เคม เช งฟ ส กส I Physical Chemistry I 014034 สาขาว ชาเคม คณะศ ลปศาสตร และว ทยาศาสตร ม.ก. กพส. Sc.4-309 : faaspmp@ku.ac.th Physical Chemistry The study of underlying physical principles that governs the properties and behaviors of chemical systems Physical chemistry provides the basic framework for all other branches of Chemistry-for inorganic chemistry, organic chemistry, biochemistry, geochemistry and engineering. Physical chemistry is the branch of chemistry that establishes and develops the principles of the subject. It concepts are used to explain and interpret observations on the physical and chemical properties of matter. Physical chemistry is also essential for developing and interpreting the modern techniques used to determine the structure and properties of matter, such as new synthetic materials and biological macromolecules. Part 1: Equilibrium Physical Chemistry 1. The properties of gases The perfect gas and real gases. Thermodynamics First, Second and Third Law 3. Phase diagram Part : Structure 1. Quantum theory. Molecular structure 3. Spectroscopy Part 3: Change 1. Molecular in Motion (gases, liquids). The rates of chemical reactions 3. The kinetics of complex reactions 4. Processes at solid surfaces One of the roles of physical chemistry is to establish the link between the properties of bulk matter and the behaviour of the particles-atoms, ions or molecules-of which it is composed. Bulk matter Gas is a fluid state of matter that fills the container it occupies. Liquid is a fluid state of matter that possesses a welldefined surface Solid retains its shape independent of the shape of the container it occupies. 1
Physical chemist thinks about the three states of matter: A gas is composed of particles in continuous rapid, chaotic motion. A particle travels several diameters before another particle, and for most of time the particles are so far apart that they interact t with each other only very weakly. The overriding image is one of busy chaos. A liquid consists of particles that are in contact with each other, but are able to move past each other in a restricted manner. The particles are in a continuous state of motion, but travel only a fraction of a diameter before bumping into a neighbour. The overriding image is one of movement, but with molecules jostling one another. Perfect Gases They move rapidly, and collide randomly and chaotically with one another. A solid consists of particles that are in contact with one another and unable to move past one another. Although the particles oscillate around an average location, they are essentially trapped in their initial positions. The overriding image is one of almost stationary, regularly arrayed particles. The essential difference between the three states of matter comes down to the freedom of the particles to move past one another. The physical properties of an ideal gas are completely described by four parameters which their respective SI units are: the amount of substance of which it is composed, n, in moles; the temperature of the gas, T, in Kelvin the pressure of the gas, P, in Pascal; The volume occupied by the gas, V, in m 3 The perfect gas equations Boyle s law; P. V = constant at constant temperature Charles s law; V T at constant pressure Avogadro s principle; V n at constant pressure and temperaturet The perfect gas equation PV = nrt Or ideal gas law was empirically developed from experimental observations. The kinetic theory of gases : a model of molecular nature of gases The kinetic theory of gases is an attempt to describe the macroscopic properties of a gas in terms of molecular behavior. 1. The size of the molecules which make up the gas is negligible compared to the distance between them;. There are no attractive forces between the molecules; 3. The molecules travel in straight lines, except during the course of collisions. Molecules undergo perfectly elastic collisions; i.e. the kinetic energy of the molecules is conversed in all collisions, but may be transferred between them. These are considered to be true of real systems at low pressure
The speed of molecules in gases Real gases Maxwell distribution of speeds Most probable speed = RT M 1/ The molecular origin of pressure Total energy = kinetic energy + potential energy Perfect gas is a gas in which the only contribution to the energy comes from the kinetic energy In fact all molecules do interact with one another provided they are close enough together, so the kinetic energy only model is only an approximation. p nmc 3V Intermolecular interaction Intermolecular attraction (long-range interaction) is responsible for the condensation of gases into liquids at low temperatures. Intermolecular repulsion (short-range interaction) is responsible for the fact that liquids and solids have a definite it bulk and ddo not collapse to an infinitesimal it i point. The van der Waals equation of state n p a V V nb nrt At low pressure : The intermolecular forces play no significant role. At moderate pressure : The attractive forces dominate the repulsive forces In this case, the gas can be expected to be more compressible than a perfect gas At high pressure : The repulsive forces dominate and the gas can be expected to be less compressible because now the forces help to drive the molecules apart. At low temperature: Intermolecular forces may also be important. Molecules travel with such low mean speeds that they can be captured by one another The compression factor, Z pvm Z RT V m = the molar volume For a perfect gas, Z = 1 under all conditions, deviation of Z from 1 is a measure of departure from perfect behaviour. At high pressures, all the gases have Z > 1, signifying that they are more difficult to compress than a perfect gas. At intermediate pressures, most gases have Z 1, indicating that the attractive forces are dominant and favour compression. The van der Waals Parameters b roughly the volume of a molecule, (3.5 10-9 1.7 10-8 ) m 3 ~(few Å) 3 a varies a lot [~ (8 10-51 3 10-48 ) J m 3 ] depending on the intermolecular interactions (strongest between polar molecules, weakest for inert gases). Substance a (J. m 3 /mol ) b (x10-5 m 3 /mol) Pc (MPa) Tc (K) Air.1358 3.64 3.77 133 K Carbon Dioxide (CO).3643 4.7 7.39 304. K Nitrogen (N).1361 3.85 3.39 16. K Hydrogen (H).047.65 1.30 33. K When be reduced to can Water (HO).5507 3.04.09 647.3 K Ammonia (NH3).433 3.73 11.8 406 K Helium (He).00341.34 0.3 5. K Freon (CClF) 1.078 9.98 4.1 385 K 3
V Nb V The van der Waals Isotherms 3 NkBT an 3 Nk BT an abn P V Nb V V 0 P P P The Critical Point The critical point is the unique point where both (dp/dv)t = 0 and (d P/dV )T = 0 Critical parameters: 1 a 8 a V 3 Nb P k T C C 7 b B C 7 b unstable 0 N b substance H He N CO H 0 T C (K) 33. 5. 16 304 647 P C (MPa) 1.3 0.3 3.4 7.4.1 Outlines : Thermodynamics Definitions 1 st, nd, 3 rd laws of thermodynamics Helmholtz and Gibbs free energies Partial molar properties, chemical potential and dfugacity Thermodynamics of solutions Equilibrium The branch of physical chemistry know as thermodynamics is concerned with the study of the transformation of energy, and in particular the transformation of energy from heat into work and vice versa. Thermodynamics is concerned with equilibrium i states of matter and has nothing to do with time. Thermodynamics was originally formulated by physicists and engineers interested in the efficiency of steam engines. Definitions The system is the part of the world in which we have a special interest. The surroundings are where we make our observations. Heat Adiabatic system Universe = surrounding + system Open Closed Isolated Types of system An open system is a system that can exchange matter with its surroundings. An closed system is a system that can exchange energy with its surroundings. An isolated system is a system that can exchange neither matter nor energy with its surroundings. Biochemical cell Stoppered flask Water in a vacuum The energy of a closed system can be changed: by transferring energy as work and by transferring energy as heat. 4
The measurement of work Work is a transfer of energy that can be used to change the height of a weight somewhere in the surrounding Heat is a transfer of energy as a result of a temperature difference between the system and the surroundings. Explanation of work and heat in the way of motion of atom???? In chemistry, a very important type of work is expasion work, the work done when a system expands against an opposing pressure. Work done by system = p ex V We were interested in the maximum work (Isothermal) The maximum work (Isothermal) is obtained when the external pressure is only infinitestimally less than the pressure of the gas in the system The expansion work (a) The general expression for work The calculation of expansion work starts from the definition used in Physics. dw = -Fdz Force opposing expansion is F = p ex A dw = -p ex Adz (b) Free expansion Expasion against zero opposing force p ex = 0 w sys 0dV 0 (c) Expansion against constant pressure dw = -p ex dv w sys pexdv p V ex w sys p dv p V ex ex Chongrak_Thermo 8-78 (d) Reversible expansion A reversible change in thermodynamics is a change that can be reversed by an infinitesimal modification of a variable. A reversible change Equilibrium Reversible expansion : p ex = p at each state of expansion w sys Note : In every day life reversible means a process that can be reversed; in thermodynamics it has a stronger meaning-it means that a process can be reversed by an infinitesimal modification in some property (such as pressure). Chongrak_Thermo 9-78 p dv pdv ex (b) Isothermal reversible expansion w sys P dv PdV ext nrt dv dv nrt V V Vf Vf nrt lnv nrt ln Vi V We can summarize by the following remarks : 1. A system does maximum expansion work when the external pressure is equal to that of the system (p ex = p). A system does maximum expansion work when it is in mechanical equilibrium with its surroundings. 3. A system does maximum expansion work when it is changing reversibly. i 5
Mechanical equilibrium The condition of equality of pressure on either side of a movable wall (a piston) is a state of mechanical equilibrium High temp. Low temp. Thermal equilibrium Equal temp. Low temp. High temp. High pressure Low pressure Diathermic wall Equal pressure Phase equilibrium Low High Chemical equilibrium pressure pressure Processes (Changes) Properties of matters Isothermal process : T const Isobaric process : Isochoric process: P const V const Intensive property : P, T, d,... Extensive property : V, m, S,... extensive properties intensive properties extensive properties Isenthalpic process: H const Adiabatic process : T An important classification of thermodynamic properties is whether they are state functions or path functions State functions and exact differential Such properties that are independent of how a sample is prepared are called state functions, such as pressure and temperature and internal energy that define the current state of the system. f U du i The value of U depends on the initial and final states of the system but is independent of the path between them. This pathindependent of the integral is expressed by saying that du is an exact differential. Path functions and inexact differential Properties that relate to the preparation of state are called path functions q i, f path dq When a system is heated, the total energy transferred as heat is the sum of all individual contributions at each point of the path. Work is also path function. This path-dependence is expressed by saying that dq is inexact differential. 6
Internal energy, U A fundamental parameter in thermodynamics is the internal energy. This is the total amount of energy in a system. Internal energy is the sum total of all kinetic and potential energy within the system. Uisa state function, as a specific system has a specific value at any given temperature and pressure. Thermodynamics deal with changes in U, denoted as U. U has negative (-) and positive (+) value. When a system releases 10kJ of energy to the surrounding as work, the internal energy of the system decrease by 10kJ U = -10kJ If we supply 0kJ of energy as heat,, then the internal energy increases by 0kJ U = -0kJ work U > O U > O heat The First Law of thermodynamics The total energy of an isolated thermodynamic system is constant The conservation of energy energy can not be created and destroyed In other words, energy may be lost from a system in only two ways, Either as work and as heat. U = q + w U = P.E + K.E (isothermal expansion process) Real Gas K.E = 0 PE P.E 0 U 0 Perfect Gas K.E = 0 P.E = 0 U = 0 Example: 1). Nutritionists are interested in the use of energy by the human body, and we can consider our own body as a thermodynamic system. Calorimeters have been constructed that can accommodate a person to measure their net energy output. Suppose in the course of an experiment someone does 6 kj of work on an exercise bicycle and loses 8 kj of energy as heat. What is the change in internal energy of the person? U = q + w = (-8) + (-6) = -704 kj We see that the person s internal energy falls by 704 kj. ). An electric battery is charged by supplying 50 kj of energy to it as electrical work, but in the process it loses 5 kj of energy as heat to the surroundings. What is the change in internal energy of the battery? (+5 kj) Chongrak_Thermo 41-78 Heat Capacity Adiabatic bomb calorimeter???? The internal energy of a substance increases when its temperature is raised. U Cv a partial derivative TT V Heat capacities are extensive properties ; J K -1. Molar heat capacity at constant volume is an intensive property ; J K -1 mol -1. Specific heat capacity ( specific heat) ; J K -1 g -1 Chongrak_Thermo 4-78 7
The heat capacity can be used to relate a change in internal energy to a change in temperature of a constant-volume system. du = C v dt If the heat capacity is independent of temperature over the range of temperatures of interest. U = C v T Because a change in internal energy can be identified with the heat supplied at constant volume. q v = C v T This relation provides a simple way of measuring the heat capacity of a sample; a measured quantity of heat is supplied to the sample (electrically, for example), and the resulting increase in temperature is monitored. Chongrak_Thermo 43-78 The measurement of U U = q at constant volume The heat supplied to or absorbed by a system that cannot change its volume The apparatus used in practice is called a bomb calorimeter One procedure is to heat the calorimeter electrically by passing a know current for a measured time through a heater, and record the increase in temperature. The heat supplied by the heater is given by heat supplied to system = Ivt I is the current in amperes (A), V is the potential of the supply in volts (V) T is the time in second for which the current flows. heat supplied to system =(1.3 A) x (1.0 V) x (13 s) = 1.8 kj (1 Avs = 1 J) If we observe that the temperature of the calorimeter rises by 4.47C, then the calorimeter constant, C, the ratio of the heat supplied to the increase in temperature, is C =? If in an experiment on an unknown reaction we measure a temperature rise of 3. C in the same calorimeter, then we could conclude that the heat output of the reaction is heat output = C x increase in temperature = x 3. C = 1.31 kj Similar procedures apply when reaction releases heat or absorbs heat: the temperature of a calorimeter rises if heat is released and it falls if heat is absorbed. Exercise Calculating work, heat and internal energy. Let the initial state be T, V i and the final state be T, V f. The change of state are; Path 1, in which there is free expansion against zero external pressure; Path, in which there is reversible, isothermal expansion. Homework : Compare the quantities of heat required to carry out the isothermal expansion of 1.00 mol of an ideal gas, initially occupying 8.0 L at pressure of 0.880 atm and a temperature of 300 K, to a volume of 40.0 L (a) reversible, and (b) irreversible against a pressure equal to the final pressure of the gas. (b) 7.49 x 10 J Enthalpy The change in internal energy is not equal to the heat supplied when system is free to change its volume. Under these circumstances some of the energy supplied as heat to the system is returned to the surroundings as expansion work, so du is less than dq Then, the enthalpy is defined as H = U + pv H = U + pv H = q p Only work done by the system is pv work State function 8
The enthalpy of a perfect gas is related to its internal energy by using pv = nrt H = U + n g RT n g is the change in the amount of gas molecules in the reaction. H (g) + O (g) H O(l) n g = -3 mol H - U = +n g RT H - U = (-3 mol) RT = -7.5 kj Example : 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 1 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 (373.15K) q = IVt = (0.5 A) x (1 V) x (300 s) = + 1.8 kj H =+1 0443-1 m +1.8 kj/ 0.0443 mol = + 41 kj mol H O (l) H O (g) U U m = + 38 kj mol -1 m H m = RT The molar enthalpy of vaporization of benzene at its boiling point (353.5 K) is 30.8 kj mol -1. What is the molar internal energy change? For how long would the same 1 V source need to supply a 0.50 A current in order to vaporize a 10 g sample? (+7.9 kj mol -1, 6.6 x 10 s) The variation of enthalpy with temperature The enthalpy of a substance increases as its temperature is raised. The slope of a graph of enthalpy against temperature at constant pressure is called the heat capacity at constant pressure, C p H Cp T H = C p T p q p = C p T The variation of heat capacity with temperature can sometimes be ignored if the temperature range in small. This approximation is highly accurate for a monatomic perfect gas (one of the noble gas) A convenient approximate empirical expression of C p is C p,m c a bt T The empirical parameters a, b, and c are independent of temperature. Excercise : What is the change in molar enthalpy of N when it is heated from 5C to 100C? The heat capacity information of N is: a = 8.58 b = 3.77 x 10-3 K -1 c = -0.5 x 10 5 K The Joule-Thomson effect The analysis of the Joule-Thomson coefficient is central to the technological problems associated with the liquefaction of gases. T p H Joule-Thomson coefficient They let a gas expand through a porous barrier from one constant pressure to another, and monitored the difference of temperature that arose from the expansion The change of internal energy of the gas as it move from one side of the throttle to the other is U f U i = w = p i V i -p f V f U f + p f V f = U i + p i V i or H f = H i Isenthalpic process The thermodynamic quantity measured is µ, T p The physical interpretation of µ is that it is the ratio of the change in temperature to the change in pressure when a gas expands under adiabatic conditions Real gases have nonzero Joule Thomson coefficients and, depending on the identity of the gas, the pressure, the relative magnitudes of the attractive and repulsive intermolecular forces, and the temperature, the sign of the coefficient may be either positive and negative The very useful meaning is µ is positive, then T decreases when P decreases, or the gas cools upon expansion. H 9
Thermochemistry The study of the heat produced or required by chemical reactions is called thermochemistry Thermochemistry is the branch of the thermodynamics because a reaction vessel and its contents form a system, and chemical reactions result in the exchange of energy between the system and the surroundings. We can use calorimetry to measure the heat produced or absorbed by reaction, and can identify q with a change in U and H H will be discussed in the detail for Thermochemistry Standard enthalpy changes The standard state of a substance at a specified temperature is its pure form at 1 bar For example; The standard state of liquid ethanol at 98K is pure liquid ethanol at 98K and 1 bar. The standard state of solid iron at 500 K is pure at 500K and 1 bar. Standard enthalpies may be reported for any temperature. However, the conventional temperature for reporting thermodynamic data is 98.15 K (a) Enthalpies of physical change The standard enthalpy change that accompanies a change of physical state is called the standard enthalpy of transition H O (s) H O (g) H sub H O (s) H O (l) H fus H O(l) HO(g) H vap Overall: H O (s) H O (g) H fus + H vap Because enthalpy is a state function, the enthalpy change in a direct route between two states is equal to the sum of the enthalpy changes for a sequence of changes between the same two states: H (indirect route) = H (direct route) Enthalpy is a state property, the enthalpy change in a reverse process is the negative of the enthalpy change in the forward process between the same two states: H (reverse) = -H (forward) The enthalpy of vaporization of water is +44 kj mol -1 at 98 K, its enthalpy of condensation at that temperature is -44 kj mol -1 (b) Enthalpies of chemical change The standard reaction enthalpy, H r, is the change in enthalpy when reactants in their standard states change to products in their standard states CH 4 (g) + O (g) CO (g) + H O(l) H r = -890 kj mol -1 Thermochemical equation For general: A + B 3C + D H r = {3H m (C) + H m (D) } { H m (A) + H m (B) } Molar enthalpy (c) Hess s law The standard enthalpies of individual of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. The importance of Hess s law is that information about a reaction of interest, which may be difficult to determine directly, can be assembled from information on other reactions. C(graphite) + ½O (g) CO(g) Hess s law 1. The reaction must be balanced.. Every substances in the reaction have to be defined a physical state (s, l, g) 3. When the direction of reaction is changed, the sign of the enthalpy hl have to be changed. 4. If any number is multiplied in the reaction, the enthalpy of that reaction have to multiplied by that number. 10
Example: Using the Hess s law H (g) + O (g) H O(l) H 1 = -85.8 kj H (g) + O (g) H O(g) H = -41.8 kj H of H O(g) H O(l) =? Exercise : Calculate the standard enthalpy of formation of CH 4 C(s) + H (g) CH 4 (g) C(s) + O (g) CO (g) H 1 = -393 kj H (g) + O (g) H O(l) H = -85 kj CH 4 (g) + O (g) CO (g) + H O(l) H 3 = -890 kj The Born - Haber cycle is a specific application of the first law of thermodynamics using Hess s law. The cycle allows indirect determination of the lattice enthalpy of an ionic solid. nm m+ (g) + mx n- M n X m (s) Because direct measurement of this process is generally impractical, and indirect path is created. The Born Haber cycle??? And How to calculate the standard of formation of ionic lattice. Standard enthalpies of formation, H f The standard enthalpy of formation of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states. For example, at 98 K The reference state of nitrogen is a gas of N molecules. The reference state of mercury is liquid mercury The reference state of carbon is graphite The standard enthalpy of formation of liquid benzene at 98 K refers to the reaction 6C(s, graphite) + 3H (g) C 6 H 6 (l) H f = +49.0 kj mol -1 Chongrak_Thermo 64-78 Element Arsenic Bromine Carbon Hydrogen Iodine Mercury Nitrogen Oxygen Phosphorus Sulfur Tin Reference state grey arsenic liquid graphite gas solid liquid gas gas white phosphorus rhombic sulfur white tin The reaction enthalpy in terms of enthalpies of formation Stoichiometric coefficients H r = ν H f + ν H f products Reactants The standard enthalpy of combustion, H com The standard d enthalpy of combustion is the standard d change in enthalpy per mole of combustible substance. CH 4 (g) + O (g) CO (g) + H O(l) H com = -890 kj 11
Exercise: Using Hess s law Given the thermochemical equations C 3 H 6 (g) + H (g) C 3 H 8 (g) H = -14 kj C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(l) H = -0 kj Calculate the standard enthalpy of combustion of propene The temperature dependence of reaction enthalpies The standard enthalpies of many important reactions have been measured at different temperatures. The standard reaction enthalpy at different temperatures may be estimated from heat capacities and the reaction enthalpy at some other temperature. H r (T ) Hr (T 1) CP,r dt T T1 Kirchhoff s law C p,r = ν C p,r + ν C p,r products Reactants Example : Using Kirchhoff s law The standard enthalpy of formation of gaseous water at 5 C is -41.8 kj mol -1. Estimate its value at 100 C H r (T ) Hr (T 1) CP,rdT -41.8 kj mol -1 The molar heat capacities at constant pressure : H O(g) = 33.58 J K -1 mol -1 H (g) = 8.84 J K -1 mol -1 O (g) = 9.37 J K -1 mol -1 : Assume that the heat capacities are independent of temperature. H r (373K) = -4.6 kj mol -1 T T1 1