Chemical Thermodynamics

Recall that, at constant pressure, the enthalpy change equals the heat transferred between the system and its surroundings.  H = H final – H initial = q p Thermodynamics involves enthalpy changes AND changes in order/disorder. spontaneous processes: ones that occur without outside intervention The First Law of Thermodynamics: Energy is conserved.

 E = q + w work done heat transferred changes in system’s internal energy If q is ___, system released heat. If q is ___, system absorbed heat. If w is ___, system did work. If w is ___, system had work done on it. + – – + Sign conventions are from system’s point of view.

reversible process: “undo exactly what you did and you’ve got what you started with” -- both system and surroundings go back to original states -- e.g., 0 o 0 o C add X J take away X J changes of state

irreversible process: getting back what you started with requires more than just an “undo” -- we can restore the original system, but the surroundings will have changed -- e.g., a gas expanding into an evacuated space piston vacuumgas

Points of note: 1. Whenever a chemical system is in equilibrium, we can go reversibly between reactants and products. 2. In any spontaneous process, the path between reactants and products is irreversible. 3. Thermodynamics refers to the direction of a reaction, not its speed. 4. In general, exothermic processes are more likely to be spontaneous.

Processes in which the system’s disorder increases tend to occur spontaneously. entropy, S  S = S f – S i +  S  –  S  P are more disordered than R P are more ORDERED than R large entropy = large # of microstates = very disordered small entropy = small # of microstates = less disordered

+ Entropy at the Particle Level There are three types of motion, each having kinetic energy (KE). -- translational, vibrational, rotational -- The more KE we have, the more _______ we have. entropy In general… and… S solid < S liquid < S gas As T /, S /. Processfreezingmelting condensing boiling Sign of  S + – – (i.e., the more microstates are possible)

The Second Law of Thermodynamics: The entropy of an isolated system that is NOT in equilibrium will increase over time. Sadi Carnot (1796–1832) Rudolf Clausius (1822–1888) The entropy of the universe increases in any ___________ process. spontaneous Entropy is NOT conserved; it is constantly increasing. For isolated systems… rev.   S sys = 0 irrev. (i.e., spont.)   S sys > 0

Find the change in entropy when 87.3 g of water vapor condense, given that water’s heat of vaporization is 5.99 kJ/mol K = – = –77.9 For a system in which heat is transferred at constant temperature… -- T in K -- common unit for entropy  i.e., heat (energy) temperature

Third Law of Thermodynamics: The entropy of a pure, crystalline substance at absolute zero is… -- that would be a state of perfect order ZERO. (impossible)

Entropy increases when: 1. the number of gas particles increases 2. liquids or solutions are formed from solids 3. gases are formed from liquids or solids 2 NH 3 N H 2 fewer bonds; fewer restrictions on motion of atoms; more degrees of freedom; more possible microstates +S+S

Which has the greater entropy? 1 mol 300 K or 1 mol 300 K 1 mol O K or 1 mol O K 2 mol 300 K or 4 mol 300 K (same volume) 1 mol 300 K or 1 mol 300 K (same volume) 2 mol 300 K or 2 mol 300 K (in a 5-L vessel)(in a 10-L vessel) (same volume)

Calculating Entropy Changes standard molar entropies, S o : molar entropy values of substances in their standard states (i.e., pure substances at ~1 atm) S o values typically… -- actually, at 1 bar = 10 5 Pa = atm In a chemical reaction… are NOT zero w /increasing molar mass w /increasing # of atoms in formula  S o =  nS o P –  mS o R (n and m are the coeff. for each substance)

Calculate the standard entropy change for… P 4 (s,white) + 10 Cl 2 (g)  4 PCl 5 (g) tabulated values of S o in J/mol-K  S o =  nS o P –  mS o R  S o = 4(353) – [ (223)] = –995J/K White phosphorus (or “WP”) is used in bombs, artillery shells, and mortar shells that burst into burning flakes of phosphorus upon impact. Smokescreen or flesh-fryer?

Gibbs Free Energy Enthalpy changes (  H) and entropy changes (  S) both have a “say” in whether or not a rxn is spontaneous. Spontaneity is determined using the equation for Gibbs free energy… Josiah Willard Gibbs 1839–1903  G (o) =  H (o) – T  S (o) If  G < 0… If  G > 0… If  G = 0… rxn. is spontaneous (i.e., as written) rxn. is AT equilibrium rxn. is nonspontaneous (i.e., spont. ) (THIS INCLUDES PHASE CHANGES AT NBP OR NFP) (o) = std. conditions are optional

For the Haber process… N 2 (g) + 3 H 2 (g) 2 NH 3 (g) spontaneous pure N 2 + H 2 pure NH 3 equilibrium mixture (Q = K,  G = 0) Q < K  G < 0 Q > K  G > 0

standard free energies of formation,  G f o -- are tabulated for pure solids, pure liquids, gases at ~1 atm pressure, and 1 M solutions -- For elements in their standard states…  G f o = 0 -- For a reaction, the standard free-energy change is found by…  G o =  nG f o P –  mG f o R  G says WHICH WAY a reaction will proceed, but it says NOTHING about the reaction rate.

Calculate the standard free-energy change for… PCl 3 (g) + Cl 2 (g)  PCl 5 (g) –286.30–324.6 tabulated  G f o ’s in kJ/mol  G o =  nG f o P –  mG f o R  G o = –324.6 – –286.3 = –38.3 kJ rxn. is spontaneous as written (i.e., left to right)

Free Energy and Temperature From  G =  H – T  S, we see that  G varies with temperature. -- When T changes, so does  G. --  H and  S change little with temperature. EX. (a) Calculate  H o,  G o, and  S o for… 2 NO(g) + O 2 (g)  2 NO 2 (g)  H f o (kJ/mol)  G f o (kJ/mol) S o (J/mol-K)

 H o = 2(33.2) – [2(90.3)]  G o = 2(51.8) – [2(86.7)]  S o = 2(240.0) – [ (210.7)]  H o = –114.2 kJ  G o = –69.8 kJ  S o = –146.4 J/K (b) Estimate  G at 400 K.  G =  H – T  S = –114,200 J – [400 K(–146.4 J/K)] = –55,600 J = –55.6 kJ (Remember:  H and  S vary very, VERY little w /temp.)

Estimate the normal boiling point of ethanol. At the NBP… CH 3 CH 2 OH(l) CH 3 CH 2 OH(g) Strategy: Realize that  G = 0. Find  S and  H. Solve for T in  G =  H – T  S.  H f o (kJ/mol) S o (J/mol-K) – –  S =  S o = ~  H =  H o = ~ 122 J/K 42.6 kJ = 349 K (76 o C) LG From  G =  H – T  S… 0

From  G =  H – T  S, we see that spontaneity (i.e.,  G) depends on T. We estimated 76 o C; the actual NBP of ethanol is 78.4 o C. As T increases,  G becomes (–) Assume we start at 78.4 o C, where  G = 0… As T decreases,  G becomes (+) -- spontaneous, liquid to gas -- spontaneous, gas to liquid  H = 42,600 J;  S = 122 J/K

Free Energy and the Equilibrium Constant  G o is the standard free-energy change (i.e., for a reaction at standard conditions). Under any other conditions…  G =  G o + RT ln Q R = J/mol-K Q = rxn. quotient Comparatively few reactions take place under standard conditions.

 G =  G o + RT ln Q =  G o = 2(–17)  G f o (kJ/mol) = –34 kJ  G = = –34,000 J  G = –45.6 kJ –34, (298) (ln ) 0 0 –17 (non-standard pressures) Calculate  G at 298 K, given the following: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 1.0 atm 3.0 atm0.50 atm

At equilibrium,  G = 0 and Q = K, so..  G (o) = –RT ln K and Calculate  G o and K at 298 K for… H 2 (g) + Br 2 (g) 2 HBr(g)  G f o (kJ/mol) –53.22  G o = 2(–53.22) – 3.14= – kJ = 1.6 x For gas-phase rxns, the K is K p ; for rxns. in soln, it is K c. Assume all gases are at 1.00 atm of pressure. Rxn. is spont. as written; eq. lies to the right; P are favored.

What if we are relating/trying-to-find  G and/or K, but NOT at standard conditions? -- To approx. K, given rxn eq and nonstandard T: 1. Use tabulated values to calc.  H o and  S o, which are “good” for all temps. 2. Use  G =  H – T  S and answers above to approx.  G at desired temp. 3. Use K = e –  G/RT to approx. K at desired temp. -- To approx. K, given  G and nonstandard T: Use K = e –  G/RT -- To approx.  G, given K and nonstandard T: Use  G = –RT ln K