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Chapter 19: Chemical Thermodynamics

Chapter 19: Chemical Thermodynamics. 2 KClO 3. 2 KCl + 3 O 2. Spontaneity of endothermic and exothermic reactions. 2 KClO 3  2 KCl + 3 O 2. The reaction proceeds spontaneously once the KClO 3 has been heated to a high temperature.

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Chapter 19: Chemical Thermodynamics

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  1. Chapter 19: Chemical Thermodynamics 2 KClO3 2 KCl + 3 O2 Spontaneity of endothermic and exothermic reactions. 2 KClO3 2 KCl + 3 O2 The reaction proceeds spontaneously once the KClO3 has been heated to a high temperature. If the reaction were endothermic, you would have to continuously provide heat or the reaction would stop. Thus the reverse reaction, 2 KCl + O2 2 KClO3, is not likely to occur spontaneously.

  2. Spontaneous for T > 0C Spontaneous for T < 0C water ice Recall that spontaneous ≠ a fast reaction rate Spontaneous only implies that the reaction can occur without outside assistance. A spontaneous process is one that proceeds without any outside assistance. • A process that is spontaneous in one direction is nonspontaneous in the reverse direction.

  3. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB

  4. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB Second premise: Reactants must have the correct orientation to form the products upon collision

  5. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB E < Ea Second premise: Reactants must have the correct orientation to form the products upon collision Third premise: Reactants must have sufficient energy for the collision to result in formation of products

  6. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB E > Ea Second premise: Reactants must have the correct orientation to form the products upon collision Third premise: Reactants must have sufficient energy for the collision to result in formation of products

  7. Ea Energy DHrxn Reaction progress Activated complex: an unstable transition state between reactants and products. It can either fall back down on the reactant side or go on to the product side.

  8. Ea Ea Energy DHrxn Reaction progress A catalyst speeds up the rate of a reaction by lowering the activation barrier

  9. Ea Energy DHrxn Reaction progress A catalyst speeds up the rate of a reaction by lowering the activation barrier

  10. The Ea for an endothermic reaction is: DHrxn + energy barrier Ea Energy DHrxn Reaction progress

  11. 2 1 DU is the same for both blocks (recall definition of a state function) However, q and w for both blocks will depend on how quickly the blocks are moved. If the blocks are moved using infinitesimally small steps, the amount of heat will be zero and the amount of work done will equal DU. A process is reversible when it can be restored exactly to its initial state without a loss of energy to the surroundings.

  12. Reversible processes are NOT spontaneous. • All spontaneous processes are irreversible. Entropy, S, is a measure of the degree of randomness, or disorder of a system, and has a positive, nonzero value for all substances above absolute zero. H2 (l) < H2 (g) < 2 H (g) DS, a state function, is a measure of the change in the degree of order in a system and can thus be positive, negative or equal to zero. (at constant Temp)

  13. Heating curve of water Gas warming 100 Temperature (°C) liquid + gas present liquid warming solid + liquid present 0 solid warming Heat (kJ/s)

  14. Heating curve of water 100 Temperature (°C) boiling/condensation point melting/freezing point 0 Heat (kJ/s) Temperature is constant during phase transitions!! All heat energy goes to changing the state of matter.

  15. Heating curve of water 100 Temperature (°C) DHvap (heat of vaporization) 0 DHfus (heat of fusion) Heat (kJ/s) DHfus = the amount of heat needed to covert a solid into its liquid phase DHvap = the amount of heat needed to convert a liquid into its gaseous phase

  16. Entropy and phase changes • Temperature is constant during any phase change. • The process is reversible. • The amount of heat energy involved is equal to DH. EX: Use Appendix C to calculate the boiling point of CCl4. CCl4 (l)  CCl4 (g) DH = Hf(product) – Hf(reactant) = -106.7 kJ – (-139.3 kJ) = 32.6 kJ DS = S(product) – S(reactant) = 309.4 J – 214.4 J = 95.0 J T = 3.26 x 10 4 J/95.0 J = 343 K (lit: 350 K)

  17. The Second Law of Thermodynamics: The total entropy of the Universe increases in any spontaneous process. Reversible process:DSuniv = DSsystem + DSsurroundings = 0 Irreversible process:DSuniv = DSsystem + DSsurroundings > 0 The Third Law of Thermodynamics: The entropy of a perfect crystal at absolute zero is zero.

  18. Three Laws of Thermodynamics(paraphrased): First Law: You can't get anything without working for it. Second Law: The most you can accomplish by work is to break even. Third Law: You can't break even. OR… First Law: You can't win. Second Law: You can't even break even. Third Law: You can't get out of the game. THE LAW OF ENTROPY: The perversity of the universe tends towards a maximum.

  19. Hess’s Law states that if a reaction is carried out in a series of steps, ΔH for the overall reaction will equal the sum of the enthalpy changes for the individual steps. EX: The enthalpies of reaction for the combustion of C to CO2 and for CO to CO2 are shown below: DH = -393.5 kJ C + O2 CO2 CO + ½ O2 CO2 DH = -283.0 kJ flip What is the DHrxn for C + ½O2 CO ? DH 0.5 C + O2 CO2 -393.5 kJ + + CO2 CO + ½ O2 +283.0 kJ C + ½ O2 CO -110.5 kJ

  20. C + O2 CO + ½ O2 CO2 An Enthalpy Diagram DH = —110.5 kJ DH = -393.5 kJ DH = —283.0 kJ

  21. S = k lnW Boltzman’s Equation k = Boltzman’s constant = 1.38 × 10−23 J/K W = number of microstates = number of equivalent ways that the particles in the system can be arranged. When you’re talking about 6.022 x 1023 particles…that’s a LOT of microstates!! Entropy Activity In general, W and thus entropy increase with: • Temperature • Volume • Number of independently moving particles

  22. Is DS positive or negative? (a) CO2(s) CO2(g) DS > 0 because gas is formed (b) CaO(s) + CO2(g)  CaCO3(s) DS < 0 because gas was consumed by reaction (c) HCl(g) + NH3(g)  NH4Cl(s) DS < 0 because gas was consumed by reaction (d) 2 SO2(g) + O2(g)  2 SO3(g) DS < 0 because there are fewer moles of gas on the product side than on the reactant side.

  23. Appendix C lists standard molar entropies, So (1 atm, 298 K). Some guidelines to help with predicting the sign of DS: • Unlike enthalpies of formation, the standard molar entropies of elements at the reference temperature of 298 K are not zero • The standard molar entropies of gases are much, much greater than those of liquids and solids. • Standard molar entropies generally increase with increasing molar mass. [Compare Li(s), Na(s), and K(s).] • Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance due to the greater number of modes of motion.

  24. H2O (l) H2O (g) H2O (g) DHvap = +40.7 kJ/mol H2O (l) Energy Entr o p y • Water will spontaneously evaporate at room temperature even though this process is endothermic. • What is providing the uphill driving force?

  25. ClF (g) + F2(g) ClF3(g) CH3OH (l) CH3OH (aq) Entr o p y a measure of the disorder or randomness of the particles that make up a system • Water will spontaneously evaporate at room temperature because it allows the disorder of the water molecules to increase. • The entropy, S, of gases is >> than liquids or solids. • If Sproducts > Sreactants, DS is > 0 Predict the sign of DS: DS < 0 DS > 0

  26. 2 H2O (l) 2 H2 (g) + O2 (g) This is a very nonspon- taneous process!! G = H T S Are all +DS reactions spontaneous? Free Energy is energy that is available to do work. DS is large and positive… …but DH is large and positive as well. • Gibb’s Free Energy, DG, allows us to predict the spontaneity of a reaction using DH AND DS. If –DG  spontaneous reaction

  27. 2 H2O (l) 2 H2 (g) + O2 (g) DHrxn = [2(0) + 0] - 2(-285.83 kJ/mol) = 571.66 kJ/mol DSrxn = 326.34 J/molK = 0.32634 kJ/molK What is DG for this reaction at 25C? DHrxn = SHf(products) –SHf(reactants) DSrxn = SSf(products) –SSf(reactants) DSrxn = [2(130.58 J/molK) + 205.0 J/molK] - 2(69.91 J/molK) DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK) DGrxn = +474.41 kJ/mol +DG = nonspontaneous

  28. 2 H2O (l) 2 H2 (g) + O2 (g) What is the minimum temperature needed to make this reaction spontaneous? DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - T(0.32634 kJ/molK) Set DGrxn = 0 to find minimum temperature 0 = 571.66 kJ/mol - T(0.32634 kJ/molK) T = (571.66 kJ/mol)/(0.32634 kJ/molK) = 1751 K T > 1478 C

  29. DH DS DG + Always spontaneous Spontaneous at low T + + Spontaneous at high T + Never spontaneous Remember, Free Energy is the energy that’s available to do work on the surroundings. Hence, DG = -wmax Q: What does it mean if DG = 0? Rate of forward reaction = rate of reverse reaction The system is at equilibrium!

  30. “Of shoes and ships and sealing wax and cabbages and Kings…” …or “What if conditions aren’t quite standard?” 3 H2 + N2 2 NH3 DG = 0 at equilibrium Q: What is DG when you first mix the reactants together?  Recall that the equilibrium constant for a reaction, K, is: When the reaction first starts, it is NOT at equilibrium. We use a reaction quotient, Q, to evaluate how far from equilibrium it is.

  31. EX: Is the reaction for the synthesis of ammonia at equilibrium when there are 2 moles of nitrogen and hydrogen, and 1 mole of ammonia in a 2 L flask? Step 1: Determine the concentrations of the reactants and products  Recall that the square brackets indicate that concentrations are in terms of moles/liter (M). [H2] = 2 moles/2 L = 1 M [N2] = 2 moles/2 L = 1 M [NH3] = 1 moles/2 L = 0.5 M Step 2: Calculate the reaction coefficient

  32. Step 3: Calculate the equilibrium constant from DGo IF the reaction is being done under standard conditions, you can use the tabulated DGf values in Appendix C to calculate DGo for the reaction 3 H2 + N2 2 NH3 K = 7.0 x 105

  33. DG = -3.7 x 104 J = -370 kJ Step 4: Compare Q and K Q = 0.25 K = 7.0 x 105 Nope. Is this reaction anywhere near equilibrium?? EX: Calculate DG for the synthesis of ammonia given the previous reaction conditions. DG = DG + RTlnQ  The fact that the reaction is far from equilibrium has increased the spontaneity of the reaction (increased its driving force).

  34. Au2O3 (s) + 2 Fe (s) 2 Au (s) + Fe2O3 (s) Chemical Equilibria: The basics Irreversible reactions: The reactants combine to form products and the reaction stops when the limiting reagent has been completely used up. “Only” the forward reaction occurs. Ex: Reversible reactions: Both the forward and the reverse reaction occur to a significant degree. Forward: 3 H2(g) + N2(g) 2 NH3(g) 3 H2(g) + N2(g) 2 NH3(g) Reverse: 3 H2(g) + N2(g)⇋2 NH3(g) Equilibrium: A reaction is at equilibrium when Rate forward rxn = Rate reverse rxn

  35. General form of equilibrium constant, Keq: aA + bB ⇋cC + dD If Keq > 1, then more products than reactants present at equilibrium If Keq < 1, then more reactants than products present at equilibrium

  36. [NH3] Keq = [N2] [H2] Concentration Time ⇋ N2(g) + 3 H2(g) 2 NH3(g) 2 H2 3 N2 NH3

  37. Heterogeneous equilibria • If the reactants and products are in more than one state, the equilibrium constant is for a heterogeneous equilibria. • Because the concentration of a solid does not change significantly compared to gaseous or liquid reactants or products, solids NEVER appear in equilibrium constant calculations. Ex: BaCl2 (s)⇋ Ba2+(aq) + 2 Cl-(aq) • Since this particular type of equilibrium involves the solubility of the product, it is given a special designation: • Ksp = solubility product constant

  38. [CO][H2] Keq = [H2O] [CO2] Keq = [CO] Heterogeneous equilibria (cont.) Since the concentration of a liquid solvent that the reaction takes place in also doesn’t change significantly, it also doesn’t appear in the equilibrium calculation. Ex: H2O (l)⇋ H2O (g) • What are the equilibrium constants for the following: • C10H8(s)⇋ C10H8(g) • CaCO3(s)⇋ CaO (s) + CO2(g) • C (s) + H2O (g)⇋ CO (g) + H2(g) • FeO (s) + CO (g)⇋ Fe (s) + CO2(g) Keq = [C10H8] Keq = [CO2]

  39. Le Chatelier’s Principle: When a stress is applied to an equilibrium, the equilibrium will shift to alleviate the stress. Fe+3(aq) + SCN-1(aq) ⇋ FeSCN+2(aq) Colorless ⇋ Dark red Reactants Products Left shift = lighter color Right shift = darker color Initial color

  40. Other ways to cause a Le Châtelier Shift: 3 H2(g) + N2(g) + heat ⇋ 2 NH3(g) What kind of shift would you see and how would G (temporarily) be affected if: Pressure increased? Right shift  G more negative Volume increased? • Left shift • G more positive Right shift  G more negative Heating temperature increased?

  41. CH4(g) + 2 Cl2(g) ⇋ CCl4(g) + 2 H2(g) + heat What kind of shift would you see if: Pressure increased? No Change G remains zero Heating temperature increased? • Left shift • G more positive

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