Physical Chemistry 1, Cheat Sheet of Physical Chemistry

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Physical Chemistry I Andrew Rosen August 19, 2013 Contents 1 Thermodynamics 5 1.1 Thermodynamic Systems and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1.1 Systems vs. Surroundings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1.2 Types of Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1.3 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1.4 Thermodynamic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 The Mole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4 Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4.1 Boyle’s and Charles’ Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4.2 Ideal Gas Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2 The First Law of Thermodynamics 7 2.1 Classical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 P-V Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.3 Heat and The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.4 Enthalpy and Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.5 The Joule and Joule-Thomson Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.6 The Perfect Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.7 How to Find Pressure-Volume Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.7.1 Non-Ideal Gas (Van der Waals Gas) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.7.2 Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.7.3 Reversible Adiabatic Process in a Perfect Gas . . . . . . . . . . . . . . . . . . . . . . . 11 2.8 Summary of Calculating First Law Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.8.1 Constant Pressure (Isobaric) Heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.8.2 Constant Volume (Isochoric) Heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.8.3 Reversible Isothermal Process in a Perfect Gas . . . . . . . . . . . . . . . . . . . . . . 12 2.8.4 Reversible Adiabatic Process in a Perfect Gas . . . . . . . . . . . . . . . . . . . . . . . 12 2.8.5 Adiabatic Expansion of a Perfect Gas into a Vacuum . . . . . . . . . . . . . . . . . . . 12 2.8.6 Reversible Phase Change at Constant T and P . . . . . . . . . . . . . . . . . . . . . . 12 2.9 Molecular Modes of Energy Storage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.9.1 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1 2.9.2 Classical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.9.3 Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.9.4 Classical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.9.5 Determining Number of Atoms in a Molecule Given Cv,m,class. . . . . . . . . . . . . . 14 3 Heat Engines 14 3.1 The Carnot Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2 Carnot Refrigerators, Freezers, Air Conditioners, and Heat Pumps . . . . . . . . . . . . . . . 16 3.3 The Otto Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.4 Historical Perspective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4 The Second Law of Thermodynamics 17 4.1 Definition of the Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.3 Calculation of Entropy Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.3.1 Cyclic Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.3.2 Reversible Adiabatic Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.3.3 Reversible Isothermal Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4.3.4 Reversible Phase Change at Constant T and P . . . . . . . . . . . . . . . . . . . . . . 17 4.3.5 Cautionary Note on Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.3.6 Constant Pressure Heating with No Phase Change . . . . . . . . . . . . . . . . . . . . 18 4.3.7 Change of State of a Perfect Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.3.8 General Change of State Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.3.9 Irreversible Phase Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.3.10 Mixing of Different Inert Perfect Gases at Constant P and T . . . . . . . . . . . . . . 19 4.3.11 Joule Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 4.4 Entropy, Reversibility, and Irreversibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 5 Material Equilibrium 20 5.1 Entropy and Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 5.2 The Gibbs and Helmholtz Energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 5.2.1 Derivation of A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 5.2.2 Derivation of G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.2.3 Connection with Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.3 Thermodynamic Relations for a System in Equilibrium . . . . . . . . . . . . . . . . . . . . . . 21 5.3.1 Basic Thermodynamic Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5.3.2 The Gibbs Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.3.3 The Maxwell Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5.3.4 Dependence of State Functions on T, P, and V . . . . . . . . . . . . . . . . . . . . . . 23 5.3.5 Remaining Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.4 Calculation of Changes in State Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 5.4.1 Calculation of ∆S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 5.4.2 Calculation of ∆H and ∆U . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 5.4.3 Calculation of ∆G and ∆A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 5.5 Chemical Potentials and Material Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 5.6 Reaction Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 2 1 Thermodynamics 1.1 Thermodynamic Systems and Properties 1.1.1 Systems vs. Surroundings • System: The macroscopic part of the universe under study in thermodynamics • Surroundings: The parts of the universe that can interact with the system • Open System: One where transfer of matter between system and surroundings can occur • Closed System: One where transfer of matter between system and surroundings cannot occur • Isolated System: One in which no interaction in any way with the system can occur 1.1.2 Types of Walls 1. Rigid or nonrigid 2. Permeable or impermeable 3. Adiabatic (does not conduct heat) or nonadiabatic/thermally conductive (conducts heat) 1.1.3 Equilibrium • An isolated system is in equilibrium when its macroscopic properties remain constant with time • A nonisolated system is in equilibrium when the the system’s macroscopic properties remain constant with time but also need to have no change with removal of the system from contact with its surroundings ◦ If removal of the system does change the macroscopic properties, it is in a steady state • Mechanical Equilibrium: No unbalanced forces act on or within the system • Material Equilibrium: No net chemical reactions are occurring in the system nor is there any net transfer of matter from one part of the system to another • Thermal Equilibrium: No change in the properties of the system or surroundings when they are separated by a a thermally conductive wall • Thermodynamic Equilibrium: Must be in mechanical, material, and thermal equilibrium 1.1.4 Thermodynamic Properties • The definition of pressure, P , is as follows and relates to force, F , and area, A: P ≡ F A • Pressure is uniform and equal to the surroundings in mechanical equilibrium • An extensive property is equal to the sum of its values for the parts of the system (eg: mass) • An intensive property does not depend on the size of the system (eg: density) • If the intensive macroscopic properties are constant in the system, it is homogeneous; otherwise, it is heterogeneous 5 ◦ A homogeneous part of a system is a phase • The definition of density, ρ, is as follows and relates to mass, m, and volume, V : ρ ≡ m V • State Functions: Values that are functions of the system’s state that doesn’t depends on the system’s past history 1.2 Temperature • Temperature is common for systems in thermal equilibrium and is symbolized by θ • Zeroth Law of Thermodynamics: Two systems that are found to be in thermal equilibrium with a third system will be found to be in thermal equilibrium with each other • A reference system (thermometer), r, is used to create a temperature scale 1.3 The Mole • The molar mass, Mi, of a substance is as follows and relates to mi, the mass of a substance i in a sample and where ni is the number of moles of i in the sample: Mi ≡ mi ni • The number of molecules in species i, Ni, is as follows and relates to Avogadro’s Constant, NA, and ni: Ni = niNA • The mole fraction, xi, is defined as follows and relates to ni and the total moles, nt: xi ≡ ni ntot 1.4 Ideal Gases 1.4.1 Boyle’s and Charles’ Laws • Boyle’s Law is as follows when θ and m are constant: PV = k • Gasses are ideal in the zero-density limit • The SI units for pressure can be expressed as either of the following 1Pa ≡ 1 N m2 • Some alternate rearrangements of the pressure equation are as follows: P = F A = mg A = ρV g A = ρgh • Charles’ Law is as follows when P and m are constant: V = a1 + a2θ ◦ Alternatively, it can be written as V T = k • The absolute ideal-gas temperature is represented as T 6 1.4.2 Ideal Gas Equation • The ideal-gas law is PV = nRT • This can be rearranged to make the following two equations where M is molar mass: PV = mRT M → P = ρRT M • The partial pressure, Pi, of a gas i in a gas mixture is defined as: Pi ≡ xiP • For an ideal gas mixture, Pi = niRT V 1.5 Equations of State • The van der Waals equation accounts for intermolecular forces, with a and b as constants defined for each gas: ( P + an2 V 2 ) (V − nb) = nRT • An approximate equation of state for most liquids and solids is1: Vm = c1 + c2T + c3T 2 − c4P − c5PT • The thermal expansivity, α, is defined as2 α(T, P ) ≡ 1 Vm ( ∂Vm ∂T ) P • The isothermal compressibility, κ, is defined as κ(T, P ) ≡ − 1 Vm ( ∂Vm ∂P ) T • These can be combined as, ( ∂P ∂T ) Vm = α κ 2 The First Law of Thermodynamics 2.1 Classical Mechanics • Work is simply a force acting over a distance, which can be mathematically expressed as follows if considering the displacement in the x direction: dw ≡ Fx dx 1All variables that are not listed can be found in Ira N. Levine’s Physical Chemistry, 6th edition 2Calculus Note: The total differential, dz, for z(x, y) is defined as dz = ( ∂z ∂x ) y dx + ( ∂z ∂y ) x dy 7 • The goal of the Joule-Thomson Experiment was to find ( ∂H ∂P ) T by measuring ( ∂T ∂P ) H • The Joule coefficient is defined as µJ ≡ ( ∂T ∂V ) U • The Joule-Thomson coefficient is defined as µJT ≡ ( ∂T ∂P ) H ◦ Resultantly, ( ∂H ∂P ) T = −CPµJT ◦ Also, ( ∂U ∂V ) T = −CV µJ • ∆T = 0 for a perfect gas in the Joule Experiment because µJ = 0 for a perfect gas 2.6 The Perfect Gas • A perfect gas is defined as one that follows PV = nRT and ( ∂U ∂V ) T = 0 ◦ This is to assure that U is dependent only on temperature. Since H is also dependent only on temperature for a perfect gas, ( ∂H ∂P ) T = 0 = ( ∂U ∂V ) T • For a perfect gas, CP − CV = nR (perf. gas) • Additionally3, CP,m − CV,m = R (perf. gas) • For a perfect gas, dU = CV dT (perf. gas) • For a perfect gas, dH = CP dT (perf. gas) 2.7 How to Find Pressure-Volume Work 2.7.1 Non-Ideal Gas (Van der Waals Gas) • Rearrange the van der Waals equation to solve for P = nRT V − nb − a n2 V 2 and substitute this into w = ´ V2 V1 P dV 2.7.2 Ideal Gas • Rearrange the ideal-gas equation to solve for P = nRT V and substitute into the work equation to get w = −nR ˆ V2 V1 T V dV (perf. gas) ◦ If it’s isothermal, temperature is constant, so w = −nRT ln ( V2 V1 ) (isothermal) 3C ≡ nCm 10 2.7.3 Reversible Adiabatic Process in a Perfect Gas • Assuming that CV,m does not change much with temperature, T2 T1 = ( V1 V2 )R/CV,m (adiabatic) • Alternatively, P1V γ 1 = P2V γ 2 (adiabatic) • In the above equation, γ is the heat-capacity ratio and is defined γ ≡ CP CV • If you’re still not happy, you can use( P1 P2 )R = ( T1 T2 )CP,m (adiabatic) • Since dU = CV dT , it is safe to use ∆U = w = CV ∆T 2.8 Summary of Calculating First Law Quantities • Always start with writing these three equations down4: 1. w = − ´ V2 V1 p dV 2. ∆U = q + w 3. ∆H = ∆U + ∆ (PV ) • If it’s a perfect gas, write these three down as well: 1. dU = CV dT 2. dH = CP dT 3. CP − CV = nR • If the pressure is equal to zero, w = 0 • If the volume change is equal to zero, w = 0 2.8.1 Constant Pressure (Isobaric) Heating 1. P is constant, so w = −P∆V 2. ∆H = qP = ´ T2 T1 CP dT 2.8.2 Constant Volume (Isochoric) Heating 1. w = 0 2. ∆U = ´ T2 T1 CV dT = qV 3. ∆H = ∆U + V∆P (a) Alternatively, ∆H = qP = ´ T2 T1 CP dT 4Whenever you compute work, make sure the units work out. For instance, at constant pressure and using w = −P∆V , one might obtain units of L · atm. However, this is not a Joule, so a conversion factor needs to be set up. 11 2.8.3 Reversible Isothermal Process in a Perfect Gas 1. ∆U = ∆H = 0 2. Rearrange the ideal-gas equation to solve for P = nRT V and substitute into the work equation to get w = −nRT ln ( V2 V1 ) = nRT ln ( P2 P1 ) 3. q = −w 2.8.4 Reversible Adiabatic Process in a Perfect Gas 1. q = 0 and ∆U = w 2. ∆U = ´ T2 T1 CV dT 3. ∆H = ´ T2 T1 CP dT 4. The final state of the gas can be found by P1V γ 1 = P2V γ 2 2.8.5 Adiabatic Expansion of a Perfect Gas into a Vacuum 1. q = w = ∆U = ∆H = 0 2.8.6 Reversible Phase Change at Constant T and P 1. q is the measured latent heat of the phase change 2. w = −P∆V (a) ∆V can be calculated from the densities of the two phases (b) If one phase is a gas, PV = nRT can be used 3. ∆H = qp 4. ∆U = q + w 2.9 Molecular Modes of Energy Storage 2.9.1 Degrees of Freedom • Every free particle has three degrees of freedom manifested in each dimension of space • Bound particles have some changes with respect to degrees of freedom 2.9.2 Classical Mechanics 1. Translation (a) E = 1 2 mv2 in each dimension 2. Rotation (a) E = 1 2 Iω2 12 • Alternatively, e = 1− TC TH • For a cycle, ∆U = 0, so −w = qH + qC ◦ Therefore, efficiency is also defined as: e = −w qH = 1 + qC qH ◦ Because qC is negative and qH is positive, efficiency is always less than 1 • Combining the last two definitions of efficiency, we get −TC TH = qC qH • All reversible heat engines have the same efficiencies if the temperatures are the same • If two Carnot cycle heat engines operating reversible between the same two temperature can have different efficiencies, then they can be linked together in such a way as to transfer heat from a cold object to a hot object without any work being done from the outside to make the flow occur. This is goes against Clausius’ version of the Second Law • A Carnot cycle can be graphed as follows: Step 1 to 2: • This is an isothermal process, so ∆U = 0, q = nRTH ln ( V2 V1 ) , and w = −nRTH ln ( V2 V1 ) Step 2 to 3: • This is an adiabatic process, so ∆U = CV (TC − TH), q = 0, and w = CV (TC − TH) ◦ Note: Since TC is the second temperature state, ∆T is equal to TC − TH Step 3 to 4: • This is an isothermal process, so ∆U = 0, q = nRTC ln ( V4 V3 ) , and w = −nRTC ln ( V4 V3 ) ◦ Alternatively, q = −nRTC ln ( V2 V1 ) and w = nRTC ln ( V2 V1 ) 15 Step 4 to 1: • This is an adiabatic process, so ∆U = CV (TH − TC), q = 0, and w = CV (TH − TC) Overall Cycle: • This is a cycle, so ∆U = 0, q = nR (TH − TC) ln ( V2 V1 ) and w = −nR (TH − TC) ln ( V2 V1 ) • For a closed system undergoing a Carnot cycle,˛ dqrev T = qC TC + qH TH = 0 3.2 Carnot Refrigerators, Freezers, Air Conditioners, and Heat Pumps The following equation holds true, where U is conductance and A is area, dq dt = UA∆T ηref,AC ≡ Coeff. of Performance = dq/dt dw/dt = qc w = qc − (qH + qc) = Tc TH − Tc ηHeat Pump = −qH w = TH TH − TC 3.3 The Otto Engine 1. Step 1→ 2 is adiabatic compression 2. Step 2→ 3 is isochoric heating 3. Step 3→ 4 is adiabatic expansion 4. Step 4→ 1 is isochoric cooling • There are no isotherms For an Otto engine, e = −w Qin = 1− T1 T3 = 1− ( V2 V1 )R/CV,m Compression Ratio ≡ cr ≡ V1 V2 Efficiency is maximized with an infinite compression ratio, but of course there are practical volume limits 3.4 Historical Perspective • Knocking or pinging is detonation of a fuel charge that occurs too early • Straight-chain hydrocarbons knock at low compression ratios and vice versa • To prevent knocking, antiknock components are added ◦ From the 1930s to 1970s, Pb(Et)4 was used, but it caused mental disorders and death due to the lead ◦ From the 1970s to the 1990s, MTBE was used, but it was carcinogenic ◦ Currently, EtOH is used, which was ironically used before Pb(Et)4 16 4 The Second Law of Thermodynamics 4.1 Definition of the Second Law of Thermodynamics • According to the Kelvin-Planck statement of the second law, it is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings • According to the Clausius statement, it is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a cold reservoir and the flow of an equal amount of heat out of the system into a hot reservoir 4.2 Entropy • The definition of entropy, S, is the following for a closed system going through a reversible process ds ≡ dqrev T • According to the fundamental theorem of calculus, ∆S = S2 − S1 = ˆ q2 q1 dqrev T • The molar entropy of a substance is Sm = S n 4.3 Calculation of Entropy Changes 4.3.1 Cyclic Process • ∆S = 0 since it is a state function 4.3.2 Reversible Adiabatic Process • Since dqrev = 0, ∆S = 0 4.3.3 Reversible Isothermal Process ∆S = qrev T (isothermal) 4.3.4 Reversible Phase Change at Constant T and P • At constant temperature, ∆S = qrev T • qrev is the latent heat of the transition in this case • Since P is constant, qrev = qP = ∆H. Therefore, ∆S = ∆H T (rev. phase change at const. T and P ) 17 5 Material Equilibrium 5.1 Entropy and Equilibrium • Material equilibrium means that in each phase of a closed system the number of moles of each substance present remains constant • Thermodynamic equilibrium in an isolated system is reached when the system’s entropy is maximized • The condition for material equilibrium in a system is the maximization of the total entropy of the system plus its surroundings • For the surroundings, dSsurr = dqsurr T . Also, dqsurr = −dqsyst ◦ For the system, dSsyst > dqsyst T • At material equilibrium, dS = dqrev T • For a material change in a closed system in mechanical and thermal equilibrium, dS ≥ dq T , where the equality only holds when the system is in material equilibrium • For an irreversible chemical reaction in thermal and mechanical equilibrium, dS > dqirrev/T • For a closed system of material change in mechanical and thermal equilibrium, dU ≤ T dS+dw, where the equality only holds at material equilibrium 5.2 The Gibbs and Helmholtz Energies 5.2.1 Derivation of A 1. If we rearrange dU , we can get dU ≤ T dS+S dT −SdT +dw, where the first summed terms are d(TS) 2. d (U − TS) ≤ −S dT + dw (a) This is a new state function, where A is Helmholtz Free Energy A ≡ U − TS 3. Substituting yields, dA ≤ −S dT + dw 4. If only PV work is done at constant temperature and volume, dA = 0 (eq., const. T ,V ) A is a kind of potential where the system is in equilibrium when A is a minimum For a reversible process at constant temperature, dA = dw. For an irreversible process, ∆A < wirrev 20 5.2.2 Derivation of G 1. If we consider material equilibrium for constant T and P , we can substitute dw = −P dV (a) dU ≤ T dS + SdT − S dT − P dV + V dP − V dP 2. Grouping gets dU ≤ d(TS)− S dT − d(PV ) + V dP 3. Algebra yields d(U + PV − TS) ≤ −S dT + V dP (a) This is a new state function, where G is Gibbs’ Free Energy G ≡ PV − TS = H − TS 4. Substituting yields, dG ≤ −S dT + V dP 5. If temperature and pressure are constant, dG = 0 G is a kind of potential where the system is in equilibrium when G is a minimum For a reversible process at constant temperature and pressure, dG = dWnon−PV . For an irreversible process, ∆G < wnon−PV 5.2.3 Connection with Entropy In a closed system capable of doing only PV work, the constant T and V material equilibrium condition is the minimization of A, and the constant T and P material equilibrium condition is the minimization of G Since −∆Hsys T = ∆Ssurr at constant T and P , ∆Suniv = −∆Gsyst T Due to the Second Law, entropy of the universe must increase for an irreversible process, so ∆Gsyst must decrease 5.3 Thermodynamic Relations for a System in Equilibrium 5.3.1 Basic Thermodynamic Quantities The basic thermodynamic relationships are: H ≡ U + PV A ≡ U − TS G = H − TS CV = ( ∂U ∂T ) V CP = ( ∂H ∂T ) P 21 Additionally, for a closed system in equilibrium, CV = T ( ∂S ∂T ) V (Closed, Eq.) CP = T ( ∂S ∂T ) P (Closed, Eq.) 5.3.2 The Gibbs Equations dU = T dS − P dV dH = T dS + V dP dA = −P dV − S dT dG = V dP − S dT ( ∂U ∂S ) V = T and ( ∂U ∂V ) S = −P( ∂G ∂T ) P = −S and ( ∂G ∂P ) T = V Furthermore, α (T, P ) ≡ 1 V ( ∂V ∂T ) P and κ (T, P ) ≡ − 1 V ( ∂V ∂P ) T Procedure: 1. Write out the corresponding Gibbs Equation 2. Set the designated variable as constant 3. Solving for the desired relation 5.3.3 The Maxwell Relations The Maxwell Relations can be derived by applying the basic Euler’s Reciprocity to the derivative forms of the equations of state. The Euler Reciprocity is6, d2z dx dy = d2z dx dy For instance, ∂2G ∂T ∂P = ( ∂ ∂T ( ∂G ∂P ) T ) P = ( ∂ ∂T V ) P = ( ∂V ∂T ) P This must equal ∂2G ∂P ∂T = − ( ∂S ∂P ) T via the Euler Reciprocity 6It is important to note that the operator in the denominator of the derivative is performed right to left 22 5.4.3 Calculation of ∆G and ∆A To calculate ∆G directly, we can use the following formula for isothermal conditions ∆G = ∆H − T∆S (Const.T ) Alternatively, here are three ways to calculate ∆G: ∆G = ˆ P2 P1 V dP = ˆ P2 P1 m ρ dP = ˆ P2 P1 nVm dP (Const. T, V ) At constant T and P , ∆G = 0 (rev. proc. at const. Tand P ) To calculate ∆A directly, we can use the following formula for isothermal conditions ∆A = ∆U − T∆S (Const. T ) Alternatively, ∆A = − ˆ V2 V1 P dV (Const. T ,P ) If the phase change goes from solid to liquid, ∆A can be calculated by using densities to find ∆V . If the phase change goes from liquid to gas, one can assume ∆V ≈ Vgas 5.5 Chemical Potentials and Material Equilibrium The Gibbs Equations previously defined are not useful for a system with interchanging of matter with the surroundings or an irreversible chemical reaction The chemical potential (an intensive state function) of a substance is defined as µi ≡ ( ∂G ∂ni ) T,P,nj 6=i (one-phase) Alternatively, µi = ( ∂A ∂ni ) T,V,nj 6=i (one-phase) Gibbs’ Free Energy can now be defined as, dG = −S dT + V dP + ∑ i µdni (one-phase, Thermal/Mech. Eq.) Substituting this equation for dG into dU yields, dU = T dS − P dV + ∑ i µi dni We can now write two more extensions to the Gibbs Equations: dH = T dS + V dP + ∑ i µi dni dA = −S dT − P dV + ∑ i µi dni 25 For a multiple-phase system, let α denote one of the phase of the system. Therefore, µαi ≡ ( ∂Gα ∂nαi ) T,P,nαj 6=i dG = −S dT + V dP + ∑ α ∑ i µαi dn α i At material equilibrium, dG = ∑ α ∑ i µαi dn α i = 0 (mat. eq., const. T/P ) dA = ∑ α ∑ i µαi dn α i = 0 (mat. eq., const. T/V) For a pure substance, µi is the molar Gibbs free energy µ = Gm ≡ G n (One-phase pure substance) 5.6 Reaction Equilibrium • An equilibrium will shift to go from a location of higher chemical potential to a lower one • Let νi be the unitless stoichiometric number, which are negative for reactants and positive for products • The extent of reaction is given by the symbol ξ ξ = ∆ni νi • For a chemical-reaction in equilibrium in a closed system,∑ i νiµi = 0 • Gibbs’ Free Energy can be expressed as the following, which is zero at equilibrium, dG dξ = ∑ i νiµi (const. T/P ) 6 Standard Thermodynamic Functions of Reaction 6.1 Standard Enthalpy of Reaction • The standard state of a pure substance is defined at pressure of 1 bar (P ◦ ≡ 1 bar) ◦ Gases are assumed to have ideal behavior and partial pressures of 1 bar • The standard enthalpy of formation, ∆fH ◦ T , is the change of enthalpy for the process in which one mole of the substance in its standard state is formed from the corresponding separated elements, each element being in its reference form • For an element in its reference form, the enthalpy of formation is zero 26 6.2 Hess’ Law • One can oxidize the reactants completely to CO2 and H2O and then make products by the reverse of oxidation • One can also convert all reactants to elements in their standard states and then make products from elements in these standard states • However, the most efficient and accepted method is to use the previously defined heat of formation • The standard enthalpy change is given as the following for the reaction aA+ bB → cC + dD ∆H◦ = ∑ i νi∆fH ◦ T,i 6.3 The Six-Step Program for Finding ∆H◦ f 1. If any of the elements involved are gases at T and 1 bar, we calculate ∆H for the hypothetical transformation of each gaseous element from an ideal gas to a real gas under the same conditions 2. We measure ∆H for mixing the pure elements at these conditions 3. We utilize ∆H = ´ T2 T1 CP dT+ ´ P2 P1 (V − TV α) dP to find ∆H for bringing the mixture from the original T and 1 bar to the conditions used to carry out the experiment 4. A calorimeter is used to measure ∆H of the reaction 5. ∆H is found for bringing the compound from the step in which it is formed back to T and 1 bar 6. If there is a compound that is a gas, we calculate ∆H for the hypothetical transformation from a real gas back to an ideal gas 6.4 Calorimetry • If there are conditions of constant volume, ∆U can be measured. If there are conditions of constant pressure, ∆H can be measured ∆rU298 = −Cavg∆T • We know ∆H = ∆U + ∆ (PV ), and the following assumption can be used by ignoring volume changes of liquids and solids ∆ (PV ) ≈ ∆ngas ·RT • Therefore, ∆H◦T = ∆U◦T + ∆ngRT • For the above equation, be careful of what you’re solving for. Look at the example below as cautionary measure 6.4.1 Cautionary Calorimetry Calculation Question: If the standard enthalpy of combustion at 25◦C of liquid (CH3)2CO to CO2 gas and H2O liquid is −1790 kJ/mol, find ∆fH ◦ 298 and ∆fU ◦ 298 of (CH3)2CO liquid Solution: 1. Write out the reaction with correct stoichiometry: (CH3)2CO(l) + 4O2(g)→ 3CO2(g) + 3H2O(l) 27 6.8 Standard Gibbs Energy of Reaction • The standard Gibbs energy change can be found as, ∆G◦T = ∑ i νiG ◦ m,T,i • An easier way to calculate this is, ∆G◦T = ∑ i νi∆fG ◦ T,i ◦ ∆fG ◦values can be obtained from ∆G = ∆H − T∆S for an isothermal process • Using the Gibbs equations, one can state that( d∆Grxn dT ) P = −∆Srxn 7 Reaction Equilibrium in Ideal Gas Mixtures 7.1 Chemical Potentials in an Ideal Gas Mixture ∆µ = µ(T, P2)− µ(T, P1) = RT ln ( P2 P1 ) (Pure Ideal Gas, Const. T ) • We know that ( ∂G ∂P ) T = V , so ( ∂µ ∂P ) T = Vm = RT P for an ideal gas • At constant T , we can say that ´ P P◦ dµ = ´ P P◦ Vm dP . This yields, µ = µ◦ +RT ln ( P P ◦ ) • The chemical potential of component i of an ideal gas mixture at T and P equals the chemical potential of pure gas i at T and Pi ◦ This is true for U , H, S, G, and CP for an ideal gas mixture 7.2 Ideal-Gas Reaction Equilibrium • Assume the following definition of the standard equilibrium constant for the reaction aA + bB cC + dD, K◦P ≡ ( PC,eq P ◦ )c( PD,eq P ◦ )d ( PA,eq P ◦ )a( PB,eq P ◦ )b • Standard change in Gibbs Free Energy can then be expressed as, ∆G◦ = −RT ln (K◦P ) • Assume the following mathematical definition, n∏ i=1 ai ≡ a1a2...an 30 • With this, K◦P can be defined as, K◦P ≡ ∏ i ( Pi,eq P ◦ )νi • Using the rules of logarithms, K◦P = e −∆G◦ RT • K◦P is only a function of temperature and is independent of all other states • If ∆G◦  0, then KP is very small. Conversely, if ∆G◦  0, then KP is very large • It is typically easier to write this without the standard restriction as, KP ≡ ∏ i (Pi,eq) νi • The molar concentration, ci, of a species is, ci ≡ ni V • Therefore, Pi = ciRT • Assuming c◦i = 1mol/L = 1M , K◦c = ∏ i (ci,eq c◦ )νi • Therefore, K◦c = K◦P ( P ◦ RTc◦ )∆n/mol • A mole fraction equilibrium constant, Kx, is defined as, Kx ≡ ∏ i (xi,eq) νi • A helpful relationship is simply, K◦P = Kx ( P P ◦ )∆n/mol • Kx depends on P and on T unless ∆n = 0, so it is not as useful 7.3 Qualitative Discussion of Chemical Equilibrium • While K◦P is always dimensionless, KP is only dimensionless when ∆ngas = 0 and typically has units of Pressure∆ngas • For a reaction not necessarily at equilibrium, QP ≡ ∏ i (Pi) νi • If QP < Kp, the reaction will proceed to the right: ξ > 0 • If QP = KP , the reaction is already at equilibrium: ξ = 0 31 • If QP > KP , the reaction must go in the reverse: ξ < 0 • If ∆G◦ is large and negative, K◦P is very large and little reactant is left at equilibrium • If ∆G◦ is large and positive, K◦P is very small and there is little product present at equilibrium • Typically, values of e−12 are considered very small and e12 are considered very large. With this approximation, reactions at 298K with ∆G◦ < −30 kJ/mol go to completion, and reactions at 298K with ∆G◦ > 30 kJ/mol don’t proceed at all • Remember, at the molecular level kT is used for energy comparison, and RT is used at the molecular level • At low temperature ∆G◦ ≈ ∆H◦, and at high temperature, ∆G◦ ≈ −T∆S◦ 7.4 Temperature Dependence of the Equilibrium Constant • Differentiation of the equation for K◦P will yield, d ln (K◦P ) dT = ∆G◦ RT 2 − 1 RT d (∆G◦) dT • Using various mathematical equalities that I will not write out, one yields the van’t Hoff equation d ln (K◦P ) dT = ∆H◦ RT 2 • This can be rearranged to d ln (K◦P ) d (1/T ) = −∆H◦ R • Therefore, for a plot of ln(K◦P ) against 1/T , the slope is −∆H◦ R , and if ∆H◦ is approximately tem- perature independent, the plot produces a straight line • Using the equation for CP in 6.6 yields, ∆H◦T = A+BT + CT 2 +DT 3 + ET 4 • Consequently, if ∆H◦ 6= constant, then ∆H◦ = ∆H◦(T1) + ∆a(T − T1) + ∆b 2 (T 2 − T 2 1 ) + ∆c 3 (T 3 − T 3 1 ) • Integrating the van’t Hoff equation yields, ln ( K◦P (T2) K◦P (T1) ) = ˆ T2 T1 ∆H◦T RT 2 dT • An easier to calculate quantity is the following where ∆H◦ is assumed to be independent of temperature, ln ( K◦P (T2) K◦P (T1) ) ≈ ∆H◦ R ( 1 T1 − 1 T2 ) • If the K◦P at some arbitrary temperature is needed, one can easily find K◦P,298 with corresponding T = 298.15K and ∆H◦ from tabulated values with Hess’ Law • If ∆H◦ cannot be assumed to be independent of temperature but C◦P can, one can use the equation in 6.6 to find ∆H◦T by finding ∆H◦298 and ∆C◦P from tabulated values with Hess’ Law 32 • Rearranging this equation yields, dP dT = ∆Sm ∆Vm = ∆S ∆V • The Clapeyron Equation states that for one component two-phase equilibrium system, dP dT = ∆Hm T∆Vm = ∆H T∆V • For phase transitions from solid to liquid, solid to gas, and liquid to gas, ∆Vm > 0, ∆Hm > 0, and( dP dT ) > 0 ◦ An exception to this is substances that have ∆Vm < 0, such as water, when going from solid to liquid. This changes it to ( dP dT ) < 0 ◦ Also, if ∆Vs→l = 0, then dP dT =∞. This implies that ∆T is very small for large changes in P 8.4.2 Liquid-Vapor and Solid-Vapor Equilibrium • If the assumption is made that ∆V ≈ Vgas,( dP dT ) vap or sub = ∆Hm TVm,gas • Using Vgas ≈ RT P , the Clausius-Clapeyron Equation is obtained ( dP dT ) vap or sub = P∆Hm RT 2 • Integration the above separable differential equation yields d lnP dT ≈ ∆Hm RT 2 • An approximation, albeit a very crude one7, is that ∆P ∆T ≈ dP dT ∴ ∆T = ∆PRT 2 P∆Hm • To obtain a better approximation, begin by substituting d(1/T ) = −(1/T 2), d lnP d(1/T ) ≈ −Hm R • If ∆Hm is independent of temperature, the above equation can be integrated to yield ln ( P2 P1 ) = −∆H R ( 1 T2 − 1 T1 ) • In handbooks, A and B constants can be found8 to make, lnP ≈ A T +B 7It is not recommended to use this approximation 8Crude form of the Antoine Equation 35 8.4.3 Effects of Pressure on Phase Transitions • Higher altitudes cause boiling point decreases since the pressure is lower • For small variations in pressure, ∆Tbp = Tboil ∆Vvap ∆Hvap ∆P • For larger variations in P , the following is true where nbp stands for normal boiling point, ln ( P 1 atm ) = ∆H R ( 1 Tnbp − 1 T ) 8.4.4 Solid-Liquid Equilibrium • Since the solid-liquid transitions doesn’t involve a gas phase, ∆V ≈ Vgas is unreasonable. • The applicable equation for solid-liquid equilibrium is9, ˆ P2 P1 dP = ˆ T2 T1 ∆fusS ∆fusV dT = ˆ T2 T1 ∆fusH T∆fusV dT • Using the approximation ∆P ∆T ≈ dP dT for small pressure changes or assuming ∆fusS ∆fusV is constant yields, ∆P = ∆T∆fusS ∆fusV = ∆T∆fusH T1∆fusV • For larger changes in P for solid-liquid transitions, assume ∆fusH and ∆fusV are constant to make, ∆P ≈ ∆fusH ∆fusV ln ( T2 T1 ) 8.4.5 Effects of Pressure on Vapor Pressure • For phase equilibrium, µg = µcond • At constant temperature, the system will go to a new vapor pressure where the gas has a higher µ and vapor pressure with applied external pressure • At constant volume and vapor pressure, T would fall with an applied external pressure 9 Solutions 9.1 Solution Composition • The definition of molarity is, ci ≡ ni V • The definition of mass concentration is, ρi ≡ mi V • The definition of molality of solute B is the following where solvent mass is wA and MA is the solvent molar mass, mB ≡ nB wA = nB nAMA 9Fusion is defined as going from solid to liquid 36 9.2 Partial Molar Quantities • From now on, V ∗ is the total volume of unmixed components, where ∗ denotes a property of pure, unmixed substances. As such, V ∗ = ∑ i niV ∗ m,i • ∆mixV = V − V ∗ = 0 for ideal solutions and ∆mixV 6= 0 for real solutions ◦ Ideality also implies that, for a solution of B and C, the B-B interactions, B-C interactions, and C-C interactions are all the same • The total differential can be applied knowing that V (T, P, ni, ..., nr) dV = ( ∂V ∂T ) P,ni dT + ( ∂V ∂P ) T,ni dP + ∑ i V̄idni • The partial molar volume is V̄j ≡ ( ∂V ∂nj ) T,P,ni6=j • Additionally, V̄i = RT P (Ideal Gas Mixture) • The partial molar volume of a pure substance is equal to its molar volume, V̄ ∗j = V ∗m,j • As a result, V = ∑ i V̄ini = ∑ i xiV̄intot • Also, ∆mixV = ∑ i ni ( V̄i − V ∗m,i ) • For a multi-component system, n1dV̄1 + n2dV̄2 + ... = 0 = x1dV̄1 + x2dV̄2 + ... • Therefore, for a two component system, x1dV̄1 = −x2dV̄2 and dV̄1 = −x2 x1 dV̄2 n1dV̄1 = −n2dV̄2 and dV̄1 = −n2 n1 dV̄2 • Moreover10, ( ∂V̄1 ∂n1 ) n2 = −n2 n1 ( ∂V̄2 ∂n2 ) n2 10I do not get this. How can you hold n2 constant and have it be changing by an infinitesimal amount, ∂n2? 37 9.8 Thermodynamic Properties of Ideally Dilute Solutions • Since µ◦i ≡ fi (T, P ), µi = RT lnxi + µ◦i • Additionally, µA = µ◦A +RT lnxA µ◦A ≡ µ∗A Pi xliP ◦ = exp ( µ◦li − µ◦vi RT ) • A new (Henry’s) constant can be defined as, Ki ≡ P ◦ exp ( µ◦li − µ◦vi RT ) • Finally, Henry’s Law can be expressed as, Pi = Kix l i • Furthermore, Raoult’s Law applies as, PA = xlAP ∗ A • For an ideal solution, the solute obeys Henry’s Law, and the solvent obeys Raoult’s Law • For an application of external pressure, P ′ = P0e V̄ P RT • For a gas dissolved in liquid, xli = Pi Ki • Therefore, solids dissolve in liquids better at higher temperatures while gases dissolved in liquids better at lower temperatures ◦ Classic example is the dissolution of carbon dioxide in soda - it goes flat at higher temperatures (less CO2 dissolved) • Since this is an ideally dilute solution, Pi = Ki,mmi = Ki,cci • Henry’s Law does not apply to strong electrolytes • If KHenry > P ∗, then Pideal > P ∗ and µ◦ = µideal > µ∗ • If KHenry < P ∗, then µ◦ = µideal < µ∗ 40 10 Nonideal Solutions 10.1 Activities and Activity Coefficients • The activity of a substance is defined as, ai ≡ exp ( µi − µ◦i RT ) • Rearranging this to a familiar form yields, µi = µ◦i +RT ln ai • Then, for an ideal or ideally dilute solution, ai = xi • The difference between the real and ideal µi is, µi − µidi = RT ln ( ai xi ) • The activity coefficient, γi, is defined so that, ai = γixi • The new µi equation is now, µi = µ◦i +RT ln (γixi) • Substitution of µideali = µ◦i +RT lnxi yields, µidi +RT ln γi • There are two conventions for the standard state Convention I The standard state of each solution component i is taken as pure liquid i at the temperature and pressure of the solution such that, µ◦I,i ≡ µ∗i (T, P ) • Additionally, γI,i → 1 as xi → 1 for each chemical species i • Use this convention if the mole fractions vary widely and both components are liquids Convention II This convention treats the solvent differently from the other components. The standard state of the solvent A is pure liquid A at the temperature and pressure of the solution such that, µ◦II,A = µ∗A (T, P ) • Additionally, γII,i → 1 as xA → 1 for each i 6= A ◦ Note that this means xi → 0 • Choose this convention if solids or gases are in dilute concentrations in a liquid 41 10.2 Determination of Activities and Activity Coefficients 10.2.1 Convention I • Raoult’s Law is now, Pi = aI,iP ∗ i • The following then can be stated, Pi = xviPtot = γI,ix l iP ∗ i • If component i has Pi > P idi , then γI,i > 1 and vice versa • If γI is less than 1, it means that the chemical potentials are less than the ideal chemical potentials. This means that G is lower than Gid, and, thus, the solution is more stable 10.2.2 Convention II • Raoult’s Law is now, PA = aII,AP ∗ A = γII,Ax l AP ∗ A • Henry’s Law is now, Pi = KiaII,i = KiγII,ix l i • γI measures deviations from ideal-solution behavior while γII measures deviations from ideally dilute solution behavior 42