Chapter 20: Thermodynamics

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1 Chapter 20: Thermodynamics Thermodynamics is the study of energy (including heat) and chemical processes. First Law of Thermodynamics: Energy cannot be created nor destroyed. E universe = E system + E surroundings E universe = E system + E surroundings If E universe = 0, then E system = E surroundings The first law of thermodynamics explains the energy exchange between the system and surroundings. Does it explain the direction a reaction proceeds toward equilibrium? For systems under constant pressure and no work, H = E. Does the sign of H tell us the direction a reaction proceeds toward equilibrium? Let s try the Second Law of Thermodynamics. 1

2 Second Law of Thermodynamics: The total entropy (S) of a system and its surroundings increases for spontaneous reactions. A spontaneous reaction is one that occurs naturally under certain conditions without a continuous input of outside energy. Note: If a reaction is spontaneous in one direction, it will be nonspontaneous in the reverse direction. Consider the following diagram: When the valve is opened some of the gas particles move spontaneously into the other side as shown in B. Will all of the molecules spontaneously move back as depicted in A? In B of the diagram above, the gas particles occupy twice the volume thus the gas particles can exist in twice as many locations. Thus greater disorder. Disorder S disorder > S order, in other words, the entropy of disordered matter is greater than the entropy of ordered matter. Examples to consider: Ice melting disorder S system increases Water freezing order S system decreases 2

3 All processes occur spontaneously in the direction that increases the total entropy of the universe. Therefore, S universe > 0, S system + S surroundings > 0 Let s consider two rxns, one exothermic and one endothermic. 4 Fe (s) + 3 O 2(g) 2 Fe 2 O 3(s) + Heat Now, let s consider another reaction. 3

4 Ba(OH) 2. 8H 2 O (s) + 2 NH 4 NO 3(s) + Heat Ba 2+ (aq) + 2NH 3(g) + 2 NO 3 - (aq) + 10H 2 O (l) Therefore changes in entropy of the system and the surroundings must be important to the spontaneity of reactions. Third Law of Themodynamics: A perfect crystalline substance at 0 K has 0 entropy. Perfect means that all of the particles are flawlessly aligned in the crystal with no imperfections. Standard Molar Entropies (S o ) at 298 K are on pg A-5 of text. Units: J/mol-K Note: Standard Molar Enthalpies ( H o ) are also located on pg A-5. Let s Make Some Comparisons Temperature versus Entropy Copper metal T(K) S o (J/mol-K) What general statement can you make about this comparison? 4

5 Physical State Compared with Entropy Values Substance S o (J/mol-K) Ba (s) 62.5 Ba (g) Br 2(l) Br 2(g) C (s) 5.7 C (g)

6 Comparison of the Solid State with Aqueous Solution Substance Entropy (J/mol-K) NaCl (s) 72.1 NaCl (aq) AlCl 3(s) 167 AlCl 3(aq) -148 CH 3 OH (l) 127 CH 3 OH (aq) 132 Note: When gases are dissolved in liquids the system decreases in entropy. Explain. 6

7 Compare Molar Mass and Entropy Substance MW (g/mol) Entropy (J/mol-K) Cu Au Ag What general statement can you make about this comparison? Compare Compound Complexity with Entropy 7

8 Substance Entropy (J/mol-K) MgCO Na 2 CO C 2 H CH What general statement can you make about this comparison? Calculating the Change in Enthalpy of a Reaction Recall the equation: H o rxn= Σ n H o products Σ n H o reactants We can use a similar equation to calculate the change in entropy of a reaction. S o rxn= Σ ns o products Σ ns o reactants Problem Solving Calculate the change in entropy of the combustion of methane. H 2 O (g) CO 2(g) CH 4(g) O 2(g) S o (J/mol-K) Did the entropy of the system increase or decrease? What does this mean for the surroundings? 8

9 Entropy of the Surroundings In order for S universe to be increasing if S system decreases for a spontaneous reaction, then S surroundings must increase. Let s consider the following diagram. Exothermic Rxn Exothermic Rxn Endothermic Rxn Let s Talk Heat Exchange ( Η) If heat is lost to the surroundings, then the entropy of the surroundings should increase. If heat is gained by the surroundings, then the KE goes up and so does the entropy. 9

10 If heat is gained by the system (lost by the surroundings), the system becomes more disordered (entropy increases). Also, If the temperature of the surroundings is low, the particles are more ordered and a change in heat of the surroundings would have an even greater effect on the change in entropy. Then S surroundings is directly related to an opposite sign of H system, and inversely related to temperature at which the heat is transferred. Mathematically Speaking S surroundings is proportional to -q system S surroundings is proportional to 1/T S surroundings = -q system /T S surroundings = - H system /T So, calculate S surroundings for the combustion of methane at 25 o C from H fo values. H fo (CH 4 ) = kJ/mol, H fo (CO 2 ) = kj/mol, and H fo (H 2 O) = kJ/mol Using your information, calculate S universe for the combustion of methane. Did the entropy of the universe increase? 10

11 Entropy and Equilibrium What is the change in entropy of the universe for rxns at equilibrium? S universe = Equilibrium S surroundings + S system = S universe Thus, S surroundings = - S Equilibrium Problem Calculate the change in entropy for the following reaction: H 2 O (l) H 2 O (g) S(J/mol-K) K If the change in enthalpy is 40,700 J/mol at 373 K, what is the change in entropy of the universe? Entropy, Free Energy, and Work One criterion for spontaneity is Gibbs Free Energy ( G) Proof: S surroundings + S system = S universe S surroundings = - H system /T - H system /T + S system = S universe H system -T S system = -T S universe Given that: -T S universe = G universe H system - T S system = G universe 11

12 Gibbs Free Energy and Spontaneity If G < 0, The rxn is spontaneous If G > 0, the rxn is nonspontaneous If G = 0, the rxn is at equilibrium Note: G o fvalues are on page A-5. These values for an element in its most stable state are 0. Also, if a rxn has a particular value for G in one direction, it will be the same value with reversed sign in the opposite direction. Just like Η. ;) Problem Determine whether or not the following reaction is spontaneous. Use TWO methods and compare: (a) use only delta G values (kj/mol), (b) use delta H o f (kj/mol) and S o (J/mol-K) values 4 KClO 3(s) 3 KClO 4(s) + KCl (s) H o f S o G o f Most exothermic reactions are spontaneous. However, temperature will affect the T S factor of the equation: H system - T S system = G universe Therefore, reaction spontaneity can change with temperature. Example: At 298 K, G o = kJ, H o = kj, and S o = J/K for a reaction. (1) Is the rxn spontaneous at 298K? (2) If temperature is increased what happens to G o? (3) Assuming Η o and S o are constant with temperature, is the rxn spontaneous at 900. o C? 12

13 Crossover Temperature At what temperature does a reaction become spontaneous? + G o == - G o At some point G o equals 0 equil) Thus, H system - T S system = G universe H system - T S system = 0 H system = T S system or H system / S system = T crossover What is the crossover temperature for the previous problem? Problem: Determine if the following reaction is spontaneous at 298K. Also determine the crossover temperature. Cu 2 O (s) + C (s) 2 Cu (s) + CO (g) S o = 165 J/K and H o = 58.1 kj 13

14 Free Energy, Equilibrium, and Rxn Direction Recall, If G < 0, rxn is spontaneous, product favored, and Q is < K. If G > 0, rxn is nonspontaneous, reactant favored, and Q > K. If G = 0, rxn is at equilibrium and Q = K. Given that G o rxn is proportional to ln (Q/K) and their signs are identical for a given reaction, we have: G rxn = RT ln (Q/K) G rxn = RTln Q - RTln K If we set [ ] s = 1.00 M for sol ns and 1.00 atm for gases (std conditions), then, G o rxn = RTln 1 - RTln K Or G o rxn = - RTln K Note: K can be determined for any rxn at any temp if we know G o rxn 14

15 Problem Solving Determine the thermodynamic equilibrium constant at 25 o C for the following rxn: N 2 O 4(g) 2 NO 2(g) G o f (kj/mol) However, most reactions are not with 1.0 M solutions and/or 1.00 atm for gases, thus we also need to remember this form of the equation: G rxn = RT lnq - RT lnk Since G o rxn = -RT lnk G rxn = G o rxn+ RTlnQ The above relationship is very important to our next chapter, you will definitely see it again. 15

16 Problem Solving Consider the following reaction: 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) G o rxn kj at 298K and kj at 973K. (a) Determine K eq at 298K. (b) Determine K eq at 973K. (c) If a vessel contains 0.500M SO 2, M O 2, and 0.100M SO 3 at 298K, what is G rxn? (d) Under the conditions in (c) which way would the reaction proceed? More Problem Solving Consider the reaction: N 2 (g) + 3H 2 (g) 2 NH 3 (g) S o (J/mol-K) H o (kj/mol) (a) Determine the equilibrium constants at 298 K and at 984 K. (b) What is the cross-over temperature for this reaction in Kelvin. (c) If, at 298K, a vessel contains 0.250M H 2, M N 2, and 12.9M NH 3, predict the direction the reaction will proceed at this temperature. 16

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