1. Dr. Lauth, University of Applied Sciences, Jlich Campus Physical Chemistry Formulary The Laws of thermodynamics hold William Thomson (Lord Kelvin; 1824 1907) when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind E=E+RT/F ln([Ox]/[Red]) k=Aexp(E/RT) G=-RTln(K) pV=nRT r=k[A]a [B]b pi=xip* dn/dt=-ADdc/dx a=a[A]/([A]+K)

2. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 1 Basic Concepts System A system is a particular segment of the world (with definite boundaries); outside the system are the surroundings (system & surroundings = universe) A system may be open (boundary permeable to matter and heat), closed (boundary permeable to heat) or isolated (boundary impermeable) The condition or state of a system is determined by a number of properties - intensives properties do not change with the quantity of matter present (p, T, , xi, Vm, Um) - extensive properties do change with the quantity of matter (m, n, V, U, S, G) These properties are state functions: they depend only on the current state of the system, not on the way in which the system got to that state. Mathematically: State functions form exact differentials. GIBBS Phase Rule F: "degrees of freedom" ; number of intensive variables (T, p, c, ...) that must be fixed in order for the condition of a system at equilibrium to be completely specified C: number of (independent) components P: number of phases Phase Diagram for a One-Component System A plot representing the phases of a substance as a function of temperature and pressure, showing one phase regions (s, l, g) two phase regions (s/l, l/g, s/g) critical point (above which vapor cannot be liquefied by pressure) triple point (three states of a substance are present) 3. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 2 Thermodynamic Process / Work and Heat Process A thermodynamic process is defined as a system preceeding from an initial state to a final state. Work and Heat are no state functions but process quantities because they describe the transition between states Work W [J] Convention: positive sign of W (W>0) indicates that the systems energy is increasing The work accompanying a change in volume is called pV work = WVol WVol = - p dVex (pex[Pa]: external pressure) Heat Q [J] Convention: positive sign of Q = endothermic process negative sign of Q = exothermic process Heat Capacity C [J/K] Heat capacity is the heat required to raise the temperature of a substance by 1 K Index p: process occuring at constant pressure Index V: process occuring at constant volume Td Q =C; Td Q =C V Vp p Specific and Molar Heat Capacity c [J/(kg K)] , Cm [J/(mol K)] c = C m c = C m p p V V C = C n C = C n p,m p V,m V (m[kg]: mass; n[mol]: moles) Examples (25C / 1 bar): Substance H2 N2 H2O C2H5OH Ag Fe Si SiO2 CDia (g) (g) (l) (l) (s) (s) (s) (s) (s) Cp,m [JK-1 mol-1 ] 28,82 29,12 75,29 111,5 25,35 25,1 28,09 44,4 6,113 Standard State (index ) definitions: - For a gas the standard state is a pressure of 1 bar (p) - For a substance present in a solution, the standard state is a concentration of 1 mol/L (c) - For a liquid or solid substance, the standard state is a molar fraction of 1 (pure substance) nitial state Final state Ti pi Vi Tf pf Vf Work Heat 4. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 3 Ideal Gases Equation of State of an Ideal Gas TRn=Vp (p [Pa]: pressure; V[m] volume; T [K] temperature; n[mol] moles; R = 8,314 J/(mol K) universal gas constant) GAY-LUSSACs Law (CHARLESs Law) The volume of a given sample of gas at constant pressure is directly proportional to the temperature in Kelvins T K273,15 V =V 0 BOYLE-MARIOTTEs Law The volume of a given sample of gas at constant temperature varies inversely with the pressure AVOGADROs Hypotheses Equal volumes of gases at the same temperature and pressure contain the same number of particles Examples (273 K; 101325 Pa = 1 atm): Gas (ideal) N2 H2 O2 He CO2 NH3 C2H6 (g) (g) (g) (g) (g) (g) (g) Vm [10-3 m3 mol-1 ] 22,414 22,40 22,43 22,39 22,43 22,26 22,40 22,17 DALTONs Law of Partial Pressures For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone py=pp=p total i total iii (y [mol/mol]: molar fraction of component i) Composition of dry air (mean molar mass M = 28,96 g/mol): N2 O2 Ar CO2 Vol.-% (Weight %) 78,09 (75,52) 20,95 (23,14) 0,93 (1,29) 0,03 (0,05) const.=Vp const.)=p(T,const.=V= n V m 5. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 4 Kinetic-Molecular Theory I Postulates of the kinetic molecular theory of ideal gases: The volume of the individual particles can be assumed to be negligible The particles are assumed to exert no forces on each other Pressure p [Pa] and Temperature T [K] of a Gas The collisions of the particles with the walls of the container are the cause for the pressure exerted by the gas p = 1 3 N V m v1 2 (N [] : number of molecules; m1 [kg] : mass of a molecule; v [m/s] : mean-square velocity) The average kinetic translational energy of a collection of gas is assumed to be directly proportional to the Kelvin temperature of the gas trans Atransm,B N=UTk 2 3 =trans EE (kB = 1,38110-23 J/K : BOLTZMANN constant; R = 8,314 J/(mol K): gas constant ; NA = 6,022 1023 1/mol : Avogadros number; N []: number of particles; Utrans [J] internal translational kinetic enrergy ; Um [J/mol] : molar translational kinetic energy) MAXWELL-BOLTZMANN Distribution Law d N N d v = 2 M R T v exp - M v 2 R T 2 2 3 2 M R = m k 1 B (N [] : number of particles; M [kg/mol]: molar mass; R = 8,314 J/(mol K): gas constant; v [m/s] : velocity) average velocity v = 8 R T M most probable velocity v = 2 R T M h square root of the mean square velocity v = 3 R T M 2 Examples: Gas N2 He H2 Cl2 CO2 NH3 O2 Hg Ar v [m s-1 ] (298 K) 475 1256 1770 298 379 609 444 177 398 6. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 5 Kinetic-Molecular Theory II Cross Section of Collision [m] = 4 r2 Mean Free Path [m] The average distance a molecule in a given gas sample travels between collisions with other molecules (r [m] : (mean) radius of particle, C = N/V [1/m]: concentration of particles; p[Pa] : pressure; T [K] : temperature; kB = 1,38110-23 J/K : BOLTZMANN constant) Collision Frequency zt [1/s] of Intermolecular Collisions (v [m/s] mean velocity, [m] mean free path) Collisions of Gas Particles with the Container Wall zW [1/(ms)] z = 1 4 C vW (v [m/s] mean velocity, C [1/m] = N/V concentration of particles ) Examples (293 K, 1 bar): Gas N2 H2 CO2 O2 Ar m [10-10 m] 600 1123 397 647 635 zt [109 s-1 ] 5,07 10,0 6,1 4,4 4,0 BOLTZMANN Distribution and Barometric Distribution Law The ratio of the populations in two energy states E1 and E2 (distribution of thermal energy on energy states) N N E E E2 1 exp k TB Applied to gas molecules in earth atmosphere p ph 0 exp M g h R T (M [kg/mol]: (mean) molar mass; h [m]: height; g = 9,81 m/s : gravitational acceleration; R = 8,314 J/(mol K): gas constant) 7. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 6 Real Gases I Compression Factor Z [] Z = p V R T = p V n R T m Virial Equation p V = A + B p + C p + D p + ...m 2 3 2. Virial Coefficient B [m/mol] and Boyle- temperature TB [K] B b RT a a T = R b B Examples : Gas N2 CO2 Ar H2 O2 B[10-6 m3 mol-1 ](273K) -10,5 -149,7 -21,7 13,7 -22,0 B[10-6 m3 mol-1 ](373K) 6,2 -72,2 -4,2 15,6 -3,7 VAN-DER-WAALS Equation of State p + V V - b = R T m 2 m a Covolume b [m/mol] b = 4 N 4 3 rA 3 Simplified VAN-DER-WAALS Equation p V = RT + - R T pm b a (a/Vm [Pa]: internal pressure; Vm [m/mol]: molar volume; b [m/mol]: covolume; R = 8,314 J/(mol K): Gas constant) ,b [m/mol]: covolume; NA = 6,022 1023 1/mol : Avogadros number; r [m] radius of gas particle) Examples: Substance Ar N2 CO2 H2O CH4 C2H6 a [m6 Pa mol-2 ] 0,1363 0,1408 0,364 0,5536 0,23 0,5562 b [10-5 m3 mol-1 ] 3,219 3,913 4,267 3,049 42,9 6,38 repulsion attraction repulsion attraction repulsion attraction repulsion attraction 8. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 7 Real Gases II Critical Constants pk, [Pa], Tk [K], Vm,k [m/mol] b3=V km, 2 27 =pk b a R27 8 =Tk b a )Z=( 8 3= TR V k k km,kp Examples: Substance N2 H2 CO2 H2O CH4 C2H6 C5H12 Tk [K] 126,2 33,3 304,2 647,4 190,7 305,4 425,2 pk [M Pa] 3,394 1,297 7,387 22,119 4,63 4,884 3,8 Vm,k [10-6 m3 mol-1 ] 90,1 65,0 94,0 56,0 98,7 148,0 255 JOULE-THOMSON Coefficient J-T [K/Pa] and Inversion Temperature Ti [K] When a non-ideal gas is forced through a membrane or porous plug, its temperature changes A gas cools on expansion at temperatures below Ti , above Ti a gas warms on expansion = 2 R T - CJ-T p,m a b T = 2 Ri a b Examples: Gas N2 He CO2 N2 H2 J-T [K bar-1 ] (300 K) 0,25 -0,059 1,1 0,25 Ti [K] 621 40 1500 621 202 9. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 8 Transport Properties I - Diffusion and Heat Transfer Diffusion (FICKs first and second Law) The rate of diffusion (the diffusive flux dn/dt) is proportional to the concentration gradient dc/dx The rate of accumulation (or depletion) of concentration is proportional to the local curvature of the concentration gradient dc/dx. dx dc D dt dc xd cd -= A n D for a perfect gas (self-diffusion) (A[m]: area; D [m/s]: diffusion coefficient) Examples: hydrogen ethanol Sugar chloride-ion sodium-ion silver in (T) air(301K) H2O(298 K) H2O (298 K) H2O (298 K) H2O (298 K) copper(900 K) D[m2 /s] 710-5 1,08. 10-9 5,21. 10-10 2,03. 10-9 1,33. 10-9 1,38. 10-15 Randow-walk Equation (EINSTEIN-SMOLUCHOWSKI Equation) x is the mean square distance traversed by a molecule in time t x = 2 D t Heat Transfer by Conduction (FOURIERs Law) The rate of flow of heat dQ/dt is proportional to the temperature gradient dT/dtx xd Td -= A Q W cCv:gasdiatomicafor mV,mW [A [m]: area; w [W/(Km)] : thermal conductivity] Examples (273 K): Substance N2 H2 Xe H2O H2O Cu glas wood soil concrete (g) (g) (g) (l) (s) (s) (s) (s) (s) (s) w [WK-1 m-1 ] 0,024 0,1682 0,0052 0,600 0,592 393 ~0,7 ~0,1 ~0,35 ~1,3 10. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 9 Transport Properties II - Viscosity Viscosity (NEWTONs Law of Viscous Flow) The frictional force F (resisting the relative layers in the fluid) is proportional to the area A and to the velocity gradient dvy/dx (applies only to laminar flow) xd vd -= A Y V F Application: rotating-plate viscosimeter v 2 1 =:gasperfectafor mV (F [Pa]: frictional force; A [m]: area; = F/A [Pa]: shear stress; dvy/dx [1/s]: shear rate) Examples: Substance N2 (g) H2 (g) Ar(g) H2O (l) EtOH (l) Hg (l) H2SO4 (l) 298K 298K 273 K 293K 273K 298K 298K 293 K 298 K V [mPas] 0,018 0,0087 0,021 0,0223 1,8 1,009 1,19 1,55 21,6 HAGEN-POISEUILLE Equation The viscosity V can be calculated from measurements of the rate of flow dV/dt in a tube of known dimensions (l, r), if the pressure difference p is known d V d t = r l p 4 V 8 Application: extrusion viscosimeter STOKESs Law The frictional force Fd opposing the motion of a large spherical particle of radius r moving at speed v through a solvent of viscosity V is given by STOKESs law F = 6 r vR V Application: falling ball viscosimeter 11. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 10 The First Law of Thermodynamics Internal Energy U [J], including The kinetic energy of motion of the individual molecules (rotation, vibration, translation) The potential energy that arises from interactions between molecules The kinetic and potential energy of the electrons within the individual molecules (chemical bonds, etc.) U = + E )kin pot (E First Law of Thermodynamics The energy of the universe is constant Applied to closed systems: change of internal energy = heat absorbed by the system + work done on the system Enthalpy H [J] H = U + p V Processes at Constant Volume Increase of internal energy U of a system at constant volume is equal to the heat QV that is supplied to it V V T T T U C TdC=U=Q 2 1 21 T VTV Processes at Constant Pressure Increase of enthalpy H of a system at constant pressure is equal to the heat Qp that is supplied to it p p T T T CH H TdC==Q 2 1 21 T pTp Phase Transitions (at constant pressure) The enthalpy change that occurs to melt a solid at its melting point = Molar enthalpy of fusion: fH = Qp,sl / n Molar enthalpy of vaporization:vH = Qp,lg / n Examples: Substance Ar H2O C6H6 Ag fH [kJ/mol] (at TE) 1,188 (83,8 K) 6,008 (273,1 K) 10,59 (278,6 K)) 11,3 (1234 K vH [kJ/mol] (at TS) 6,51 (87,3 K) 40,656 (373,1 K) 30,8 (353,2 K) 250,6 (2436 K) 12. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 11 Thermochemistry I - Basics Stoichiometric Equation of a Chemical Reaction A B C DA + B C + D A, B: reactants C, D: products Stoichiometric Coefficients i reactants < 0 products > 0 i i = 0 Extent of Reaction [mol] d = d ni / i ( = 0: reactants only; =1 products only) Enthalpies of Formation Hf,i [J/mol] Change in enthalpy (Heat) that accompanies the formation of 1 mole of a compound from its elements (with all substances in their standard state) Examples (298 K, 1 bar): compound N2 H2O H2O CO2 CO NH3 NO C2H4 C6H6 (g) (l) (g) (g) (g) (g) (g) (g) (l) H0 B [kJ mol-1 ] 0 - 285,84 -241,83 -393,51 -110,52 -46,19 90,37 52,3 49,03 more examples see appendix Enthalpy of Reaction RH [J/mol] Heat associated with a chemical reaction at constant pressure rH can be calculated from the enthalpies of formation of the reactants and products Enthalpy and energy changes are considered with reference to the extent of reaction; stoichiometric equation must be specified i if, o i o H=H r 0 reactantsf, 0 productsf, H-H= Energy of Reaction RU [J/mol] Heat associated with a chemical reaction at constant volume; rU can be calculated from rH r r r r r rH = U + (p V) H = U + R T ngaseous (R = 8,314 J/(mol K) : gas constant Rngaseous: change in moles of gaseous products and reactants) HESSs Law The enthalpy of reaction is not dependent on the reaction pathway RH = 0 d H = 0 13. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 12 Thermochemistry II - Calorimetry Enthalpy of Combustion cH [J/mol] Amount of heat produced for the complete combustion of a substance (at constant pressure) Examples : compound H2 CH4 C2H4 C4H10 CH3OH C8H18 C12H22O11 (g) (g) (g) (g) methanol(l) i-octan(l) saccharose(s) cH [kJ mol-1 ] - 286 -890 -1411 - 2877 -726 -5461 -5645 Measurement of Enthalpy Changes (Calorimetry) Direct measurement of rH or indirect calorimetry (application of Hes law) Indirect calorimetry: determination of enthalpies of formation HB,i from enthalpies of combustion cH icproductscombustionf,, H-H=ifH KIRCHHOFFs Law Temperature dependence of enthalpies of reaction can be calculated from the heat capacities of the reactants and products H T = C H = H + C dT i p,i i T T T T i p,i i 2 1 1 2 R R R Enthalpy of Atomization - Hf,at [J/mol] Heat to be supplied at constant pressure in order do dissociate all the molecules into gaseous atoms atomsfgifiatf HHH ,)(,,, Bond Enthalpy Hbond The bond enthalpie or bond strenth is an average quantity bondiatf HH ,, Examples: Bond O-H C-H C-C C=C CC C-O C-N C=O N-H (ketone) Hbond [kJ mol-1 ] -463 -416 -340 -615 -815 -340 -296 -750 -390 14. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 13 Second Law of Thermodynamics - Entropy Second Law of Thermodynamics The entropy of the universe tends towards a maximum (while the energy of the universe is a constsant) Suniverse 0 In any spontanous irreversible process (occuring without outside intervention) there is always an increase in the entropy of the universe Suniv., irrev. > 0 In reversible processes the entropy of the universe remains constant Suniv., rev. = 0 Entropy S [J/K] measure of randomness or disorder; increase in entropy means an increase in disorder Microscopic Definition of Entropy (BOLTZMANNs Equation) entropy is related to thermodynamic probabiltiy W, the number of microstates corresponding to a given state (including both position and energy) S = k ln(W)B (kB = 1,38110-23 J/K: BOLTZMANN constant; W []: thermodynamic probability) Macroscopic Definition of Entropy Change S (CLAUSIUS) S = Q T d S = Q T rev rev (Qrev [J] heat absorbed or released during a reversible process; T [K] temperture) Third Law of Thermodynamics (NERNSTs Heat Theorem) The entropies of a perfect crystal at 0 K is zero S = 0crystalline, pure, 0 K Absolute Entropies Si [J/(mol K)] The more complex the molecule, the higher the standard entropy value Examples: (1 bar, 298 K) Substance Si N2 N2O4 H2O H2O SiCl4 SiF4 CO2 C6H6 (s) (g) (g) (l) (g) (l) (g) (g) (l) S0 i [J K-1 mol-1 ] 18,82 191,5 304,3 69,94 188,72 239,32 282,14 213,64 172,8 more examples see appendix 15. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 14 Entropy and Energy Conversion Temperature Dependence of Entropy S = C T d TT T p T T 1 2 1 2 (Cp [J/(mol K)] : molar heat capacity) Entropy Changes Associated with Changes of Aggregation State entropy of fusion: f H S = T f .E ; entropy of vaporization: v v S S = T H Examples: Substance Ar N2 O2 Cl2 H2O CH3OH C6H6 CH3COOH fS [J/(K mol)] (at TE) 14,17 11,39 8,17 37,22 22,00 18,03 38,00 40,4 vS [J/(K mol)] (at TS) 74,53 75,22 75,63 85,38 109,0 104,6 87,19 61,9 Concentration Dependence of Entropy (Expansion; Dillution) S = n R ln V V V V 2 1 1 2 (n [mol] :moles; R = 8,314 J/(K mol): gas constant) Entropy Changes in Chemical Reactions RS [J/(K mol)] rS can be calculated from the absolute entropies of the reactants and products Stoichiometric equation must be specified RS = S = S -0 i i 0 i product 0 reactant 0 CARNOT Cycle The efficiency C of the reversible CARNOT engine can be defined as the work done by the system during the cycle Wrev, divided by the heat absorbed at the higher temperature Q1 Q W = 1 rev C T T-T = Q W 1 12 1 rev 0WQQ rev21 16. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 15 Free Enthalpy and Equilibrium Free Enthalpy (GIBBS Energy) G [J] At constant temperature and pressure, systems tend to move toward a state of minimum GIBBS energy G H - TS (= second law of thermodynamics for constant T and p) rG = G/ is a measure for the tendency of a reaction to occur; the more negative the value of rG, the further a reaction will go to the right to reach equilibrium rG can be calculated from standard reaction free enthalpy rG and reaction ratio Qr R R RQG = G + RT ln0 i iR a= ][Reactants [Products] =Q i Reactants Products The standard reaction free enthalpy rG can be calculated from standard reaction enthalpy rH and standard reaction entropy rS (sometimes called GIBBS-HELMHOLTZ equation) R R RG = H - T S Equilibrium of a reaction is established when rG = 0 (minimum GIBBS energy); the reaction ratio has a particular value - the thermodynamic equilibrium constant KGG i i GG GG a= ][Reactants [Products] =K i Reactants Products GG The equilibrium constant K can be calculated from the standard reaction free enthalpy rG R G = - R T ln Ko K = exp - G R T o R Examples: (298 K) reaction N2 + 3 H2 2 NH3 H2O H+ + OH - K 6,8105 bar-2 1,010-14 mol/l reaction CH3COOH CH3COO- + H+ AgCl Ag+ + Cl - K 1,810-5 mol/l 1.610-10 mol/l 17. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 16 Shifts of Equilibrium Exergonic Reations Endergonic Reactions RG < 0 RG > 0 KGG > 1 KGG < 1 Temperature Dependence of Equilibrium Constants ULICHs approximation : rG for temperatures other than 298 K can be calculated from rH- and rS- values for 298 K R R R G (T) = H (298 K) - T S K)o o o (298 Possible Combinations of H and S for a Process RH < 0 RS > 0 exergonic (spontaneous) at all temperatures RH < 0 RS < 0 exergonic at T < RH/RS RH > 0 RS > 0 exergonic at T > RH/RS RH > 0 RS < 0 not exergonic at any temperature VANT HOFFs Equation A plot of ln{K} against 1/T will have a slope equal to H/R ln K = - H R 1 T + S R o o R R (T [K]: temperature; R = 8,314 J/(K mol) : gas constant) Examples: Reaction N2 + 3 H2 2 NH3 N2 + O2 2 NO T 298 K 400 K 500 K 1000 K 3000 K K 6,8105 bar-2 41 bar-2 3,510-2 bar-2 6,810-9 0,017 LE CHATELIER/BRAUN Principle If a change in conditions (a stress) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions process stress shift RH > 0 increase in temperature product yield higher RH < 0 increase in temperature product yield lower RV < 0 increase in pressure product yield higher RV > 0 increase in pressure product yield lower 18. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 17 Phase Equilibria of One-Component Systems Vapor Pressure [Pa] Vapor pressure = the pressure of the vapor over a liquid at equilibrium At equilibrium, the rates of condensation and evaporation are equal (dynamic equilibrium) Examples: Substance H2O(s) (s/l) H2O(l) (vH = 41 kJ/mol ) | Hg(l) (vH = 59 kJ/mol) T [C] -10 0,0 10 25 100 150 200 | -10 25 100 p [mbar] 2,5 6,0 12,3 31,7 1 013 4 760 15 551 | 8,810-5 110-3 0,35 ANTOINEs Equation Empirical equation for vapor pressure change with temperature (p [bar] : vapor pressure in bar; 1 bar = 100 000 Pa; T [C] temperature in C ; A, B, C ANTOINE constants) examples: Substance benzene ethanol toluol water A 4,2144 3,9251 4,0899 4,9513 B 1314,90 912,01 1348,21 1575,61 C 232,40 154,35 219,55 218,62 CLAUSIUS-CLAPEYRON Equation The CLAUSIUS-CLAPEYRON equation represents the vapor pressure change with temperature Enthalpy (heat) of vaporization vH is assumed to be independent of temperature A CLAUSIUS-CLAPEYRON-plot (ln(p) versus 1/T) results in a straight line with a (-vH /R) - slope [ ( )] (T1 / T2 [K] temperature corresopoding to vapor pressure p1 / p2; vH [J/mol]: enthalpy of vaporization; R = 8,314 J/(mol K): gas constant) 19. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 18 Surface Tension Definition of Surface Tension [J/m = N/m] Surface Tension : the work dW necessary to increase the surface area by dA Ad Wd = L F = Examples (20 C) Substance H2O (l) Ethanol (l) glycerol (l) Hg (l) [mN m-1 ] 72,8 22,3 63,0 480 YOUNG Equation and Meniscus Angle LG SL- =)(cos SG ([]: meniscus angle ; SG [N/m] surface tension between solid phase and gas) Examples (20 C) (l) C6H6 H2O H2O Hg Hg (s) glas glas wax steel glas 6 0 105 154 140 LAPLACE Equation The pressure on the inside of a bubble (radius r) must be higher than on the outside r 2=p convexconcave p-p=p ( [N/m] surface tension; p [Pa] Pressure difference; r [m] radius of the surface curvature) Capillarity Spontaneous rising of a wetting liquid in a narrow tube to the height h rg )(cos2 =h (: meniscus angle; [kg/m] density; g = 9,81 m/s : gravitattional acceleration; r [m]: radius of the capillary) 20. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 19 Two-Component Systems - Ideal Solutions Ideal Solutions Mixture of component (A) = solvent and component (B) = solute Ideal solution: Inter molecular attractions of (A) for (A) and of (B) for (B) are the same as the attraction of (A) for (B) { interaction (A B) interactions (AA / BB) } RAOULTs Law Vapor pressure of solvent p is equal to its mole fraction in the solution x multiplied by the vapor pressure of the pure solvent p (ideal solutions). (p A [Pa] : vapor pressure of component A in sulution; p A * [Pa] : vapor pressure of pure component A; x 1 []: mole fraction of component 1) HENRYs Law The molar fraction of a gas dissolved by a given volume of solvent at constant temperature, is propotional to the pressure of the gas in equilibrium with the solution. (pB [Pa]: vapor pressure of component B (gas) in solution; kH [Pa] : Henrys constant; x2 []: molare fraction of component 2) Examples (25 C) Component 2 (gas) N2 O2 CO2 H2 CH4 N2 CO2 H2 CH4 Solvent 1 H2O H2O H2O H2O H2O C6H6 C6H6 C6H6 C6H6 kH [GPa] 8,68 4,4 0,167 7,12 0,0419 0,239 0,0114 0,367 0,0569 Vapor Pressure and Boiling Point of two Liquids obeying Raoults Law The total vapor pressure in a binary solution will be intermediate between the vapor pressures of the two pure components (no maximum or minimum in p-x-diagram) The boiling points of all the mixtures are inter- mediate between the boiling points of the pure components 21. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 20 Colligative Properties A colligative propery is a property of dilute solutions that depends on only the number of solute molecules and not on the type of species present Boiling Point Elevation Boiling point elevation is a colligative propery The molality b is the number of moles of solute per kilogram of solvent in a solution (bB [mol/kg] : molality ; i[]: VANT HOFF factor; Keb [K kg/mol]: ebullioscopic constant) Freezing Point Depression Freezing point depression is a colligative property (RAOULTs second law) (bB [mol/kg] = nB/mA: molality ; i []: VANT HOFF factor; kkr [K kg/mol]: cryoscopic constant) Examples : Solvent H2O C6H6 cyclohexane camphen acetic acid CS2 CCl4 keb [K kg/mol] 0,514 2,64 2,75 6,09 3,07 2,37 4,95 Kkr [K kg/mol] 1,86 5,07 20,2 40,0 3,90 3,8 30 Osmotic Pressure [Pa] (VANT HOFF Equation) Osmosis: the flow of solvent into a solution through a semipermeable membrane (permeable for solvent molecules only) Osmotic pressure : the pressure that must be applied to stop osmosis Molarity c: moles of solute / volume of solution (i[] : vant Hoff factor; cB [mol/m]: molarity; R = 8,314 J/(mol K) : gas constant) Examples: Solution 0,4 mol sugar in 1 kg water/ 20 human blood at 37 C [bar] 10,3 bar 7,7 bar Dissociation (VANT HOFF Factor i) vant Hoff Factor i: the number of particles in solution formed from one molecule ( ) (+,- []: number of cations/anions formed from one molecule; z+,- []: charge of cation / anion; []: degree of dissociation.) 22. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 21 Binary Phase Diagrams / Boiling and Melting Negative Deviations from RAOULTs Law (example: acetone-chloroform) Interaction (A B) > interactions (AA / BB) Solution more stable than components Negative sign of mH and mV In case of large deviations: Maximum in boiling point curve (azeotrope) Positive Deviations from RAOULTs law (example: ethylene chloride - ethanol) Interaction (AB) < interactions (AA / BB) Solution less stable than components Positive sign of mH and mV In case of large deviations: Minimum in boiling point curve (azeotrope) Azeotrope (minimum or maximum in boiling point curve) Liquid and vapor are of the same composition; separation by direct distillation impossible Examples: boiling point of (1) boiling point of (2) boiling point of azeotrope TS(H2O): 100 C TS(HNO3) 87C TS (69 % HNO3): 120 C TS(H2O): 100 C TS(EtOH): 78,4C TS (96 % ethanol): 78,2 C Eutectic (minimum in liquidus curve) Composition which has the lowest melting point Mixture of eutectic composition (heterogeneous !) will melt sharply at the eutectic temperature to form a homogeneous liquid phase of the same composition Invariant point: A(s) + B(s) A/B (l) (Other invariant points: Peritectic / eutectoid / peritectoid) Examples: melting point of (1) melting point of (2) melting point of eutectic TE (Si): 1685 C TE(Al): 930 C TE(xSi = 0,11): 851C TE (Sn): 232 C TE(Pb): 327,5 C TE(33% Pb): 183C TE (H2O): 0 C TE(NaCl): 801 C TE(23% NaCl): -21C 23. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 22 Ternary Phase Diagrams / Triangle Diagram General Remarks on Phase Diagrams Phase Diagrams are are descriptions of the state of a system on a graph of temperature versus composition Binodal Curve: Boundary between homogeneous and heterogeneous regions in a phase diagram (Examples: boiling , liquidus curve, solidus curve) Conodal Curve (tie line in 2-phase regions): connection line between equibilibrium phase Using the lever rule one can determine quantitatively the relative composition of a mixture in a two-phase region ln=ln GIBBSs Triangle Diagram In ternary phase diagrams the composition of a phase containing up to three components is represented on triangular axes. Within the two-phase regions only specific compositions can be in equilibrium; the lines joining these compositions are known as tie lines . Triangle Diagram and NERNST Distribution Law If two liquids A (raffinate R) and B (extractant E) are partially immiscible and if there is a third component C present in both phases (which behaves individually as an ideal solute), the ratio of its concentrations is constant K= c c NernstR GGC, E GGC, 24. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 23 Adsorption Adsorption: the collection of one substance on the surface of another Surface Concentration a [mol/g] and Fraction of Surface covered [ ] adsorbent speciesadsorbed m n =a layermomoa a LANGMUIR Adsorption Isotherm Formation of a unimolecular layer at saturation [ ] [ ] (K [Pa or M]: Langmuir constant; [A] [Pa] Pressure or [M] : molarity) Surface areas of the sorbent can be calculated from a (spec. area = a . NA . Aadsorbed molecule) In order to test the Langmuir isotherm it is best to use a reciprocal plot (1/a versus 1/c or 1/p) BET Adsorption Isotherm Extension of the Langmuir treatment (by BRUNAUER, EMMETT AND TELLER) to allow for the physisorption of additional layers of adsorbed molecules monad, * monad,ad * * VK p/p1)-(K + VK 1 = V)p/p-1( p/p (K []: BET-constante; p* :saturation vapor pressure; Vmon: volumen that can be adsorbed as a monolayer) FREUNDLICH Adsorption Isotherm [ ] In order to test the Freundlich isotherm it is best to use a logarithmic plot (ln(a) versus ln(c)) )ln()ln(ln(a) c 25. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 24 Chemical Kinetics Rate of Consumption and Formation ri [mol/(s m)] The change in concentration of a reactant or product per unit time [ ] Rate of Reaction r [mol/(s m)] Rate Law, Rate Constant and Half Life Rate law: an expression, which shows how the rate depends on the concentrations of reactants r = k f(c ,c ,c ...)A B C often ( ) [ ] [ ] [ ] The proportionality constant k is called the rate constant; the exponents a,b,c.. are called the order of the reactant The form of the rate law (the values for k, a, b, c ...) must be determined by experiment The integrated rate law relates concentration to reaction time: cA = f(t) The time required for a reactant to reach half of its original concentration is called the half life of a reaction and is designated by the symbol t 0st order reaction 26. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 25 Simple-Order Reactions First-Order Reaction (A P) Doubling the concentration of A doubles the reaction rate Half-life does not depend on concentration A plot of ln(cA) versus t is a straight line A1 A ck= td cd -=r tk-expc=c 1A,0A tk-cln=cln 1A,0A t1 2 1 = ln 2 k Second-Order Reaction (A P) Doubling the concentration of A quadruples the reaction rate Half-life is dependent of cA,0 A plot of 1/cA versus t is linear 2 A2 A ck= td cd -=r c k t A = c 1+ c A,0 A,0 2 1 c = 1 c + k t A A,0 2 t1 2 2 A,0 = 1 k c Examples: Reaction 2 N2O5 2NO2 + O2 Cyclopropane Propene C2H6 2 CH3 2 NO2 2 NO + O2 H+ + OH- H2O 25 C (gas phase) 500 C (gas phase) 700 C (gas phase) 300C (gas phase) 25 C (in water) k 3,1410-5 1/s 6,7110-4 1/s 5,46 10-4 1/s 0,54 dm/(mol s) 1,5. 1011 dm/(mol s) Kinetics of simple order reactions Reaction Rate Law Integrated Rate Law Half Life A P r = k0 k0 t = cA,0 cA t = cA,0/(2k0) A P r = k1 cA k1 t = ln(cA,0) - ln(cA) t = ln(2)/k1 A P r = k2 cA k2 t = 1/cA 1/cA,0 t = 1/(cA,0k2) A + B P (cA,0 = cB,0) r = k2 cA cB k2 t = 1/cA 1/cA,0 t = 1/(cA,0k2) A + B P (cA,0 cB,0) r = k2 cA cB k2t = 1/(cA,0 cB,0)ln(cB,0cA/cA,0cB) 27. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 26 More Complicated Mechanisms Equilibrium Reaction ( A P) A reversible chemical reaction will be at dynamic equilibrium when the rate of forward reaction is equal to the rate of the reverse reaction. GGA GGp c c , , k k =K AA EEH R Consecutive Reaction (A I P) The steady-state approximation (SSA) assumes that, after an initial induction period the rates of change of concentrations of all unstable intermediates are negligibly small, i.e.: unstable intermediate The rate-determining step approximation states that in a series of consecutive elementary rections the rate of production of the final products depends only the rate coefficient for the slowest step in the sequence stable intermediate Competitive Reaction ( P A P) Reaction 1 (-----) has a more stable transition state and therefore a lower activation barrier. So diamond is the kinetic product. Reaction 2 (_____ ) generates the more stable product since graphite it is at lower energy than diamnond. So graphite is the thermodynamic product. Kinetic control is favoured with mild and low temperature conditions. Thermodynamic reaction control takes place with vigorous reaction conditions or when the reaction is allowed to continue over a long time 28. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 27 Influence of Temperature and Catalyst on Reaction Rates ARRHENIUS Equation The rate constant k shows an exponential increase with temperature k = A exp E R T A (A: frequency factor; EA [J/mol] activation energy; R = 8,314 J/(mol K) gas constant) A plot of ln(k) against (1/T) gives a straight line (ARRHENIUS plot); the value of the activation energy can be obtained from the slope of the line, which equals -EA/R; the intercept can be used to determine A. ln( )k = ln(A) - E R 1 T A Activation energy EA : for a reaction , mole- cules must come together with at least EA in order to surmount the energy barrier (threshold energy) Frequency factor A (preexponential factor): rate constant at very high temperture Examples: Reaction 2 N2O5 2NO2 + O2 cyclopropane propene sucrose fructose + glucose 1. order, gas phase 1. order, gas phase 2. order; aqueous solution EA 88 kJ/mol 272 kJ/mol 107,9 kJ/mol A 6,311014 1/s 1,581015 1/s 1,5 1015 dm/(mol s) Collision Theory and Activated Complex Molecules must collide to react (reaction rate = number of successfull collisons) The relative orientation of the reactants must allow formation of any new bonds necessary to produce products (steric factor = fraction of collisions with effective orientation) The collision must involve enogh energy to form the activated complex (transition state) Catalysis A catalyst is a substance that increases the rate of reaction without being consumed itself A catalyst does not modify the overall thermodynamics (H, G, S) of the reaction A catalyst provides a new pathway with lower activation energy for the reaction Reaction 2 HI H2 + I2 catalyst (none) Au(s) Pt(s) EA [kJ/mol] 184 105 59 29. Physical chemistry formulary Dr. Lauth, University of Applied Sciences, Jlich Campus 28 Electrochemistry - Solutions of Electrolytes Dissociation of Electrolytes Strong electrolytes occur almost entirely as ions in solution; degree of dissociation 1 Weak electrolytes are present only partially as ions;