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The Basic Concepts: the system Thermodynamics The First Law Work, Heat, Energy System Matter System Matter Energy Energy Surroundings Thermodynamics is the study of the transformations of energy. Surroundings (b) closed (a) open System Matter Energy Oxtoby, Chapter 10 (10.1-10.4) Surroundings (c) isolated Stanford Chemistry Summer Session 2005 Chem 31 The Basic Concepts: state and path fns A state function, X, is a property that depends only on the current state of the system, and not on how it was prepared. Changes in a state function depend only on the start and end points of an experiment The Conservation of Energy: Energy can neither be created not destroyed (experimental observation) DX = Xfinal – Xinitial e.g. the duration of this lecture depends only when I start and when I finish A path function, Y, is a property that depends on the history of the system. e.g. the boredom/interest factor for this lecture depends on a lot more than just my first and last sentence fl Energy is a state function The principle of conservation of energy can be used to assess the energy changes that accompany physical and chemical processes. Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Work, Heat and Energy ICE + SOLID C + + heat heat O2 Melting Liquid water Work, w, is done when an object is moved against an opposing force Liquid CO2 + heat Work is a path function (how strong is gravity?) Expanding gas pushes out piston against (e.g, atmospheric) pressure Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 1 Work, Heat and Energy Heat, q The Energy, E of a system is its capacity to do work. When the energy of a system changes as a result of a temperature difference between it and its surroundings, it is said that energy has been transferred as Heat, q When spring unwinds, work will be done Processes that release energy as heat are exothermic; Increasing its capacity to work A B + heat Combustion reactions When piston moves out against higher pressure, work will be done Processes that absorb energy as heat are endothermic; A + heat Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Hence, for an isolated system: The First Law In thermodynamics, the “total” energy of a system is called the internal energy, E. ∆E = w + q = 0 + 0 = 0 Energy Surroundings ∆E = Ef - Ei The change in internal energy ∆E is the sum of work done on a system and the energy transferred as heat to a system according to; ∆Euniverse = 0 System Matter Experiments measure the change in energy between start and finish: and a closed system: ∆E = w + q System Matter Heat must be a path function: the system can gain/lose energy either as heat or as work. Energy Surroundings SI unit for work, heat and energy is the Joule (or J/mol, kJ/mol, …) the change in internal energy of a closed system is equal to the energy that passed through its boundary — either as heat or as work. Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Work, w In general: B Melting, Vaporization of Water Work and pressure w=F. d work Pressure Volume Work: distance External pressure, p Force along path Area A More specifically, for thermodynamics: w=- F. d ∆h w = - Fext . d w = - Fext ∆ h We focus on the force the system has to push against w = - pext A ∆ h Work is done on system by the surroundings w = - pext ∆ V Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 2 Work, and pressure +/-? The Sign Convention w = - pext ∆ V Work done by the system on its surroundings is “lost” by the system (a negative w). Work done on the system by the surrounding increases the internal energy (positive w) – energy is stored in the system For pext = 0 Expansion ∆ V > 0 Compression ∆ V < 0 w < 0, hence system performs work on surroundings w > 0, hence surroundings perform work on system A positive heat flow (positive q) corresponds to heat gained by the system. Hence, a heat loss is to be understood as a negative gain (negative q). Thermal Equilibrium is established when two bodies have the same temperature. Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Example: Recall: Calculate the work done when 25g of iron reacts with Hydrochloric Acid in (a) a closed cell, (b) in a open beaker at 25oC? (a) ∆ V = 0 ∆E = w + q & w=0 w = - pext ∆V If the system is kept at constant volume: (b) Fes + 2 HCl(aq) w = - pext ∆ V ∆E = qv FeCl2 (aq) + H2(g) ∆ V = Vf-Vi =Vgas pV = nRT n= m/M 25g mol 55.845 g w = - pext ∆ V = - pext Vgas = - n R T w= - 0.45mol x 8.314 J x 298 K K mol = 0.45 mol = - 1.2 kJ Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Examples Calorimetry Adiabatic Bomb Calorimeter We measure ∆E usually using Calorimetry (the measurement of amounts of heat flowing into or out of a system and the accompanying temperature changes) Adiabatic means no heat is transferred from calorimeter to surroundings The change in temperature, ∆T, of the calorimeter is proportional to the heat that the reaction releases or absorbs. q = Ccal ∆T Calorimeter Constant Calibrated using a process of known energy output (eg, burning of a substance of known mass) or from an electrical current, I, of known potential, V q=VIt (a) When 100 g of Napthalene was burned in a bomb calorimeter releasing heat of 3.91 kJ, the temperature rose by 3.01 K. What is the calorimeter constant? q = Ccal ∆T 3.91 kJ = Ccal µ 3.01 K Ccal = 1.30 kJ / K (b) 10.0 A from a 12 V battery were passed through the heater of a calorimeter for 150 s. The temperature rose by 5 K. What is the calorimeter constant? q = (10.0 A) µ (12 V) µ (150 s)=1.8 µ 104 A V s = 18 kJ 18 kJ = Ccal µ 5 K Ccal = 3.6 kJ / K Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 3 Heat Capacity Heat Capacity Generally: C = q ∆T Units: J K-1, cal K-1 1 cal = 4.184 J Molar Heat Capacity (J K-1 mol-1 or cal K-1 mol-1) q C Heat capacity of a sample divided by the c= = n chemical amount of substance, n ∆T n The heat capacity C of a substance is the heat required to change its temperature by one Kelvin, and has units of energy per Kelvin. The heat capacity is an extensive variable: the quantity is proportional to the amount of matter present. Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Heat Capacity Example: measured at constant pressure measured at constant volume (eg, conduct experiments in closed container) qp Cp = ∆T cp = qp = ∆T n Cp n Specific Heat Capacity (in J K-1 kg-1 or cal K-1 kg-1) q C Heat capacity of a sample divided by the cs = ∆T m = m mass of substance, m c = M cs molar qv Cv = ∆T qv cv = = ∆T n Cv n If reaction contains only solids and/or liquids, cp ≈ cv A new calorimeter is filled with 200g of room temperature water. Adding 1212 J of heat to the contents, raises the interior temperature by 1.42 K. Calculate the calorimeter constant given that the molar heat capacity of water, cp(H2O) is 75.38 J K-1 mol-1. 1212 J = qwater+ qcalorimeter = Ccal ∆T + cp(H2O) n(H2O) ∆T = Ccal ∆T + cp(H2O) m(H2O) ∆T M(H2O) Ccal = 1212 J - 75.38 J K-1 mol-1 200 g 1.42 K 1.42 K 18 g mol-1 1.42 K = 16 J K-1 Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Recall: ∆E = w + q & w = - pext ∆V If the system is kept at constant volume: Enthalpy ∆E = qv But if it is not kept at constant volume? ∆E = qv Another type of energy, the ENTHALPY, is useful in these circumstances Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 4 The Enthalpy, H Example: Relating ∆H and ∆U For a system kept at constant pressure: ∆H = qp where ∆H = Hf - Hi The Enthalpy, like the internal Energy, is a state function The Enthalpy is related to the Energy by: H = U + pV ∆H = ∆(U+pV)=Hf - Hi = (U+pV)a - (U+pV)c = Ua - Uc + (pV)a -(pV)c as p = const. = ∆ U + p ∆V m Va = ρ a ∆H- ∆U = p ∆V a H = U + pV = U + nRT Va = 17.0 cm3 Vc = 18.5 cm3 Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Enthalpy of Reaction - ∆Hr The Reaction Enthalpy, ∆Hr A B HA HB ∆Hr = Hproducts - Hreactants In General: ∆Hr = Σ ν Hproducts - Σ ν Hreactants Enthalpy of Reaction: ∆Hr = HB - HA Example 2) Ma = Mc = 100 g/mol ma = mc = 50.0 g ∆H- ∆U = 0.152 J for an ideal gas Example 1) The internal energy change when 0.500 mol calcite (ρc =2.71 g cm-3) converts to aragonite (ρa =2.93 gcm-3) is 0.100 kJ. Calculate the difference between the changes in enthalpy and in internal energy at a pressure of 1 atm. [calcite and aragonite are different forms of CaCO3] Sum over all standard molar enthalpies taking into account their stochiometric factors, ν 2C + D 2 HC HD E +3F HE 3 HF Enthalpy of Reaction: ∆Hr = HE + 3 HF - HD - 2 HC Example: 2 C2H6 + 7 O2 ν= 2 ν= 7 4 CO2 ν= 4 + 6 H2O ν= 6 ∆Hr = 6H(H2O) + 4H(CO2) - 7H(O2) - 2H(C2H6) Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Standard Enthalpies Enthalpies of phase change Enthalpy are normally tabulated for substances in their standard state; Denotes standard condition these are called the standard enthalpy, Ho. Enthalpy changes also occur when a substance melts/freezes or condenses/evaporates The standard state of a substance at a specified temperature is its pure form at 1 atm (should now be 1 bar). Fusion (melting): H2O(s) kJ H2O(l) ∆Hofus(273)=+6.0 mol For dissolved species, the standard state is the concentration of 1M under a pressure of 1 atm at a specified temperature. Vaporisation (boiling): H O 2 (l) kJ H2O(g) ∆Hovap(373)=+40.7 mol Standard enthalpies at T = 298 K are often denoted Hʅ endothermic! Hence, the standard enthalpy of reaction is ∆Hro =Σν Ho Sublimation products - Σν Ho reactants H2O(s) kJ H2O(g) ∆Hosub(298.15)=+46.7 mol (direct conversion from solid to gas) Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 5 Rules for enthalpy changes (1) Rules for enthalpy changes (2) A change in enthalpy is independent of the path taken between two states: Final state Initial state Enthalpy is a state function fl can derive some rules ∆Hovap(373 K)= + 40.7 kJ mol–1 (1) H2O(l) vaporisation condensation Sublimation H2O(g) solid ∆Hofus gas ∆Hovap gas ∆Ho(A→B) = - ∆Ho(B→A) ∆Hosub = ∆Hofus + the enthalpy of a forward process differs from that of the backwards process only in sign. ∆Hovap Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 One could also have written … Hess’s Law H2O(s) H2O(l) ∆Hofus = +6.0 kJ mol–1 H2O(l) H2O(g) ∆Hovap = +40.7 kJ mol–1 Overall: H2O(s) H2O(g) ∆Hosub = +6.0 + 40.7 kJ mol–1 = + 46.7 kJ mol–1 Because enthalpy is a state function, these rules apply to every type of reaction or change The enthalpy of an overall reaction is the sum of the enthalpies of the individual reactions into which the reaction can be divided. Example: Calculate the enthalpy of combustion of benzene (C6H6) from its enthalpy of hydrogenation (-205 kJ/mol) to cyclohexane, and the enthalpy of combustion of cyclohexane (∆Ho (C6H12) = –3920 kJ/mol)). The enthalpy of the combustion for H2 is -286 kJ/mol. Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Standard Molar Enthalpies of Formation, ∆Hof Enthalpy of Combustion of Benzene We want to know: C6H6 + 7.5 O2 6 CO2 We do know: C6 H6 + 3 H2 C6H12 C6H12 + 9 O2 6 CO2 + 6 H2O C6 H6 + H2O(g) ∆Hosub solid ∆Hocond(373 K)= - 40.7 kJ mol–1 In general: H2O(s) + 3 H2 O -205 kJ/mol 3 H2 O 3H2 + 1.5 O2 7.5 O2 6 CO2 + 3 H2O -3920 kJ/mol -(3 µ -286) kJ/mol -3268 kJ/mol The enthalpy of formation is the enthalpy change when a compound is formed from its elements, and those elements are in their most stable form under the prevailing conditions. When the prevailing conditions are the standard state, this is called the standard enthalpy of formation, ∆Hof H2(g) + 0.5 O2(g) 6 C(s, graphite) + 3 H2(g) H2O(l) ∆Hof = - 286 kJ/mol C6 H6(l) ∆Hof = + 49 kJ/mol The standard enthalpies of elements in their reference states are zero at all temperatures (graphite is the reference state of carbon!). Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 6 Bond enthalpies A–B(g) Applications of bond enthalpies A(g) + B(g) ∆Hrxn = energy of the A–B bond But bond enthalpies are affected by the neighbouring bonds CH4(g) CH3(g) + H(g) ∆H = 439 kJ mol–1 C2H6(g) C2H5(g) + H(g) ∆H = 410 kJ mol–1 CHCl3 (g) CCl3(g ∆H = 380 kJ mol–1 ) + H(g) fl usually tabulate average bond enthalpies (determined from A–B bond enthalpies in many different A–B containing molecules) Average bond enthalpies can be used to estimate the enthalpy of a compound: just count the number and type of bonds involved. Using average bond enthalpies, estimate the enthalpy change for C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) Step 1: what bonds are in the reactants (draw Lewis structures) C3H8(g) H H C H H H C C H H H 1 µ (8 C-H bonds), 1 µ (2 C-C bonds) O2 O O 5 µ (1 O=O double bond) nb: angles are not 90o Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Step 2: what bonds are in the products (draw Lewis structures) 4 H2O 3 CO2 O C O 3 µ (2 µ CO double bonds) H O H 4 µ (2 µ OH single bonds) Step 3: calculate the net breakage / formation of bonds: energy is absorbed in breaking reactant bonds, and released it in making them DHrxn º 8H(C–H) + 2H(C–C) + 5H(O=O) – 6H(C=O) – 8H(O–H) Finally… DHrxn º 8 µ 412 + 2 µ 348 + 5 µ 497 – 6 µ 743 – 8 µ 463) kJ mol–1 = –1685 kJ mol–1 Value from calorimetry experiments: ∆Hco = -2220 kJ/mol Bond enthalpies appear to give a 30% error! • Works better when not all bonds are being broken (1st, 2nd, 3rd and 4th C–H bonds in methane have very different strengths) • better to use thermochemical groups bonded to at least two other atoms rather than individual bonds Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Spontaneous Changes: Why do some processes happen spontaneously? •Why does a hot body get cooler (rather than hotter) when surrounded by a cooler medium? Entropy •Why does a gas expand into all available volume of a container rather than contract? The driving force for spontaneous change (change that happens without intervention — doing work or heating) is described in the second law of Thermodynamics Chapter 11 Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 7 No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Energy is not accumulated in ball and thermal motion is not directional Kinetic energy converted into thermal motion Entropy, S, and the second law These spontaneous changes happen because they increase the randomness with which energy is spread through an isolated system The Entropy, S, a thermodynamic state function, is a measure of “molecular disorder”, or “freedom of movement” molecules have, and helps us to define the direction of spontaneous change The Entropy of an isolated system increases in the course of a spontaneous change ∆Ssystem + ∆Ssurroundings = ∆Stotal > 0 Hence, in a spontaneous process: ∆Suniverse > 0 Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Entropy and Equilibrium Entropy and Heat Nothing changes when a system is at equilibrium, including the entropy of the system So, for any process ∆Ssystem + ∆Ssurroundings = ∆Suniverse r 0 Entropy measures dispersal of energy in a system Heat changes kinetic energy of molecules, i.e. disperses energy by increasing the velocities of all the molecules fl heat and entropy are related? But, equilibrium is dynamic at a microscopic level For a reversible process reactants Ý products This introduces the idea that changes can be reversible, i.e. a change can be made, and then exactly undone (“reversed”), so ∆Suniverse = 0 DS = qrev / T = DH / T (at constant pressure) For an irreversible process DS > qrev / T for a reversible process In practice, reversible changes are an idealised limit in which changes happen infinitely slowly via a series of imperceptible shifts in the equilibrium Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Entropy and Disorder But … Water freezes (spontaneously) on a cold night! The Entropy of the System Entropy increases ?? Solid Gas Liquid Liquid (water molecules able to move) Entropy of system increases Solid (molecules fixed in crystal) Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 8 Spontaneous Freezing Entropy of Phase Changes System at transition temperature Water freezing is an exothermic process H2O(l) ∆H = - 6.88 kJ/mol H2O(s) So heat is put into the surroundings –∆ Hsystem = + ∆ Hsurroundings= qp,surroundings b ∆ Ssurroundings decreases Solid increases ∆Ssystem + ∆Ssurroundings = ∆Suniverse > 0 Entropy is not the final determinant of spontaneity — see tomorrow ∆SSysfus = similarly, for vaporization ∆Hfus Liquid TFus ∆SSysvap = ∆Hvap Tvap Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Entropy of Reaction Trends in entropy Increasing Temperature As entropy is a state function, the entropy of reaction is defined (by analogy to Hess’ law for the enthalpy) as the difference between the molar entropies for the pure, separated products and the pure, separated reactants: ∆Sr = Σ ν Soproducts - Σ ν Soreactants Where o as before indicates that all substances are in their standard states at the specified temperature. rotation translation vibration More modes to distribute energy become available ENTROPY INCREASES Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Entropy and Disorder: the microscopic view The Statistical Entropy Let’s say you are tossing a coin with your mates to check who will have to go to the next lecture to take notes (if handouts are not provided). You decide on head and your mate on tails. As you toss the coin for the first time, the likelihood that of head:tail is 50:50, or 1/2:1/2. Tail Head # of combinations 1 1 Increase in Entropy Example: mixing of two gases (pure O2 and pure N2) To quantify S, see microstate considerations below The chance that you win heads twice is ¼, as it is to win tails twice; the mixed result has a probability of ½ # Tail/Tail Tail/Head or Head/Tail Head 1 2 1 Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 9 Similarly for particles in boxes For three throw # T/T/T TTH,THT,HHT 1 3 HHT,HHT,THH H/H/H 1 3 Imagine you have four balls and can put them into two containers (T or H) then you can distribute them as follows; Most likely outcome T/T/T/T TTTH,TTHT, THTT, HTTT # 1 4 TTTH,TTHT, THTT, HTTT T/T/T/T …. and for four throws # TTHH, THHT, THTH, HHTT, HTHT, HTTH HHHT,HHTH, HTHH, THHH 6 1 TTHH,THHT,HTTH, THTH, HTHT,HTTH 4 HHHT,HHTH, HTHH, THHH 6 H/H/H/H 1 4 H/H/H/H 4 1 T H T H T H T H T H Even split most favored Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 The Boltzmann Formula Third Law of Thermodynamics The entropy change for any equilibrium between pure substance becomes zero as the temperature approaches absolute zero S = k ln W The Entropy of any crystalline substance approaches zero as T | 0 K. W = weight of most probable configuration of the system! k = Boltzmann constant = R/N0 = 1.308630 x 10-23 J K-1 Walther Nernst 1864-1941 Nobel Laureate in chemistry in recognition of his work in thermochemistry Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 The third law Let’s remember when processes are spontaneous S = k ln W As T | 0 K, every atoms is located at the perfect crystal location, so W = 1 But … •Removing every defect is difficult (if not impossible) ∆Suni > 0 Spontaneous ∆Suni = 0 Equilibrium; also true for idealised changes that are so slow they never disturb equilibrium (reversible) ∆Suni < 0 Does not happen (reverse reaction will be spontaneous) •QM ensures atoms still have zero-point energy, and so some entropy persists (see last week) fl can’t actually achieve absolute zero Temperatures < 0.000001 K have been achieved Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 10 The Gibbs Free Energy, G ∆Suni = ∆Ssystem + ∆Ssurroundings Requires knowledge of both system and surroundings Gibbs Free Energy, G At constant pressure and temperature, and for reversible changes: T∆Ssurroundings = qp,rev = –∆Hsystem Define a new form of the energy function G = H – TS Called the Gibbs free energy. G is a state function Stanford Chemistry Summer Session 2005 Chem 31 The Gibbs Free Energy: criterion for spontaneity At constant T: ∆Gsys = ∆ (Hsys - Tsys Ssys) = ∆Hsys - T ∆Ssys ∆G = ∆H - T ∆S But also: ∆Hsys = - ∆Ssurr T ∆Gsys = - T ∆Ssurr - T ∆Ssys = - T (∆Ssurr + ∆Ssys) = - T (∆Suni) < 0 for spontaneous pr. and as ∆Suni > 0 Resulting in: Criterion for Spontaneity ∆Suni > 0 Spontaneous ∆Gsys < 0 ∆Suni = 0 Equilibrium ∆Gsys = 0 ∆Suni < 0 Not spontaneous (reverse reaction is spontaneous) ∆Gsys > 0 N.B. left hand describes the whole universe, but right hand is just the system. We can drive the system in the wrong direction (hence can have reactions where DG is positive, but cannot force the universe to show negative entropy Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Examples ∆G = ∆H - T ∆S ∆G < 0 for spontaneous Remember ∆Gsys < 0 for spontaneous processes 2) An endothermic process where entropy decreases ∆G = ∆H - T ∆S 1) An exothermic process which also increase entropy ∆H = + ve (>0) ∆H = - ve (<0) ∆S = + ve ; -T ∆S = - ve (<0) ∆S = - ve ; -T ∆S = + ve (>0) ∆G = -ve (<0) spontaneous ∆G = +ve (>0) NOT spontaneous (reverse reaction would be spontaneous as G is a state function) Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 11 ∆G = ∆H - T ∆S ∆G < 0 for spontaneous ∆G = ∆H - T ∆S ∆G < 0 for spontaneous 3) An exothermic process in which entropy decreases (freeze water) Depends on relative magnitudes of ∆H and -T ∆S ∆H = - ve (<0) ∆S = - ve ; -T ∆S = + ve (>0) 4) An endothermic process in which entropy increases (melting ice) ∆S = + ve ; -T ∆S = - ve (<0) |∆H| < |T ∆S| |∆H| = |T ∆S| |∆H| > |T ∆S| |∆H| < |T ∆S| ∆G > 0 ∆G = 0 ∆G < 0 ∆G < 0 ∆G = 0 Not spontaneous Equilibrium Spontaneous Equilibrium Spontaneous For fixed ∆H and ∆S , spontaneity depends on T Depends on relative magnitudes of ∆H and -T ∆S ∆H = + ve (>0) |∆H| = |T ∆S| |∆H| > |T ∆S| ∆G > 0 Not spontaneous For fixed ∆H and ∆S , depends on T Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 Driving forces in chemistry Gibbs Energy for Phase Transitions Two driving forces underpin Chemistry: P = 1atm ¾ Systems tend to a state of minimum enthalpy H2O(s) H2O(l) ¾ Systems tend to a state of maximum entropy H2O(l) H2O(s) The Gibbs free energy expresses the balance between these two driving forces DG = DH – T DS (b 0 ?) ∆G = ∆H - T ∆S kJ ∆Hofus(273.15)= + 6.0 mol kJ ∆Hofrze(273.15)= – 6.0 mol ∆Hfrze, 273.15 ∆Sfrze, 273.15 = Tfrze=273.15 ∆G273.15 = ∆H273.15 - Tfrze=273.15 Equilibrium ∆S273.15 = 0 Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 If ∆H and ∆S are independent of temperature (a good, but not perfect, approximation), then: T < Tfrze T > Tfrze As the Gibbs energy is a state function, the (standard) Gibbs energy of reaction is defined (in analogy to Hess’ law for the enthalpy) as: ∆G = ∆H - T ∆S ∆G < 0 Spontaneous Freezing ∆G = ∆H - T ∆S ∆G > 0 Gibbs Energy for Chemical Reaction Spontaneous Melting of Ice Clearly, the spontaneity of freezing (negative ∆S and negative ∆H) depends on a balance of the entropy and enthalpy ∆Gro = Σ ν Goproducts - Σ ν Goreactants Where o as before indicates that all substances are in their standard states at the specified temperature. ∆Gro = ∆ Hro - T ∆ Sro Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 12 Standard Molar Gibbs Energy of Formation Example: By analogy with the definition of the standard molar Enthalpy of formation we define Pt (NH3)2I2, cis Pt (NH3)2I2, trans ∆Gfo = ∆ Hfo - T ∆ Sfo as the standard molar Gibbs energy of formation when 1 mole of a substance forms in a standard state at a specified temperature from the most stable forms of its constituent elements in standard states at the same temperature. ∆Gfo (298.15K) ∆Hfo(298.15K) cis -286.6 kJ/mol trans -316.9 kJ/mol -130.2 kJ/mol -161.5 kJ/mol Calculate the standard entropy, Pt(s) + 2 NH3,(g) Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Gibbs Free Energy & Equilibrium Constants Pt(NH3)2I2 (s) cis trans ∆ Sfo = (-316.9+161.5) kJ/mol 298.15K = -0.5212 kJ/(K mol) = -0.5246 kJ/(K mol) ∆ Sfo = So(Pt(NH3)2I2) - So(Pt(s)) - 2 So(NH3,(g)) - So(I2,(s)) -0.5425 kJ/(K mol) So(Pt(NH3)2I2)cis = 17.9 J/(K mol) Now remember, at equilibrium, ∆Gr = 0 and Q = K G So(Pt(NH3)2I2)trans = 21.3 J/(K mol) cCg + bBg dDg For the “standard” reaction all reactants and products in standard states, but in general they are related by the reaction quotient. ∆Gr = ∆Gro + RT lnQ (PC)c (PD)d N.B. both ∆G and Q can be used to determine the direction of change: ∆G | 0 Q | K Q= (PA)a (PB)b Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Temperature and the Equilibrium Constant ∆Gr = ∆Gro + RT lnQ ∆Gro = - RT lnK Assume ∆S and ∆ Hro don’t change with temperature! ∆Gr = - RT lnK + RT lnQ = RT ln Q/K RT2 ln K 2 = ∆H − T2 ∆S RT1 ln K1 = ∆H − T1∆S ⎛K ⎞ ∆H − T1∆S ∆H − T2 ∆S ln ⎜ 1 ⎟ = ln K1 − ln K 2 = − RT1 RT2 ⎝ K2 ⎠ 1852-1911 Spontaneous to products reactants aAg + Summer Session 2005 Chem 31 Equilibrium ∆G=0 Q=K ∆G<0 Q<K of both compounds at 298.15 K. Summer Session 2005 Chem 31 ∆ Sfo = ∆ Hfo - ∆ Gfo T + I2,(s) ∆ Sfo = -286.6 + 130.2 kJ/mol 298.15K So, Nobel Prize in Chemistry, 1901 The van’t Hoff Equation ∆G>0 Q>K Not spontaneous to reactants ln ∆Hro K2 K1 = R ( 1T 1 1 T2 ) products Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 13 Similarly, for liquid-vapour equilibrium The Clausius-Clapeyron Equation ln ∆Hvapo 1 P2 = (T P1 R 1 1 ) T2 assuming ∆S & ∆ Hro independent of temperature This expression is particularly useful when you want to determine the boiling points of substances at different pressures. Example: Water boils at 373.15K at an atmospheric pressure of 1.0 atm. Calculate the boiling temperature of water on top of Mt Everest (assuming a pressure of about 0.36 atm). The enthalpy of vaporization of water can be assumed to be 40.7 kJ/mol. ∆Hvapo 1 ln P2 = ( T1 P1 R 8.314 J ln 0.36 atm 1 atm K mol 40.7 103 J mol-1 1 T2 P1 =1.0 atm P2 = 0.36atm ) T1 = 373.15 K ∆Hvapo = 40.7 kJ/mol 1 373.15 K = 1 T2 = 1 346 K Stanford Chemistry Stanford Chemistry Summer Session 2005 Chem 31 Summer Session 2005 Chem 31 14