energy

53 spacings between electronic energy levels vibrational rot. trans. spacings between translational energy levels are very small.can be treated as a continuum as in classical mechanics Why is this important? How does this relate to HEAT CAPACITY? when a system absorbs heat..the energy is used to promote various kinds of motion remember: the heat capacity of a system tells us it s ability to store energy energy may be stored partly in vibrational, rotational motion etc...i.e. the molecules may be promoted to a higher vibrational energy level or a higher rotational energy level etc.

54 since spacings between rotational energy levels are less than those between vibrational energy levels (Fig. 3.13)..it s easier to promote to a higher rotational energy level than to a higher vibrational energy level The ratio of the populations of molecules in any two energy levels (E 1 and E 2 ) is given by the Boltzmann Distribution Law: N 2 = e (E)/(k B T) N 1 where E = E 2 E 1 (where we assume E 2 E 1 ) k B = Boltzmann Constant = x JK -1 N 2 /N 1 is always 1.meaning fewer molecules in upper level (at thermal equilibrium at a finite temperature)

55 N 2 = e (E)/(k B T) N 1 Using the Boltzmann Equation we can make estimates of populations in lower and higher energy levels If E is large compared with k B T molecules will crowd into lower energy level E

56 OK.let s estimate how many molecules are in lower versus higher energy level for translational motion N 2 = e (E)/(k B T) N 1. (E)/(k B T) at 298 K is J = 2.4 x (1.381 x JK -1 )(298K) thus N 2 /N 1 1 Therefore, for translational motion the number of molecules in the higher energy level is same as the number in the lower energy level. Meaning.the Energy is not Quantized.

57 for rotational motion E is also small compared to k B T so ratio of N 2 /N 1 is close to but smaller than 1 Energies are quantized for rotational (but not translational) motion for vibrational motion, spacings between energy levels is large so E is k B T.and ratio of N 2 /N 1 is much smaller than 1 at 298 K most molecules are in lower energy level and only a small fraction in the higher levels forelectronic motion the spacings between levels are so large that almost all molecules are in lowest energy level at 298 K

58 What should we understand from this.. 1) Remember using classical mechanics the equation for Total Energy (U) was: U = E TRANSLATION + E ROTATION + E VIBRATION + E ELECTRONIC heat capacity = C v = [U / T ] v.but when we used this equation the values calculated for C v did not match experimental values for C v (for diatomic or polyatomic molecules) 2) So we tried to use quantum mechanics to explain this discrepancy and to calculate new values for C v - from quantum mechanics approach we see that at room temperature only the translational and rotational motions would contribute to the heat capacity

59 So the equation for Total Energy (U) becomes: U = E trans + E rot = (5/2) RT C v = (5/2)R = JK -1 mol -1..this value is close to the experimental value of JK -1 mol -1 (see Table 3.3 in Chang text) **this discrepancy means that vibrational motion does make a small contribution If we increase the temperature we see that vibrational motions makes even more of a contribution T(K) C v

61 Sample Problem What is the total energy of CO 2 and NH 3? (Calculate using classical mechanics.) Solution CO 2 is a linear molecule bond angles = 180 Total Energy = (3N 5/2)RT = (9 5/2)RT = 6.5 RT N NH 3 is a non-linear molecule H H bond angles = H Total Energy = (3N 3)RT = (12 3)RT = 9 RT

62 e.g. For H 2 at room temperature only translational motion and rotational motion contribute to the heat capacity at elevated temperatures vibrational energy must be taken into account and at even higher temperatures electronic motion must be accounted for

63 Our Progress to DATE.. GOALS (1) to understand how molecules store and use energy (2) to define heat capacity (3) to test equipartition of energy theorem using heat capacity measurements PROGRESS (1) molecules store heat or energy by translational, rotational vibrational and electronic motion (2) heat capacity is a measure of the ability of a substance to store energy (3) must use quantum mechanics to estimate Cv only translational and rotational motion make contribution at room temperature

Handout 11: Ideal gas, internal energy, work and heat. Ideal gas law

Handout 11: Ideal gas, internal energy, work and heat. Ideal gas law Handout : Ideal gas, internal energy, work and heat Ideal gas law For a gas at pressure p, volume V and absolute temperature T, ideal gas law states that pv = nrt, where n is the number of moles and R