Thermodynamics II The Spontaneous Process We want to

Thermodynamics II

The Spontaneous Process • We want to know if a reaction will occur or not, i. e. , we want to know if the reaction is spontaneous or nonspontaneous • We know that process (a) is spontaneous, i. e. , it will happen naturally • We know that process (b) is nonspontaneous, i. e. , it doesn’t happen naturally • We want a general method to predict if a reaction is spontaneous or nonspontaneous

The Spontaneous Process • Some examples of spontaneous processes: • In a waterfall, the water always falls, it never rises • A sugar cube dissolves in coffee, but dissolved sugar does not reform into a cube of sugar • Heat flows from a hot object to a cold object , and never from a cold object to a hot object • Iron exposed to water and oxygen forms rust, but rust does not spontaneously return to iron • N. B. A spontaneous reaction is not necessarily fast

The Spontaneous Process • Spontaneity is not a question of energy – ex. ; In figure (a) on the second slide, the energy of the gas does not change during the expansion into the empty side (kinetic energy does not change if the temperature is constant) – ex. ; During a transfer of heat from a hot object, H, to a cold object , C, the energy of H decreases, the energy of C rises, and the energy of the universe does not change (the First Law of Thermodynamics )

Entropy • Entropy (S) is a quantity that directly measures the disorder in a system • The greater the disorder in the system, the greater the entropy will be • Entropy and probability are connected – ex. ; In the figure on the second slide, the probability of finding all of the (say 100) molecules of gas on the same side is (1/2)100 = 8 x 10 -31 – An ordered system (low entropy) is unlikely – A disordered system (high entropy) is likely

Entropy • Another definition of entropy comes from the work of Carnot DS = Q / T where DS is the change in entropy in the system during a transfer of heat, Q, at a temperature T • The units for entropy are J/K • The Third Law of Thermodynamics (we do not touch on this law in this course) allows us to have an absolute entropy, So, for each compound (as compared to enthalpy where each compound has a DHfo which is relative to an arbitrary standard)

Entropy • Consisttent with the idea that entropy corresponds to disorder, it is observed that – S(solid) < S(liquid) < S(gas) i. e. , The disorder increases when going from solid to liquid to gas – ex. ; S(diamond) < S(graphite) i. e. , Diamond has less disorder than graphite – For a reaction where the disorder increases, DS > 0 – For a reaction where the disorder decreases, DS < 0 • The value of DS is independent of the path chosen, i. e. , S is a state function (reasonable, because the difference in the inherent disorder of the initial and final states should not depend on the chosen path to join them)

Entropy

The Second Law of Thermodynamics • The Second Law of Thermodynamics: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. • The entropy of the universe can never decrease

The Second Law of Thermodynamics • Mathematically, the second law says: – Spontaneous Process: – Equilibrium Process: DSuniv = DSsyst + DSext > 0 DSuniv = DSsyst + DSext = 0 • N. B. For a spontaneous process, the change in entropy of the system, DSsyst, can be negative as long as the entropy change of the environment, DSext, is positive enough such that the change in the entropy of the universe, DSuniv, is positive • If a given reaction has a negative DSuniv, the inverse reaction will be spontaneous

The Calculation of DSreaction • Under standard conditions, the entropy change for the reaction a A + b B c. C+d. D is given by or, in general, where m and n are the stoichiometric coefficients of the balanced chemical equation

The Gibbs Function • The second law says that for a spontaneous reaction: DSsyst + DSext > 0 • However, we just want to focus on the system being studied • Carnot demonstrated that, at constant temperature, DS = Q/T, and therefore DSext = -Q/T where Q is the heat that enters the system (and therefore -Q is the heat that flows out of the system and into the environment)

The Gibbs Function • The second law becomes (DSsyst will now simply be called DS) • At constant pressure, Q = DH, and therefore • We define the change in the Gibbs free energy (DG) at constant temperature as DG = DH - T DS < 0

The Gibbs Function • G, the Gibbs function, is a state function • The previous derivation shows that at constant temperature and pressure – If DG < 0 : the reaction is spontaneous – If DG > 0 : the reaction is nonspontaneous (the reaction is spontaneous in the opposite direction) – If DG = 0 : the system is at equilibrium • One can predict whether a reaction is spontaneous or non-spontaneous by considering a function focused entirely on the system

Free Enthalpy • The value of DG corresponds to the energy that is available (“free”) to do work, i. e. , if a reaction has a DG value of -100 k. J, a maximum of 100 k. J of work can be done • Consider the combustion of hydrogen at a pressure of 1 atm at 298 K 2 H 2(g) + O 2(g) 2 H 2 O(l) DH = -571. 6 k. J DS = -327. 2 J/K DG = -474. 2 k. J • 571. 6 k. J of heat is released during the combustion at constant pressure, but theoretical limit for the conversion of this heat into work is only 474. 2 k. J

Free Enthalpy • Why can’t we have 100% effectiveness and convert all the heat (one form of energy) into work (an alternate form of energy)? • N. B. The entropy of the system decreased by 327. 2 J/K, thus the entropy of the environment must increase by a minimum of 327. 2 J/K as the entropy of the universe cannot decrease • The only way to increase the entropy of the environment is to transfer heat from the system to the environment (again, Q in the formula below is the heat that enters the system, therefore -Q is the heat that leaves the system and enters the environment)

Free Enthalpy • In our example, 97. 5 k. J of heat must absolutely remain in the environment in order to respect the second law • Therefore, of the 571. 6 k. J of heat released by the reaction, only (571. 6 - 97. 5 =) 474. 1 k. J is available (“free”) to do work • The “free” enthalpy is therefore -474. 1 k. J (the system does work, so the sign is negative) – The difference of 0. 1 k. J between this value and the value of DG at the beginning is due to rounding errors

Standard Free Enthalpy Changes • Just like there is a DHo, there is also a DGo, i. e. , the variation of standard free enthalpy where the reactants in their standard state are converted into products in their standard state under standard conditions (pression = 1 atm, concentrations = 1 M) For the reaction a. A + b. B c. C + d. D In general: where m and n are the stoichiometric coefficients of the balanced chemical equation

Standard Free Enthalpy of Formation • The standard free enthalpy of formation, DGfo, of a compound is the free energy change that occurs when one mole of a compound is synthesized from its elements in their standard state • By definition, DGfo of an element in its standard state is zero • ex. ; The value of DGfo(CO 2, g) is the value of DGo for the reaction: C(graphite) + O 2(g) CO 2(g)

Standard Free Enthalpy Changes • Example: Calculate the standard free enthalpy changes for the following reactions, at 25 o. C: (a) H 2(g) + Br 2(l) 2 HBr(g) (b) C 2 H 6(g) + 2 O 2(g) 4 CO 2(g) + 6 H 2 O(l) DGfo(O 2, g) = 0 k. J DGfo(H 2, g) = 0 k. J DGfo(Br 2, l) = 0 k. J DGfo(HBr, g) = -53. 2 k. J DGfo(C 2 H 6, g) = -32. 9 k. J DGfo(CO 2, g) = -394. 4 k. J DGfo(H 2 O, l) = -237. 2 k. J • Solution: (a) DGo = [(2)(-53. 2) - (1)(0)] k. J = -106. 4 k. J (b) DGo = [(4)(-394. 4) + (6)(-237. 2) - (1)(-32. 9) - (1)(0)] k. J = -2967. 9 k. J

• For the combustion of one mole of methane in a closed system under a constant pressure of 1. 00 atm and a temperature of 25 o. C: CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(l) Calculate the values of Q, W, ΔUo, ΔHo, ΔSo, ΔGo, ΔSenv, et ΔSuniv • Data (all at 25 o. C): ΔHof (CH 4, g) = -74. 8 k. J/mol ΔHof (CO 2, g) = -393. 5 k. J/mol ΔHof (H 2 O, g) = -285. 8 k. J/mol ΔGof (CH 4, g) = -50. 7 k. J/mol ΔGof (CO 2, g) = -394. 4 k. J/mol ΔGof (H 2 O, g) = -237. 1 k. J/mol

Temperature and Chemical Reactions • The free enthalpy change is given by DG = DH – TDS • If DH is negative and DS is positive, DG is always negative and the reaction occurs spontaneously at all temperatures • If DH is positive and DS is negative, DG is always positive and the reaction is spontaneous in the opposite direction at all temperatures • If DH is positive and DS is positive the reaction takes place spontaneously at elevated temperatures, but at low temperatures, the reaction is spontaneous in the opposite direction • If DH is negative and DS is negative, the reaction occurs spontaneously at low temperatures, but at elevated temperatures, the reaction is spontaneous in the opposite direction

Temperature and Chemical Reactions • With the exact values of DHo and DSo, we can predict the temperature at which the sign of DGo changes, i. e. , the temperature at which the reaction becomes spontaneous (or non-spontaneous) under standard conditions • ex. ; For the reaction Ca. CO 3(s) Ca. O(s) + CO 2(g) • DHo = [DHfo(Ca. O, s) + DHfo(CO 2, g)] - [DHfo(Ca. CO 3, s)] = [(-635. 6)+(-393. 5)] k. J - [-1206. 9] k. J = +177. 8 k. J • DSo = [DSo(Ca. O, s) + DSo(CO 2, g)] - [DSo(Ca. CO 3, s)] = [39. 8+213. 6] J/K - [92. 9] J/K = +160. 5 J/K

Temperature and Chemical Reactions • For our example, at 25 o. C, • The reaction is not spontaneous at 25 o. C • We want to find the temperature where DGo equals zero • In our example, T = (177 800 J)/(160. 5 J/K) = 1108 K or 835 o. C

Temperature and Chemical Reactions • In our example: – The reaction is spontaneous (DGo < 0) under standard conditions at temperatures above 835 o. C – The reaction is non-spontaneous (DGo > 0) under standard conditions at temperatures below 835 o. C N. B. in the above, we are only speaking of spontaneity under standard conditions, i. e. , when the CO 2(g) is present at a pressure of 1 atm

Temperature and Chemical Reactions • In our example, the CO 2(g) will always have a certain pressure that increases with temperature • At 835 o. C, where DGo = 0, the pressure of CO 2(g) finally attains a value of 1 atm • N. B. the equation is only valid if the changes in enthalpy and entropy do not vary with temperature – This is a very good approximation

Phase Changes • during a phase change, there is an equilibrium between the two phases, so • ex. ; For the melting of ice, DH = 6. 01 k. J/mol • ex. ; For the boiling of water, DH = 40. 79 k. J/mol • N. B. for the freezing of water and condensation of water vapour, DS is equal to -22. 0 J/(K mol) and -109. 3 J/(K mol), respectively

Free Enthalpy and Chemical Equilibrium • N. B. DGo is the change in the standard free enthalpy, i. e. , the change when we start with reactants in their standard state (ex. ; gases at 1 atm and solutes at 1 mol/L (or M)) and we also end with products in their standard state • We are often interested in processes that occur under conditions other than the standard conditions • When the conditions are not standard, the enthalpy change is simply DG (rather than DGo)

Free Enthalpy and Chemical Equilibrium • The relation between ΔG and ΔGo is ΔG = ΔGo + RT ln Q where R is the gas constant in SI units (8. 3145 J/(K mol)), T is the absolute temperature, and Q is the reaction quotient • The reaction quotient for the reaction is given by b. B+c. C Where ax = 1 if X is a solid or a liquid = Px (in atm) if X is a gas = [X] (in mol/L) if X is a solute d. D+e. E

Free Enthalpy and Chemical Equilibrium • For arbitrary conditions DG = DGo + RT ln Q • Thus, it is possible to push a reaction to occur spontaneously (in either direction) by playing with the value of Q – ex. ; If DGo is positive, we can make Q << 1 so that RT ln Q is very negative and DG < 0 • An equilibrium is established between the reaction and the reverse reaction when ΔG = 0 because RT ln Q = - DGo

Free Enthalpy and Chemical Equilibrium • At equilibrium, DG = 0, and Q = K so DGo + RT ln K = 0 DGo = -RT ln K • If DGo is negative, K > 1 and the products are more dominate at equilibrium • If DGo is positive, K < 1 and the reactants are more dominate at equilibrium

Free Enthalpy and Chemical Equilibrium • The formula DGo = -RT ln K is one of the most important equations in chemistry because it connects DGo (thermodynamics) with the equilibrium constant • For reactions where either the reactants or the products clearly dominate at equilibrium, we can use DGo to calculate K – ex. ; For the reaction K = 4. 0 x 10 -31 at 25 o. C, therefore it is not really possible to measure PNO in such a situation (because it is very small), so we use DGo to determine K, and afterwards, PNO

Free Enthalpy and Chemical Equilibrium • Example: Calculate the equilibrium constant of the following reaction, at 25 o. C: • Solution: DGfo(O 3, g) = +163. 4 k. J

Free Enthalpy and Chemical Equilibrium • Example: The value of ΔGo for the reaction is 2. 60 k. J at 25 o. C. Over the course of an experiment, the initial pressures are PH 2= 4. 26 atm, PI 2= 0. 024 atm, and PHI = 0. 23 atm. Calculate ΔG for the reaction and determine the direction of the reaction. • Solution: Therefore HI(g) decomposes spontaneously to reform the H 2(g) and I 2(g) under these particular conditions

Van't Hoff Equation • We can determine the values of DHo and DSo from the equilibrium constants (K 1 and K 2) at two different temperatures (T 1 and T 2, respectively)

Van't Hoff Equation • Taking the difference, • The above equation is the Van't Hoff equation • Once we know DHo (from the Van't Hoff equation) and DGo (from DGo = - RT ln K), we can calculate DSo (where T is the temperature that corresponds to the DGo value that we calculated)

• For the reaction A(aq) + B(aq) D 2 C(aq) the equilibrium constant is 10. 3 at 25 o. C and 17. 7 at 45 o. C. Making the approximation that ΔHo and ΔSo do not vary with temperature, calculate the values of ΔHo, ΔSo, ΔGo at 75 o. C, and the equilibrium constant at 75 o. C.