Advanced physical chemistry notes

Advanced topics in physical chemistry

1. Advanced Physical Chemistry CHEM540 Dr. Fateh Eltaboni Assistant Professor of Physical Chemistry at the University of Benghazi Omer Al Mokhtar University Faculty of Science Chemistry Department     CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 1

2. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 2 Instructor Profile

3. 3 Recommended Text Authors Profile CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) Recommended Notes

4. Physical Chemistry CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 4

5. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 5 Gas Laws ❶

6. 6 CHEM540 Course Syllabus Kinetic molecular theory of gases Thermochemistry Thermodynamics Chemical Kinetics Nuclear chemistry and Radio chemistry Electrochemistry Symmetry and group theory Quantum mechanics and atomic structure Spectroscopy and Photochemistry Course General Information CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

7. The properties of gases The simplest state of matter is a gas , a form of matter that fills any container it occupies. The perfect (ideal) gas: Ideal gas molecules are widely separated from one another and move in paths that are largely unaffected by intermolecular forces. The states of gases: The physical state of a sample of a substance, its physical condition, is defined by its physical properties . Two samples of a substance that have the same physical properties are in the same state. The state of a pure gas, for example, is specified by giving its volume, V , amount of substance (number of moles), n pressure, p , and temperature, T. 7 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

8. The properties of gases Experimentally, is sufficient to specify only three of these variables, for then the fourth variable is fixed. That is, it is an experimental fact that each substance is described by an equation of state , an equation that interrelates these four variables. The general form of an equation of state is: This equation tells us that, if we know the values of n , T , and V for a particular substance, then the pressure has a fixed value. One very important example is the equation of state of a ‘perfect gas’, which has the form p = nRT / V , where R is a constant. 8 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

9. The states of gases: (a) Pressure Pressure , p , is defined as force, F , divided by the area, A , to which the force is applied: That is, the greater the force acting on a given area, the greater the pressure. The origin of the force exerted by a gas is the continuous collisions of the molecules on the walls of its container. The SI unit of pressure, the pascal ( Pa, 1 Pa = 1 N m −2 ) . But there are several other units are still widely used (Table 1.1). 9 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

10. Pressure Fig. 1.1 When a region of high pressure is separated from a region of low pressure by a movable wall ( piston ), the wall will be pushed into one region or the other, as in (a) and (c) . However, if the two pressures are identical, he wall will not move (b) . The latter condition is one of mechanical equilibrium between the two regions. This condition of equality of pressure on either side of a movable wall is a state of mechanical equilibrium between the two gases. 10 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

11. Pressure units 11 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

12. The measurement of pressure: The pressure exerted by the atmosphere is measured with a barometer . The original version of a barometer (which was invented by Torricelli, a student of Galileo) was an inverted tube of mercury sealed at the upper end. When the column of mercury is in mechanical equilibrium with the atmosphere, the pressure at its base is equal to that exerted by the atmosphere. It follows that the height of the mercury column is proportional to the external pressure. 12 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

13. The measurement of pressure: Example 1.1 Calculating the pressure exerted by a column of liquid Derive an equation for the pressure at the base of a column of liquid of mass density ρ (rho) and height h at the surface of the Earth. The pressure exerted by a column of liquid is commonly called the ‘ hydrostatic pressure ’. Solution: where and The pressure at the base of the column is therefore: ‘ hydrostatic pressure ’. Hydrostatic pressure is independent of the shape and cross-sectional area of the column. The mass of the column of a given height increases as the area ( F= mg ), so the two cancel. 13 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

14. The measurement of pressure: 14 The pressure of a sample of gas inside a container is measured by using a pressure gauge , which is a device with electrical properties that depend on the pressure. For instance, a Bayard–Alpert pressure gauge is based on the ionization of the molecules present in the gas and the resulting current of ions is interpreted in terms of the pressure. In a capacitance manometer , the deflection of a diaphragm relative to a fixed electrode is monitored through its effect on the capacitance of the arrangement. Certain semiconductors also respond to pressure and are used as transducers in solid state pressure gauges. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

15. (c) Temperature The concept of temperature springs from the observation that a change in physical state ( for example, a change of volume ) can occur when two objects are in contact with one another, as when a red-hot metal is plunged into water. The temperature , T , is the property that indicates the direction of the flow of energy through a thermally conducting, rigid wall. If energy flows from A to B when they are in contact, then we say that A has a higher temperature than B (Fig. 1.2). 15 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

16. (c) Temperature 16 Fig. 1.2 Energy flows as heat from a region at a higher temperature to one at a lower temperature if the two are in contact through a diathermic wall (Boundary), as in (a) and (c) . However, if the two regions have identical temperatures, there is no net transfer of energy as heat even though the two regions are separated by a diathermic wall (b) . The latter condition corresponds to the two regions being at thermal equilibrium . Two types of boundary: Diathermic (thermally conducting; ‘ dia ’ is from the Greek word for ‘through’) if a change of state is observed when two objects at different temperatures are brought into contact. A metal container has diathermic walls. Adiabatic (thermally insulating) if no change occurs even though the two objects have different temperatures. A vacuum flask is an approximation to an adiabatic container. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

17. (c) Temperature The temperature is a property that indicates whether two objects would be in ‘thermal equilibrium’ if they were in contact through a diathermic boundary. Zeroth Law of thermodynamics : 17 Fig. 1.3 The experience summarized by the Zeroth Law of thermodynamics : if an object A is in thermal equilibrium with B and B is in thermal equilibrium with C , then C is in thermal equilibrium with A . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

18. thermometer The Zeroth Law justifies the concept of temperature and the use of a thermometer , a device for measuring the temperature. Suppose that B is a glass capillary containing a liquid , such as mercury , that expands as the temperature increases. When A is in contact with B . According to the Zeroth Law, if the mercury column in B has the same length when it is placed in thermal contact with another object C , then we can predict that no change of state of A and C will occur when they are in thermal contact. Moreover, we can use the length of the mercury column as a measure of the temperatures of A and C . 18 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

19. Thermometry In the thermometry : temperatures are related to the length of a column of liquid, and the difference in lengths shown when the thermometer was first in contact with melting ice and then with boiling water was divided into 100 steps called ‘ degrees ’, the lower point being labelled . This procedure led to the Celsius scale of temperature. The Celsius scale are denoted θ (theta) and expressed in degrees Celsius ( °C ). Because different liquids expand to different extents, thermometers constructed from different materials showed different numerical values of the temperature between their fixed points. 19 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

20. Thermodynamic temperature scale The pressure of a gas can be used to construct a perfect-gas temperature scale that is independent of the identity of the gas. Thermodynamic temperature scale : in which temperatures are denoted T and are normally reported in kelvins (K; not °K). Thermodynamic and Celsius temperatures are related: 20 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

21. (c) Temperature Example: Express 25.00°C as a temperature in kelvins. Answer: T / K = (25.00 °C )/ °C + 273.15 = 25.00 + 273.15 = 298.15 Note how the units (in this case, °C) are cancelled like numbers. This is the procedure called ‘quantity calculus’ in which a physical quantity (such as the temperature) is the product of a numerical value (25.00) and a unit (1°C). Multiplication of both sides by the unit K then gives T = 298.15 K . 21 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

22. The gas laws 22 (a) The perfect gas laws: Avogadro’s principale : V α n at constant p , T Boyle’s law: p α 1/ V at constant n , T Charles’s law: V α T at constant n , p p α T at constant n , V Avogadro’s principle is a principle rather than a law because it depends on the validity of a model, in this case the existence of molecules. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

23. The gas laws Boyle’s and Charles’s laws are examples of a limiting law , a law that is strictly true only in a certain limit, in this case p → 0. The empirical observations summarized by (A,B, and C) can be combined into a single expression pV = constant × nT Perfect gas law: 23 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

24. The gas laws 24 Linear relationship CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

25. The gas laws 25 The perfect gas law (or perfect gas equation of state ). It is the approximate equation of state of any gas, and becomes increasingly exact as the pressure of the gas approaches zero. A gas that obeys perfect gas law exactly under all conditions is called a perfect gas (or ideal gas ). A real gas , an actual gas, behaves more like a perfect gas at lower the pressure. The gas constant R can be determined by evaluating R = pV / nT for a gas. More accurate value for R can be obtained by measuring the speed of sound in a low-pressure gas (argon is used in practice) and extrapolating its value to zero pressure. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

26. The gas laws 26 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

27. The gas laws Example 1.2 Using the perfect gas law In an industrial process, nitrogen is heated to 500 K in a vessel of constant volume. If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it behaved as a perfect gas? Method : We expect the pressure to be greater on account of the increase in temperature. The perfect gas law in the form pV / nT = R Combined gas law: 27 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

28. 28 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

29. Perfect gas law The perfect gas law is important in physical chemistry because it is used to derive a wide range of relations that are used in thermodynamics. It is also used to calculate the properties of a gas under a variety of conditions. For instance, the molar volume, V m = V / n , of a perfect gas under the conditions called standard ambient temperature and pressure . ( SATP ), which means 298.15 K and 1 bar (that is, exactly 10 5 Pa), is easily calculated from V m = RT / p to be 24.789 dm 3 mol −1 Standard temperature and pressure (STP), was 0°C and 1 atm ; at STP, the molar volume of a perfect gas is 22.414 dm 3 mol −1 . 29 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

30. The kinetic model of gases A gas may be pictured as a collection of particles in ceaseless , random motion (Fig. 1.6). The ‘kinetic model’ (or the ‘kinetic molecular theory’, KMT) of gases: is a technique proposes a qualitative model and then expressing that model mathematically. The perfect gas law is quantitative model very simple, and experimentally confirmable. 30 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

31. KMT The kinetic model of gases is based on three assumptions : A gas consists of molecules in ceaseless random motion. The size of the molecules is negligible in the sense that their diameters are much smaller than the average distance travelled between collisions. The molecules do not interact, except during collisions. Assumption 3 suggests that the potential energy of the molecules (their energy due to their position) is independent of their separation and may be set equal to zero. The total energy of a sample of gas is therefore the sum of the kinetic energies (the energy due to motion) of all the molecules present in it. It follows that the faster the molecules travel (and hence the greater their kinetic energy), the greater the total energy of the gas. 31 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

32. The pressure of a gas according to the kinetic model The kinetic model accounts for the steady pressure exerted by a gas in terms of the collisions the molecules make with the walls of the container. The pressure exerted by a gas of mass m in a volume V is: 32 Compare it to pV = nRT . This conclusion is a major success of the kinetic model, for the model implies an experimentally verified result. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

33. KMT: Root mean square speed Here c (or μ ) is the root-mean-square speed (rms speed) of the molecules. This quantity is defined as the square root of the mean value of the squares of the speeds, v , of the molecules. That is, for a sample consisting of N molecules with speeds v 1 , v 2 , . . . , v N , The rms speed might at first sight seem to be a rather peculiar measure of the mean speeds of the molecules, but its significance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass m travelling at a speed v is E k = ½ mv 2 , which implies that the mean kinetic energy, 〈 E k 〉, is the average of this quantity, or mc 2 . It follows from the relation ½ mc 2 = 〈 E k 〉 that 33 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

34. KMT: rms rms its a measure of the mean kinetic energy of the molecules of the gas. The rms speed is a measure of molecular speed. The mean speed , ć, of the molecules: For samples consisting of large numbers of molecules, the mean speed is slightly smaller than the rms speed. The precise relation is: 34 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

35. rms speed of gas molecules Which gives: The ns now cancel to give: Rearrange: rms = Substitution of the molar mass of O 2 (32.0 g mol −1 ) and a temperature corresponding to 25°C (that is, 298 K) gives an rms speed for these molecules of 482 m s −1 . The same calculation for nitrogen molecules gives 515 m s −1 . Both these values are not far off the speed of sound in air (346 m s −1 at 25°C). 35 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

36. Speeds of gas molecules That similarity is reasonable, because sound is a wave of pressure variation transmitted by the movement of molecules, so the speed of propagation of a wave should be approximately the same as the speed at which molecules can adjust their locations. The rms and mean speeds of molecules in a gas is proportional to the square root of the temperature. Therefore, doubling the thermodynamic temperature (that is, doubling the temperature on the Kelvin scale) increases the mean and the rms speed of molecules by a factor of 2 1/2 = 1.414. . . . 36 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

37. Maxwell distribution of speeds Not all molecules travel at the same speed: some move more slowly than the average (until they collide, and get accelerated to a high speed, like the impact of a bat on a ball), and others move at much higher speeds than the average , but be brought to a sudden stop when they collide. There is a ceaseless redistribution of speeds among molecules as they undergo collisions. Each molecule collides once every nanosecond ( 1 ns = 10 −9 s ) or so in a gas under normal conditions. The mathematical expression that tells us the fraction of molecules that have a particular speed at any instant is called the distribution of molecular speeds . 37 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

38. The Maxwell distribution of speeds The distribution tell us that at 20°C 19 out of 1000 O 2 molecules have a speed in the range between 300 and 310 m s −1 , that 21 out of 1000 have a speed in the range 400 to 410 m s −1 , and so on. The precise form of the distribution was worked out by James Clerk Maxwell towards the end of the 19 th century, and his expression is known as the Maxwell distribution of speeds . According to Maxwell, the fraction f of molecules that have a speed in a narrow range between s and s + Δ s (for example, between 300 m s −1 and 310 m s −1 , corresponding to s = 300 m s −1 and Δ s = 10 m s −1 ) is 38 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

39. Maxwell equation One of the skills to develop in physical chemistry is the ability to interpret the message carried by equations . Equations carry information, and it is more important to be able to read that information than simply to remember the equation . Let’s read the information in Maxwell equation piece by piece . It will be useful to know the shape of exponential functions . Here, we deal with two types, e − ax and e − ax 2 . An exponential function , a function of the form e − ax , starts off at 1 when x = 0 and decays toward zero , which it reaches as x → ∞ (Fig. 1.7). This function approaches zero more rapidly as a increases . 39 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

40. Maxwell equation 40 • A Gaussian function , a function of the form e - ax 2 , also starts off at 1 when x = and decays to zero as x increases, however, its decay is initially slower but then plunges down to zero more rapidly than an exponential function (Fig. 1.7). The illustration also shows the behaviour of the two functions for negative values of x . The exponential function e - ax rises rapidly to infinity, but the Gaussian function falls back to zero and traces out a bell-shaped curve . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

41. Maxwell equation Maxwell equation includes a decaying exponential function , the term: Its presence implies that the fraction of molecules with very high speeds will be very small because e − s 2 becomes very small when s 2 is large. The factor Ms 2 /2 RT in the exponent, is large when the molar mass , M , is large , so the exponential factor goes most rapidly towards zero when M is large. That tells us that heavy molecules are unlikely to be found with very high speeds. 41 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

42. Maxwell equation The opposite is true when the temperature, T , is high : then the factor M /2 RT in the exponent is small, so the exponential factor falls towards zero relatively slowly as s increases. This tells us that at high temperatures , a greater fraction of the molecules can be expected to have high speeds than at low temperatures. A factor s 2 goes to zero as s goes to zero, so the fraction of molecules with very low speeds will also be very small. The remaining factors (the term ) simply ensure that when we add together the fractions over the entire range of speeds from zero to infinity , then we get 1 . 42 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

43. Maxwell equation: T & M 43 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

44. Maxwell Equation: Experiment The Maxwell distribution has been verified experimentally by: 44 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

45. Diffusion and effusion Diffusion is the process by which two or more molecules of different substances mix with each other. The atoms of two solids diffuse into each other when the two solids are in contact, but the process is very slow . The diffusion of a solid through a liquid solvent is much faster but mixing normally needs to be encouraged by stirring or shaking (the process is then no longer pure diffusion ). Gaseous diffusion is much faster . It accounts for the largely uniform composition of the atmosphere , for if a gas is produced by a localized source (such as CO 2 from the breathing of animals, O 2 from photosynthesis by green plants, and pollutants (oxides) from vehicles and industrial sources), then the molecules of gas will diffuse from the source and will be distributed throughout the atmosphere ( CO 2 +O 2 +Oxides ). 45 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

46. Diffusion and effusion The process of effusion is the escape of a gas through a small hole, as in a puncture in an inflated balloon or tyre (Fig. 1.10). 46 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

47. Diffusion The rates of diffusion and effusion of gases increase with increasing temperature , as both processes depend on the motion of molecules, and molecular speeds increase with temperature ( Maxwell ). The rates also decrease with increasing molar mass , as molecular speeds decrease with increasing molar mass ( Maxwell ). The dependence on molar mass, however, is simple only in the case of effusion. The experimental observations on the dependence of the rate of effusion of a gas on its molar mass are summarized by Graham’s law of effusion , proposed by Thomas Graham in 1833: 47 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

48. Diffusion At a given pressure and temperature , the rate of effusion of a gas is inversely proportional to the square root of its molar mass: Rate in this context means the number (or number of moles) of molecules that escape per second. The rate of effusion of gases was used to determine molar mass by comparison of the rate of effusion of a gas or vapour with that of a gas of known molar mass. 48 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

49. Molecular collisions The average distance that a molecule travels between collisions is called its mean free path , λ (lambda). The mean free path in a liquid is less than the diameter of the molecules , because a molecule in a liquid meets a neighbour even if it moves only a fraction of a diameter. In gases, the mean free paths of molecules can be several hundred molecular diameters. If we think of a molecule as the size of a tennis ball, then the mean free path in a typical gas would be about the length of a tennis court. 49 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

50. Collision The collision frequency , z , is the average rate of collisions made by one molecule. Specifically, z is the average number of collisions one molecule makes in a given time interval divided by the length of the interval. The inverse of the collision frequency, 1/ z , is the time of flight , the average time that a molecule spends in flight between two collisions: For instance, if there are 10 collisions per second, so the collision frequency is 10 s −1 , then the average time between collisions is 0.1 of a second and 0.1 the time of flight is s). The collision frequency in a typical gas is about 10 9 s −1 at 1 atm and room temperature, so the time of flight in a gas is typically 1 ns. 50 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

51. Collision The rms speed c: 51 To find expressions for λ and z we need a decorative version of the kinetic model. The basic kinetic model supposes that the molecules are point-like; however, to obtain collisions, we need to assume that two ‘points’ score a hit whenever they come within a certain range d of each other, where d can be thought of as the diameter of the molecules (Fig. 1.11). The collision cross-section , σ (sigma), the target area presented by one molecule to another, is therefore the area of a circle of radius d , so σ = π d 2 . When this quantity is built into the kinetic model, it is possible to show that: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

52. Collision 52 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni)

53. Thermodynamics❷ CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 53

54. Thermodynamics: First Law The basic concepts Work, heat, and energy The internal energy Expansion work Heat transactions Enthalpy Adiabatic changes Thermochemistry CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 54 Standard enthalpy changes Standard enthalpies of formation The temperature dependence of reaction enthalpies The Joule–Thomson effect

55. System and Surrounding The basic concepts: The universe is divided into two parts, the system and its surroundings . System is the part of the world exists under study Examples: a reaction vessel, an engine, a biological cell, …… 2. Surroundings is the region outside the system and are where we make our measurements. Types of System: 1.Open 2. Closed 3.Isolated CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 55

56. Types of Systems Open system : exchange matter and energy with its surroundings. B. Closed system: exchange energy with its surroundings but it cannot exchange matter. C. Isolated system: exchange neither energy nor matter with its surroundings. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 56

57. Work, heat, and energy Work: is done to achieve motion against an opposing force. Energy: is the capacity to do work. Types of thermodynamic processes: An exothermic process: is a process that releases energy as heat into its surroundings (combustion reactions). 2. An endothermic process: is a process that absorbs heat from its surroundings. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 57 When an endothermic process occurs in an adiabatic system, the temperature falls; if the process is exothermic, the temperature rises. (c) When an endothermic process occurs in a diathermic container, the system remains at the same temperature. (d) If the process is exothermic, the process is isothermal.

58. The molecular interpretation of heat and work CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 58 Transfer heat to surroundings stimulates random motion Transfer work to surroundings stimulates orderly motion

59. The internal energy Total energy of a system is called its internal energy , U . The internal energy is the total kinetic and potential energy of the molecules in the system. The change in internal energy ( Δ U ): when a system changes from an initial state i with internal energy U i to a final state f of internal energy U f : Δ U = U f − U i The internal energy is a state function, its value depends only on the final and initial states of the system. The internal energy is an extensive property of a system (depends on amount) and is measured in joules (1 J = 1 kg m 2 s −2 ). The molar internal energy( U m ) is the internal energy divided by the amount of substance in a system U m = U / n: it is an intensive property (independents on amount) its unit (kJ mol −1 ). CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 59

60. (a) Molecular interpretation of internal energy Modes of motion of a molecule: The ability to translate (the motion of its centre of mass through space) Rotate around its centre of mass Vibrate (as its bond lengths and angles change, centre of mass unmoved). Many physical and chemical properties depend on the energy associated with each of these modes of motion. For example, a chemical bond might break if a lot of energy becomes concentrated in it, for instance as strong vibration. Total energy of a monatomic perfect gas free to move in three dimensions is   CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 60 where U m (0) is the molar internal energy at T = 0,

61. Modes of motions When the gas consists of molecules, we need to take into account the effect of rotation and vibration: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 61 (a) A linear molecule can rotate about two axes perpendicular to the line of the atoms. (b) A nonlinear molecule can rotate about three perpendicular axes.

62. First Law of thermodynamics First Law of Thermodynamics: The internal energy of an isolated system is constant. Mathematical Statement of the First Law: Δ U = q + w (q: heat & w: work) w and q are positive if energy is transferred to the system as work or heat and are negative if energy is lost from the system. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 62

63. Expansion work Expansion work: is the work arising from a change in volume. (a) The general expression for work: dw = -F x dz (opposing force x distance) F = p ex A dw = - p ex A x dz (A x dz = dV ) dw = - p ex dV ( Expansion work ) CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 63 To obtain the total work done when the volume changes from an initial value V i to a final value V f we integrate this expression between the initial and final volumes:

64. Types of Work CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 64

65. (b) Free expansion: It occurs when p ex = 0. w = 0 J (c) Expansion against constant pressure: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 65

66. (d) Reversible expansion: To achieve reversible expansion we set p ex equal to p (gas pressure) . (e) Isothermal reversible expansion ( T is constant) : CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 66

67. Example 2.1 Calculating the work of gas production Calculate the work done when 50 g of iron reacts with hydrochloric acid to produce FeCl 2 ( aq ) and hydrogen in (a) a closed vessel of fixed volume , (b) an open beaker at 25°C . Answer: (a) Fixed volume ∆V = 0, so no expansion work is done and w = 0. (b) Gas drives back the atmosphere and therefore w = − p ex Δ V . Neglect V i because V f (after the production of gas) is so much larger Δ V = V f − V i ≈ V f = nRT / p ex , where n is the amount of H 2 produced. Therefore, CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 67 n can be taken as the amount of Fe atoms that react. Because the molar mass of Fe is 55.85 g mol −1

68. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 68

69. Heat transfer The heat transfer as at constant volume is equal to the change in internal energy of the system. ∆U = q v Calorimetry is the measurement of heat transfer. Calorimeter constant (C): q = C Δ T At constant volume C is the slope of the internal energy with respect to temperature. Heat capacity at constant V: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 69

70. Heat capacities are extensive properties: 100 g of water, for instance, has 100 times the heat capacity of 1 g of water. The molar heat capacity at constant volume , C V ,m = C V / n (intensive property) Specific heat capacity: C V ,s = C V / m C p − C V = nR Enthalpy: H = U + pV Energy transferred as heat at constant pressure is equal to the change in enthalpy of a system: Δ H = q p CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 70

71. Justification: The relation ΔH = q p For a general infinitesimal change : H changes from U + pV to H + dH = (U + dU ) + (p + dp )(V + dV ) H + dH = U + dU + pV + pdV + Vdp + dpdV ( dpdV is small and is neglected) After recognizing U + pV = H H + dH = H + dU + pdV + Vdp CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 71 and hence that d H = d U + p d V + V d p If d U = d q + d w d H = d q + d w + p d V + V d p If d w = − p d V d H = d q + V d p If the heating occurs at constant pressure ( d p = 0). d H = d q

72. The measurement of an enthalpy change An enthalpy change measured by monitoring the temperature change that accompanies a physical or chemical change occurring at constant pressure. Isobaric calorimeter: used for studying processes at constant pressure. Adiabatic flame calorimeter: used for a combustion reaction to measure Δ T. Bomb calorimeter: used to measure Δ U and convert it to Δ H . Differential scanning calorimeter (DSC) Noncalorimetric technique Δ H = Δ U for solids and liquids Because solids and liquids have small molar volumes: H m = U m + pV m ≈ U m CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 72

73. Example 2.2 Relating Δ H and Δ U The change in molar internal energy when CaCO 3 (s) as calcite converts to another form, aragonite, is +0.21 kJ mol −1 . Calculate the difference between the molar enthalpy and internal energy changes when the pressure is 1.0 bar given that the densities of the polymorphs are 2.71 g cm −3 and 2.93 g cm −3 , respectively. Answer: Δ H m = H m (aragonite) − H m (calcite) = { U m (a) + pV m (a)} − { U m (c) + pV m (c)} = Δ U m + p { V m (a) − V m (c)} by substituting V m = M /ρ CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 73

74. Substitution of the data, using M = 100 g mol −1 , gives = −2.8 × 10 5 Pa cm 3 mol −1 = −0.28 Pa m 3 mol −1 Because 1 Pa m 3 = 1 J , Δ H m − Δ U m = −0.28 J mol −1 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 74

75. The enthalpy of a perfect gas (PV= nRT ) H = U + pV = U + nRT Δ H = Δ U + Δ n g RT ( Δ n g is the change in moles of gas ) Example: In the reaction 2 H 2 (g) + O 2 (g)→2 H 2 O(l) Δ n g = ( 0- 3 ) = −3 mol. Therefore, at 298 K, when RT = 2.48 kJ mol −1 , the enthalpy and internal energy changes taking place in the system are related by: Δ H m − Δ U m = (−3 mol ) × RT ≈ −7.4 kJ mol −1 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 75

76. Example 2.3 Calculating a change in enthalpy Water is heated to boiling under a pressure of 1.0 atm . When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporized. Calculate the molar internal energy and enthalpy changes at the boiling point (373.15 K). Answer The enthalpy change is Δ H = q p = I ΔΦ t = (0.50 A ) × (12 V ) × (300 s ) = 0.50 × 12 × 300 J In the process H 2 O(l) → H 2 O(g): Δ n g = +1 mol , so Δ U m = Δ H m − RT = (+41 kJmol -1 ) – (8.314x10 -3 kJmol -1 K -1 x 373.15K) = +38 kJ mol −1 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 76

77. The variation of enthalpy with temperature Heat capacity at constant pressure: Δ H = C p Δ T (at constant pressure) q p = C p Δ T ( Δ H = q p ) C p , m /(J K −1 mol −1 ) = a + bT + c / T 2 ( a,b,c constants) CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 77

78. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 78 Answer: 2.20 kJ mol -1 Method: ∆H

79. Adiabatic changes For the reversible adiabatic expansion of a perfect gas, pressure and volume are related by an expression that depends on the ratio of heat capacities . Δ U = C V ( T f − T i ) = C V Δ T Because the expansion is adiabatic, q = 0 And because Δ U = q + w , So work done during an adiabatic expansion w ad = C V Δ T T i and T f for reversible adiabatic expansion: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 79 c = C V ,m / R

80. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 80

81. Adiabatic Changes V i and V f for reversible adiabatic expansion: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 81

82. Thermochemistry Thermochemistry : is a branch of thermodynamics that study of the heat transferred during the chemical reactions. If we know Δ U or Δ H for a reaction, we can predict the heat the reaction can produce Exothermic process: Δ H < 0 ( - ) Endothermic process: Δ H > 0 ( + ) Standard enthalpy change , Δ H o , the change in enthalpy for a process in which the initial and final substances are in their standard states : Standard state of a substance at a specified temperature is its pure form at 1 bar (1 atm ). Standard enthalpy of vaporization ( Δ H o vap ): CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 82

83. Enthalpies of physical changes CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 83

84. Enthalpy is a state function: Enthalpy of sublimation: single step: Multi steps Inverse enthalpy: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 84 For example, because the enthalpy of vaporization of water is + 44 kJ mol -1 at 298 K, its enthalpy of condensation at that temperature is - 44 kJ mol -1 .

85. Enthalpies of chemical change Thermochemical equation , a combination of a chemical equation and the corresponding change in standard enthalpy: Standard reaction enthalpy , Δ r H o : For a reaction of the form 2 A + B→3 C + D Example, slandered enthalpy of combustion: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 85

86. Hess’s law: The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. Example 2.5 Using Hess’s law From the following data: CH 2 =CHCH 3 (g) + H 2 (g)→CH 3 CH 2 CH 3 (g) ∆H 1 = −124 kJ mol −1 CH 3 CH 2 CH 3 (g) + 5 O 2 (g)→3 CO 2 (g) + 4 H 2 O(l) ∆H 2 = −2220 kJ mol −1 H 2 O(l)→H 2 (g) + O 2 (g) ∆H 3 = +286 kJ mol −1 Calculate the standard enthalpy of combustion of propene . Answer: combustion reaction we require is CH 2 =CHCH 3 (g) + O 2 (g) → 3 CO 2 (g) + 3 H 2 O(l) (balance CHO) ---------------------------------------------------------------------------------------------------------------------------------------------------------- CH 2 =CHCH 3 (g) + H 2 (g) → CH 3 CH 2 CH 3 (g) ∆H 1 = −124 kJ mol −1 CH 3 CH 2 CH 3 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O (l) ∆H 2 = −2220 kJ mol −1 H 2 O (l) → H 2 (g) + O 2 (g) ∆H 3 = +286 kJ mol −1 -------------------------------------------------------------------------------------------------------------------------------------------------------- CH 2 =CHCH 3 (g) + 4.5 O 2 (g) → 3 CO 2 (g) + 3 H 2 O(l) ∆H = ∆H 1 + ∆H 1 + ∆H 1 = ( −124 – 2220 +286) kJ mol −1 = -2058 kJ mol −1   CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 86

87. Standard enthalpies of formation: The standard enthalpy of formation , Δ f H o , of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states : The reference state of an element is its most stable state at the specified temperature and 1 bar. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 87

88. The Joule–Thomson effect: The Joule–Thomson effect is the change in temperature of a gas when it undergoes isenthalpic expansion. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 88

89. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 89 Fig. 2.30 The sign of the Joule–Thomson coefficient, μ, depends on the conditions. Inside the boundary, the blue area, it is positive and outside it is negative. The temperature corresponding to the boundary at a given pressure is the inversion temperature’ of the gas at that pressure. For a given pressure, the temperature must be below a certain value if cooling is required but, if it becomes too low, the boundary is crossed again and heating occurs. Reduction of pressure under adiabatic conditions moves the system along one of the isenthalps , or curves of constant enthalpy. The inversion temperature curve runs through the points of the isenthalps where their slope changes from negative to positive. Fig. 2.32 The principle of the Linde refrigerator is shown in this diagram. The gas is recirculated, and, so long as it is beneath its inversion temperature, it cools on expansion through the throttle. The cooled gas cools the high-pressure gas, which cools still further as it expands. Eventually liquefied gas drips from the throttle.

90. Thermodynamics : Second Law The direction of spontaneous change The dispersal of energy Entropy Entropy changes accompanying specific processes The Third Law of thermodynamics Concentrating on the system The Helmholtz and Gibbs energies Standard molar Gibbs energies CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 90 Combining the First and Second Laws The fundamental equation Properties of the internal energy Properties of the Gibbs energy

91. The dispersal of energy During a spontaneous change in an isolated system the total energy is dispersed into random thermal motion of the particles in the system. Spontaneous: some things happen naturally. The energy of the system that tends towards a minimum CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 91 Fig. 3.2 The direction of spontaneous change for a ball (system) bouncing on a floor (surrounding) . On each bounce some of its energy is degraded into the thermal motion of the atoms of the floor, and that energy disperses. The reverse process is nonspontaneous .

92. Entropy Entropy: is a measure of the energy dispersed in a process. The Second Law of thermodynamics ( entropy): The entropy of an isolated system increases in the course of a spontaneous change: Δ S tot > 0 (where S tot is the total entropy of the system and its surroundings) Thermodynamically irreversible processes like: Cooling to the temperature of the surroundings free expansion of gases are spontaneous processes, and hence must be accompanied by an increase in total entropy. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 92

93. The thermodynamic definition of entropy dS is change in entropy, q rev is the heat supplied reversibly and T is temperature in kelvin. The units of entropy are (J K −1 ). Entropy is an extensive property. The units of molar entropy are (J K −1 mol −1 ). Entropy is an intensive property. Calculating the entropy change for the isothermal expansion of a perfect gas: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 93

94. If Δ U = q + w and Δ U = 0 for Isothermal process Therefore that q rev = − w rev (Isothermal expansion work) CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 94

95. Entropy change of the surroundings T sur is constant: P sur is constant: For an adiabatic change : q sur = 0 ΔS sur = 0 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 95 ∆ H sur

96. The statistical view of entropy: Boltzmann formula for the entropy: S = k ln W where k = 1.381 × 10 −23 J K −1 and W is the number of microstates When W = 1, S =0 Increasing W leads to increase S CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 96

97. Carnot cycle: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 97

98. Carnot Cycle: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 98 P A V A T h P B V B T h P C V C T C P D V D T C To prove entropy is a state function we need to show that the integral of d S is independent of path. To do so, it is sufficient to prove that the integral of d S around an arbitrary cycle is zero, for that guarantees that the entropy is the same at the initial and final states of the system regardless of the path taken between them

99. Carnot Cycle: Step (1): Reversible isothermal expansion from A to B at T h ; dS = + q h / T h , q h is the energy supplied to the system as heat from the hot source. 2. Step (2): Reversible adiabatic expansion from B to C. No energy leaves the system as heat, dS = ( T h decreases to T c ) 3 . Step (3): Reversible isothermal compression from C to D at T c . Energy is released as heat to the cold sink; dS = - q h / T h 4. Step (4): Reversible adiabatic expansion from B to C. No energy leaves the system as heat, dS = ( T c increases to T h ) The total change in entropy around the cycle: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 99

100. Heat during the two isothermal stages are: Step (1): Step (3): From the relation between T and V for reversible adiabatic processes: Multiplication of the above gives: which, on cancellation of the temperatures, simplifies to: With this relation we can write: And therefore: Since and So = = 0   CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 100

101. Efficiency,η (eta), of a heat engine: Carnot Efficiency: The Second Law of thermodynamics suggests that all reversible engines have the same efficiency regardless of their construction!! . Coefficient of performance : CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 101

102. The greater the coefficient of performance and the more efficient is the refrigerator: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 102

103. Entropy changes accompanying specific processes: Entropy change for the isothermal expansion of a perfect gas: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 103 This change is the negative of the change in the system, so we can conclude that ΔS tot = 0, which is what we should expect for a reversible process. On the other hand the isothermal expansion occurs freely (w = 0), then q = 0 (because ΔU = 0). Consequently, ΔS sur = 0, and the total entropy change is given by: In this case, ΔS tot > 0, as we expect for an irreversible process.

104. Entropy of phase transition: If the phase transition is exothermic ( Δ trs H < 0, as in freezing or condensing ) If the phase transition is endothermic ( Δ trs H > 0, as in melting or vaporization ) CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 104 ∆ trs S = 85 JK -1 mol -1 taken from table

105. Entropy of heating CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 105

106. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 106

107. The measurement of entropy The entropy of a system at a temperature T is related to its entropy at T = 0 At T= 0 C p ,m = aT 3 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 107

108. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 108

109. The Third Law of thermodynamics Nernst heat theorem : Δ S →0 as T →0 The Third Law : entropy of all perfect crystalline substances is zero at T = 0. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 109

110. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 110

111. The Helmholtz and Gibbs energies Helmholtz energy , A , which is defined as: d A = d U − T d S Gibbs energy , G , which is defined as: d G = d H − T d S The criteria of spontaneous change as: (a) d A T , V ≤ 0 (b) d G T, p ≤ 0 The criterion of equilibrium , when neither the forward nor reverse process has a tendency to occur, is (a) d A T , V = 0 (b) d G T, p = 0 The tendency of a system to move to lower A is due to its tendency to move towards states of lower internal energy and higher entropy. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 111

112. Gibbs energy: The Gibbs energy (the ‘free energy’) is more common in chemistry than the Helmholtz energy because, at least in laboratory chemistry, we are usually more interested in changes occurring at constant pressure than at constant volume. Standard molar Gibbs energies: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 112

113. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 113

114. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 114

115. The fundamental equation: The fundamental equation, a combination of the First and Second Laws, is an expression for the change in internal energy that accompanies changes in the volume and entropy of a system. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 115 The equilibrium constant for a reaction is related to its standard reaction Gibbs energy by: ΔrG o = − RTln K

116. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 116

117. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 117 C hemical kinetics ❸

118. The Rates of Chemical Reactions  chemical kinetics, the study of reaction rates .  mechanism of reaction, the sequence of elementary steps involved in a reaction .   CHEMICAL KINETICS Experimental techniques  real-time analysis, a procedure in which the composition of a system is analysed while the reaction is in progress. flow method, a procedure in which the composition of a system is analysed as the reactants flow into a mixing chamber . stopped-flow technique, a procedure in which the reagents are mixed very quickly in a small chamber fitted with a syringe instead of an outlet tube . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 118

119. The Rates of Chemical Reactions (3) flash photolysis, a procedure in which the reaction is initiated by a brief flash of light . quenching methods, techniques based on stopping the reaction after it has been allowed to proceed for a certain time . chemical quench flow method, a technique in which the reactants are mixed as in the flow method but the reaction is quenched by another reagent . freeze quench method, a technique in which the reaction is quenched by cooling the mixture . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 119

120. The Rates of Chemical Reactions The rates of reactions (a) The definition of rate  rate of consumption of a reactant R, –d[R]/ d t .  rate of formation of a product P, d[P]/ d t .  rate of reaction, v = (1/ V ) d ξ / d t where ξ is the extent of reaction.  rate of homogeneous reaction, v = (1/ v J )d[J]/ d t . rate of heterogeneous reaction, v = (1/ v J ) d σ J / d t . (b) Rate laws and rate constants rate law, the rate as a function of concentration, v = f ([A],[B], ...) . rate constant, the constant k in a rate law. hydrogen–bromine reaction: the observed rate law is d[HBr]/ d t = k r [H 2 ][Br 2 ] 3/2 /([Br 2 ] + k r  [HBr]) . (c) Reaction order reaction order, the power to which the concentration of a species is raised in a rate law of the form v = [A] a [B] b ... .  first-order reaction, a reaction with a rate law of the form v = k r [A].  second-order reaction, a reaction with a rate law of the form v = k r [A] 2 .  overall order, the sum of the orders a + b +..., in a rate law of the form v = k r [A] a [B] b .... zero-order rate law, a rate law of the form v = k r . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 120

121. The Rates of Chemical Reactions (d) The determination of the rate law  isolation method, a procedure in which the concentrations of all the reactants except one are in large excess .  Pseudo first-order rate law, v = k r  [A] with k r  = k r [B] by maintaining B in large excess. S. W. Han et al., Chem. Lett ., 2007 , 36 , 1350. Nanoparticle Catalyst NaBH 4 AgNPs CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 121

122. The Rates of Chemical Reactions method of initial rates, a procedure in which the rate is measured at the beginning of the reaction for several different initial concentrations of reactants; v = k r  [A] a log v = log k r  + a log [A] . Example 21.2 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 122

123. The Rates of Chemical Reactions Integrated rate laws integrated rate law, the integrated form of a rate law for concentration as a function of time. (a) First-order reactions  first-order integrated rate law, - d[A]/ dt = k r [A]  ln([A]/[A] ) = – k r t , [A] = [A] e – k r t .  half life, t 1/2 = (ln 2)/ k r .  time constant, the time required for the concentration of a reactant to fall to 1/e of its initial value, τ = 1/ k r . Example 21.3 CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 123

124. The Rates of Chemical Reactions (c) Second-order reactions second-order integrated rate law, - d[A]/ dt = k r [A] 2  1/[A] – 1/[A] = k r t [A] = [A] /(1 + k r t [A] ).  half life, t 1/2 = 1/ k r [A] .  half life for n th-order reaction (n>1), t 1/2 = 2 n -1 -1/( n -1) k r [A] n-1 . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 124

125. The Rates of Chemical Reactions CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 125

126. The Rates of Chemical Reactions 21.4 Reactions approaching equilibrium 21.4(a) First-order reactions close to equilibrium CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 126

127. The Rates of Chemical Reactions 21.4(b) Relaxation methods  relaxation, the return to equilibrium.  temperature jump, a procedure in which a sudden temperature rise is imposed and the system returns to equilibrium.  pressure-jump techniques, as for temperature jump, but with a sudden change in pressure. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 127

128. The Rates of Chemical Reactions The temperature dependence of reaction rates  Arrhenius equation, ln k = ln A – E a / RT .  pre-exponential factor (frequency factor), the parameter A in the Arrhenius equation.  activation energy, the parameter E a in the Arrhenius equation; the minimum kinetic energy for reaction during a molecular encounter.  Arrhenius parameters, the parameters A and E a . generalized activation energy, E a = RT 2 (d ln k /d T ).  activated complex, the cluster of atoms that corresponds to the region close to the maximum potential energy along the reaction coordinate.  transition state, a configuration of atoms in the activated complex which, if attained, leads to products. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 128

129. The Rates of Chemical Reactions ACCOUNTING FOR THE RATE LAWS 21.6 Elementary reactions elementary reaction, a single step in a reaction mechanism. H + Br 2  HBr + Br molecularity, the number of molecules coming together to react in an elementary reaction . reaction order, the power to which the concentration of a species is raised in a rate law of the form v = [A] a [B] b ... ; an empirical quantity, and obtained from the experimental rate law.  unimolecular reaction, an elementary reaction involving a single reactant molecule. bimolecular reaction, an elementary reaction involving the encounter of two reactant molecules. CH 3 I(alc) + CH 3 CH 2 O - (alc)  CH 3 OCH 2 CH 3 (alc) + I - (alc) Mechanism: CH 3 I + CH 3 CH 2 O -  CH 3 OCH 2 CH 3 + I - , a single elementary step Rate law: v = k r [ CH 3 I ][ CH 3 CH 2 O - ] CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 129

130. Chapter 21: The Rates of Chemical Reactions 21.7 Consecutive elementary reactions  consecutive first-order reactions, a sequence of first-order reactions. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 130

131. Chapter 21: The Rates of Chemical Reactions  steady-state approximation (or quasi-steady-state approximation, QSSA) the rates of change of concentrations of all reaction intermediates are negligibly small: d[I]/d t  0 and their concentrations are low.  induction period, the initial stage of a reaction during which reaction intermediates are formed. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 131

132. Chapter 21: The Rates of Chemical Reactions  validation of steady-state approximation (QSSA) CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 132

133. Chapter 21: The Rates of Chemical Reactions  An example of steady-state approximation CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 133

134. Chapter 21: The Rates of Chemical Reactions rate-determining step, the step in a mechanism that controls the overall rate of the reaction; commonly but not necessarily the slowest step . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 134

135. Chapter 21: The Rates of Chemical Reactions  pre-equilibrium, a state in which an intermediate is in equilibrium with the reactants and which arises when the rates of formation of the intermediate and its decay back into reactants are much faster than its rate of formation of products . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 135

136. Chapter 21: The Rates of Chemical Reactions Examples of reaction mechanisms 21.8 Unimolecular reactions  Lindemann–Hinshelwood mechanism, a theory of ‘unimolecular’ reactions. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 136

137. Chapter 21: The Rates of Chemical Reactions  Test of Lindemann–Hinshelwood mechanism  activation energies of composite reactions CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 137

138. Chapter 21: The Rates of Chemical Reactions 21.9 POLYMERIZATION KINETICS  stepwise polymerization, a polymerization reaction in which any two monomers present in the reaction mixture can link together at any time and the growth of the polymer is not confined to chains that are already forming.  chain polymerization, a polymerization reaction in which an activated monomer attacks another monomer, links to it, then that unit attacks another monomer, and so on. stepwise polymerization chain polymerization CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 138

139. Chapter 21: The Rates of Chemical Reactions 21.9(a) Stepwise polymerization  degree of polymerization, the average number of monomer residues per polymer molecule,  n  = 1/(1 – p ), where p is the average number of monomers per polymer molecule;  n  = 1 + k r t [A] . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 139

140. Chapter 21: The Rates of Chemical Reactions 21.9(b) Chain polymerization rate of chain polymerization CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 140

141. Chapter 21: The Rates of Chemical Reactions  kinetic chain length, v , the ratio of the number of monomer units consumed per activated centre produced in the initiation step; v = k [M][I] –½ . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 141

142. Chapter 21: The Rates of Chemical Reactions 21.10 PHOTOCHEMISTRY  primary process, a process in which products are formed directly from the excited state of a reactant.  secondary process, a process in which products originate from intermediates formed directly from the excited state of a reactant. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 142

143. Chapter 21: The Rates of Chemical Reactions 21.10(a) The primary quantum yield  primary quantum yield, ϕ , the number of photophysical or photochemical events that lead to primary products divided by the number of photons absorbed by the molecule in the same interval, ϕ = v / I abs . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 143

144. Chapter 21: The Rates of Chemical Reactions 21.10(b) Mechanism of decay of excited singlet states CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 144

145. Chapter 21: The Rates of Chemical Reactions 21.10(c) Quenching  quenching, shortening of the lifetime of an excited state.  Stern–Volmer equation, φ f,0 / φ f = 1 + τ k Q [Q]. Stern–Volmer plot CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 145

146. Chapter 21: The Rates of Chemical Reactions  Modified Stern–Volmer equation Example 21.9 Q: Fe(OH 2 ) 6 3+ CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 146

147. Chapter 21: The Rates of Chemical Reactions  Three common mechanism of quenching 21.10(d) Resonance energy transfer Förster theory, a theory of resonance energy transfer, - efficiency ( η T =1- φ f,0 / φ f ); η T  1/R 6 [ η T = R 6 /( R 6 + R 6 )]. - h  donor  h  acceptor CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 147

148. Chapter 21: The Rates of Chemical Reactions  fluorescence resonance energy transfer (FRET), a technique used to measure distances in biological systems. η T = R 6 /( R 6 + R 6 ) 7.9 nm Protein rhodopsin η T =1- φ f,0 / φ f CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 148

149. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 149 Quantum theory

150. Quantum theory: introduction and principles The origins of quantum mechanics Energy quantization Wave–particle duality The dynamics of microscopic systems The Schrödinger equation The Born interpretation of the wavefunction Quantum mechanical principles The information in a wavefunction The uncertainty principle The postulates of quantum mechanics CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 150

151. The origins of quantum mechanics Classical Mechanics: the laws of atoms and subatomic particles motion introduced in the 17 th century by Isaac Newton , Quantum Mechanics: towards the end of the 19 th century, experimental evidence accumulated showing that classical mechanics failed when it was applied to particles as small as electrons, and it took until the 1920s to discover the appropriate concepts and equations for describing them. In classical physics, light is described as electromagnetic radiation : an oscillating electric and magnetic disturbance that spreads as a harmonic wave through the vacuum. Wave displacements that can be expressed as sine or cosine functions. The speed of light , c , The constant speed of wave traveling ( c ≈ 3 × 10 8 m s −1 ) CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 151

152. Wavelength ,λ (lambda), the distance between the neighbouring peaks of the wave. Frequency , ν (nu), the number of times per second at which its displacement at a fixed point returns to its original value. The wavelength is usually measured in nm , where 1 nm = 10 −9 m . The frequency is measured in hertz , where 1 Hz = 1 s −1 . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 152 Wavenumbers ( ῡ )are normally reported in reciprocal centimetres (cm -1 ).

153. The electromagnetic spectrum and the classification of the spectral regions: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 153

154. Energy quantization: Energy can be transferred only in separate amounts. Black-body radiation (19 th -century): Black body: is an ideal emitter object capable of emitting and absorbing all wavelengths of radiation uniformly. A good approximation to a black body is a pinhole in an empty container maintained at a constant temperature, CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 154 Energy distribution in a blackbody at several temperatures. The total energy density (the area under the curve) increases as the temperature is increased (as T 4 ).

155. The energy density ( d E , J m −3 ) is proportional to the width, dλ ρ (rho), the constant of proportionality, is called the density of states (J m −4 ). The total energy density in a region is the integral over all wavelengths: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 155

156. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 156 The electromagnetic vacuum can be regarded as able to support oscillations of the electromagnetic field. When a high frequency, short wavelength oscillator (a) is excited, that frequency of radiation is present. The presence of low frequency, long wavelength radiation (b) signifies that an oscillator of the corresponding frequency has been excited. The Rayleigh–Jeans law predicts an infinite energy density at short wavelengths. This approach to infinity is called the ultra v iolet catastrophe.

157. Quantization of energy: Planck (1900) found that the permitted energies of an electromagnetic oscillator of frequency ν are integer multiples of h ν : E = nh ν n = 0, 1, 2, . . . h is Plank’s constant = 6.626 × 10 −34 J s. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 157 The Einstein temperature, θ E = hν/k

158. Atomic and molecular spectra: Bohr theory CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 158 Spectrum of radiation emitted by excited iron atoms: consists of radiation at a series of discrete wavelengths (or frequencies). When a molecule changes its state, it does so by absorbing radiation at definite frequencies. This spectrum is part of that due to the electronic, vibrational, and rotational excitation of SO 2 molecules. This observation suggests that molecules can possess only discrete energies, not an arbitrary energy. Spectroscopic transitions, such as those shown above, can be accounted for if we assume that a molecule emits a photon as it changes between discrete energy levels. Note that high-frequency radiation is emitted when the energy change is large.

159. Wave–particle duality The particle character of electromagnetic radiation: The photoelectric effect establishes the view that electromagnetic radiation, regarded in classical physics as wave-like, consists of particles (photons). Each particle having an energy h ν , if two the energy is 2 h ν , and so on. E = Nhν Number of photons (N): P is power in watts. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 159

160. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 160

161. Wave function ( Φ ): the energy required to remove an electron from the metal. The photoelectric effect: is supposed that the incident radiation is composed of photons that have energy proportional to the frequency of the radiation. (a) The energy of the photon is insufficient to drive an electron out of the metal. (b) The energy of the photon is more than enough to eject an electron, and the excess energy is carried away as the kinetic energy of the photoelectron (the ejected electron). CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 161

162. The wave character of particles Louis de Broglie when, in 1924, linear momentum p = mv (with m the mass and v the speed of the particle) Δφ = potential difference, m = mass of electron e = charge of electron CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 162

163. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 163

164. The Schrödinger equation, 1926 The mathematical representation of the wave that in quantum mechanics replaces the classical concept of trajectory is called a wavefunction , ψ ( psi). The Schrödinger equation is a second-order differential equation used to calculate the wavefunction of a system. The factor V ( x ) is the potential energy of the particle at the point x ħ = h /2π (which is read h -cross or h -bar) is a modification of Planck’s constant with the value 1.055 × 10 −34 J s. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 164

165. The Schrödinger equation For one-dimensional systems: V ( x ) is potential energy of the particle and E is its total energy. 2. For three-dimensional systems: where V may depend on position and ∇2 (‘del squared’) is: In systems with spherical symmetry is: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 165 where

166. In the general case the Schrodinger equation is written: where Ĥ is the hamiltonian operator for the system: For the evolution of a system with time, it is necessary to solve the time-dependent Schrödinger equation: Using the Schrödinger equation to develop the de Broglie relation: For a freely moving particle in a region where its potential energy V is constant. After writing V ( x ) = V , we can rearrange into: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 166

167. cos kx is a wave of wavelength λ = 2π/k , as can be seen by comparing cos kx with the standard form of a harmonic wave, cos(2πx/λ). The quantity E − V is equal to the kinetic energy of the particle, E k , so k = (2mE k /ħ 2 ) 1/2 , which implies that E k = k 2 ħ 2 /2m . Because E k = p 2 /2m , it follows that p = kħ . Therefore, the linear momentum is related to the wavelength of the wavefunction by: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 167 which is the de Broglie relation.

168. The Born interpretation of the wavefunction: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 168 If the wavefunction of a particle has the value ψ at some point x, then the probability of finding the particle between x and x + dx is proportional to |ψ | 2 dx . |ψ | 2 is the probability density For a one-dimensional system The probability of finding the particle in the volume element dτ = d x d y d z at some location r is proportional to the product of dτ and the value of |ψ | 2 at that location. For three-dimensional space

169. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 169 The sign of a wavefunction has no direct physical significance: the positive and negative regions of this wavefunction both correspond to the same probability distribution (as given by the square modulus of ψ and depicted by the density of shading). a a constant and r the distance from the nucleus of H atom. Relative probabilities of finding the electron inside a region of volume δV : P ∝ e −2r/a δV

170. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 170 Example (Interpreting a wavefunction) Calculate the relative probabilities of finding the electron inside a region of volume δV = 1.0 pm 3 , which is small even on the scale of the atom, located at (a) the nucleus, (b) a distance a from the nucleus. Answer : P ∝ e −2r/a δV (a) At the nucleus, r = 0, so P ∝ e × (1.0 pm 3 ) = (1.0) × (1.0 pm 3 ) = 1 (b) At a distance r = a in an arbitrary direction, P ∝ e −2* a /a × (1.0 pm 3 ) = (0.14) × (1.0 pm 3 ) = 0.14 Comment: the ratio of probabilities is 1.0/0.14 = 7.1. Note that it is more probable (by a factor of 7) that the electron will be found at the nucleus. The negatively charged electron is attracted to the positively charged nucleus, and is likely to be found close to it.

171. Normalization: This freedom to vary the wavefunction by a constant factor means that it is always possible to find a normalization constant , N , such that the proportionality of the Born interpretation becomes an equality. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 171 In one dimension, the wavefunction is normalized if: In three dimensions, the wavefunction is normalized if:

172. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 172 x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ r, the radius, ranges from 0 to ∞ θ, the colatitude, ranges from 0 to π φ, the azimuth, ranges from 0 to 2 π Then d τ = r 2 sin θ drd θ d φ Spherical polar coordinates

173. Example: (Normalizing a wavefunction) Normalize the wavefunction Answer: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 173

174. Quantization of Ψ An acceptable wavefunction cannot be zero everywhere, because the particle it describes must be somewhere. Since the energy of a particle is quantized . ψ must be: continuous have a continuous slope be single-valued be square- integrable CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 174 (d) Unacceptable because it is infinite over a finite region (c) Unacceptable because it is not single-valued (b) Unacceptable because its slope is discontinuous (a) Unacceptable because it is not continuous

175. Quantum mechanical principles The information in a wavefunction: The Schrödinger equation for a particle of mass m free to move parallel to the x -axis with zero potential energy is obtained from by setting V = 0, and is CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 175 The solutions of this equation have the form and A and B are constants

176. The probability density CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 176 Suppose that B = Where is the particle? To find out, we calculate the probability density: This probability density is independent of x so, wherever we look along the x -axis, there is an equal probability of finding the particle The same would be true if A = 0; then the probability density would be | B | 2 , a constant. The square modulus of a wavefunction corresponding to a definite state of linear momentum is a constant; so it corresponds to a uniform probability of finding the particle anywhere.

177. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 177 Now suppose that the wavefunction A = B . Then The probability density now has the form The probability density periodically varies between 0 and 4 | A | 2 . The locations where the probability density is zero correspond to nodes in the wavefunction. N ode is a point where a wavefunction passes through zero.

178. Operators, eigenvalues, and eigenfunctions CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 178 with (in one dimension) Eigenvalue equation : (E)Energy is eigenvalue

179. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 179

180. The construction of operators The importance of eigenvalue equations to measure an observable Observables is measurable properties of a system, such as the momentum or the electric dipole moment . CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 180 operator (^) for location along the x -axis is multiplication (of the wavefunction) by x operator for linear momentum parallel to the x-axis is proportional to taking the derivative (of the wavefunction) with respect to x.

181. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 181

182. Construct the operator for kinetic energy CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 182 Operator for the total energy, the hamiltonian operator:

183. Hermitian operators CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 183

184. To say that two different functions ψ i and ψ j are orthogonal means that the integral (over all space) of their product is zero: CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 184

185. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 185 Example: The wavefunctions sin x and sin 2x are eigenfunctions of the hermitian operator d 2 /dx 2 , with eigenvalues −1 and −4. Verify that the two wavefunctions are mutually orthogonal. Answer: we integrate the product (sin x)(sin 2x) over all space, which we may take to span from x = 0 to x = 2π,

186. The uncertainty principle CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 186 Δp is the ‘uncertainty’ in the linear momentum parallel to the axis q, and Δq is the uncertainty in position along that axis.

187. CHEM540 Lecture Notes 1 (Dr Fateh Eltaboni) 187