Chapter 10: Phenomena Phenomena: Below is data from several - PowerPoint PPT Presentation

Chapter 10: Phenomena Phenomena: Below is data from several different chemical reactions. All reaction were started by putting some of every substance in the chemical reaction into an expandable/contractable container at 25 ˚C . If reactants/products were in a gas state, 1 atm of the gas was added to the container. If reactants/products were in an aqueous state, 100 mL of water was added to the container and then the concentrations were adjusted to 1M using solids. If the reactants/products were in a liquid or solid state, 10 g of the substance was added. Therefore, for the reaction: 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) 10 g of Fe, 10 g of Fe 2 O 3 , and 1 atm of O 2 would be put into the reaction container. Looking at the thermodynamic data that was gathered see what patterns you can identify. (You do not need to know what Δ G ˚ or Δ S ˚ are to identify the patterns.) Direction Exp. Reaction Reaction Δ E ˚ Δ H ˚ Δ G ˚ Δ S ˚ Runs -1,635 𝑙𝐾 -1,642 𝑙𝐾 -1,484 𝑙𝐾 -549 1 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) Forward 𝐾 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚∙𝐿 NH 4 NO 3 (s)  26 𝑙𝐾 26 𝑙𝐾 -6 𝑙𝐾 108 2 Forward 𝐾 NH 4 + (aq) + NO 3 - (aq) 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚∙𝐿 4AlCl 3 (s) + 9O 2 (g)  679 𝑙𝐾 686 𝑙𝐾 520 𝑙𝐾 545 3 Reverse 𝐾 2Al 2 O 3 (s) + 12ClO(g) 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚∙𝐿 4NH 3 (g) + 5O 2 (g)  -911 𝑙𝐾 -908 𝑙𝐾 -958 𝑙𝐾 181 4 Forward 𝐾 4NO(g) + 6H 2 O(g) 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚∙𝐿 -6 𝑙𝐾 -6 𝑙𝐾 0.5 𝑙𝐾 -22 5 H 2 O(l)  H 2 O(s) Reverse 𝐾 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚 𝑛𝑝𝑚∙𝐿 Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy

Chapter 10 Spontaneity, Entropy, & Free Energy o Entropy o ΔS of Physical Reactions o Isothermal Processes Big Idea: The change in free o 2 nd Law of Thermo energy of a reaction o Free Energy indicates whether a o Hess’s Law/ 3 rd Law of reaction is Thermo spontaneous. In any o Equilibrium spontaneous process there is always an increase in the entropy of the universe. 2

Entropy # of molecules # of ways of of left side arranging  Entropy (S): Entropy is (microstates) 4 a measure of how 3 energy and matter 2 can be distributed in a 1 chemical system. 0 Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 3

Entropy In General:  Entropy increases from solid to liquid to gas corresponding to an increase in positional probability.  Entropy increases when you dissolve a solid in liquid corresponding to an increase in positional probability.  The larger the volume the larger the positional probability and the greater the entropy (n constant).  The larger the pressure the smaller the positional probability and the lower the entropy (n constant).  The larger the molecule the larger the number of relative positions of the atoms resulting in a greater positional probability and a greater entropy.  The higher the temperature the greater the range of energies, therefore the larger the entropy. Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 4

Entropy Student Question Predict which of the following reactions has a negative entropy change. I. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) II. NH 3 (g) + HCl(g)  NH 4 Cl(s) III. 2KClO 4 (s)  2KClO 3 (s) + O 2 (g) a) II and III b) III c) II d) I e) I and II Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 5

Entropy  Phase Change: The condition (for a given pressure, and temperature) at which two different phases are in dynamic equilibrium.  Melting/Freezing Liquid/Solid  Evaporation/Condensation Liquid/Gas  Sublimation/Deposition Solid/Gas Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 6

Δ S of Physical Reactions Calculating ΔS for physical reaction X(s, T i )  X(g, T f )  Step 1: Calculate ΔS ∆𝑇 = 𝑛𝐷 𝑡𝑝𝑚𝑗𝑒 𝑚𝑜 𝑈 𝑁 𝑈 𝑗 to bring to melting point  Step 2: Calculate ΔS  n H   fus S involved in fusion T  Step 3: Calculate ΔS 𝑈 𝐶 ∆𝑇 = 𝑛𝐷 𝑚𝑗𝑟𝑣𝑗𝑒 𝑚𝑜 𝑈 𝑁 to bring to boiling point  Step 4: Calculate ΔS  n H   vap involved in vaporization S T  Step 5: Calculate ΔS 𝑈 𝑔 ∆𝑇 = 𝑜𝐷 𝑄 𝑕𝑏𝑡 𝑚𝑜 to bring to final temperature 𝑈 𝐶 Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 7

Δ S of Physical Reactions Student Question What is ΔS for 88.0 g of CO 2 undergoing the following reaction at constant pressure? CO 2 (s, 150. K)  CO 2 (g, 195. K) Helpful Information: 𝑈 𝑡𝑣𝑐 = 195𝐿 , ∆𝐼 𝑡𝑣𝑐 = 25.2 𝑙𝐾 𝑛𝑝𝑚 , 𝐷 𝐷𝑃 2 (𝑡) = 1.07 𝐾 𝑕∙𝐿 a) 24.9 𝐾 𝐿 b) 154 𝐾 𝐿 c) 233 𝐾 𝐿 d) 283 𝐾 𝐿 e) None of the above Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 8

Isothermal Processes Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 9

Isothermal Processes 2 Step Isothermal Expansion 6 Step Isothermal Expansion P P 1 1 Pressure Pressure P 1 2 P 1 2 P 1 4 P 1 4 V 2 V 4 V V 2 V 4 V 1 1 1 1 Volume 1 1 Volume ∞ Step Isothermal Expansion  Reversible Process: A P process that can be 1 Pressure reversed by an infinitesimal change in a variable. P 1 2 P Not ote: In order for a reversible process to 1 4 occur the system must be at equilibrium during the entire process. V 2 V 4 V 1 1 1 Volume Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 10

Isothermal Processes Student Question Calculate the ΔS associated with a process in which 5.00 mol of gas expands reversibly at constant temperature T = 25°C from a pressure of 10.0 atm to 1.00 atm. a) 28,500 𝐾 𝐿 b) 95.7 𝐾 𝐿 c) -95.7 𝐾 𝐿 d) -28,500 𝐾 𝐿 e) None of the above Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 11

2 nd Law of Thermo 2 nd Law of Thermodynamics: A spontaneous change is accompanied by an increase in the total entropy of the system and its surroundings.  ∆𝑇 𝑣𝑜𝑗 = ∆𝑇 𝑡𝑧𝑡 + ∆𝑇 𝑡𝑣𝑠 Not ote: The second law of thermodynamics applies to Δ S uni and not Δ S sys . So far we have only discussed Δ S sys . Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 12

Free Energy Can we relate spontaneity to a change in the system instead of the universe assuming we are at constant temperature and pressure?  ∆𝐻 𝑡𝑧𝑡 = ∆𝐼 𝑡𝑧𝑡 − 𝑈∆𝑇 𝑡𝑧𝑡  Divide Through by T ∆𝐻 𝑡𝑧𝑡 ∆𝐼 𝑡𝑧𝑡 𝑈∆𝑇 𝑡𝑧𝑡 = −  𝑈 𝑈 𝑈 Δ S univ >0 then Δ G sys <0 ∆𝐻 𝑡𝑧𝑡 = −∆𝑇 𝑡𝑣𝑠 − ∆𝑇 𝑡𝑧𝑡 spontaneous process  𝑈 ∆𝐻 𝑡𝑧𝑡 Δ s univ <0 then Δ G sys >0  − = ∆𝑇 𝑣𝑜𝑗𝑤 non spontaneous process 𝑈 Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 13

Free Energy Student Question Hold the rubber band a short distance from your lips. Quickly stretch it and press it against your lips carefully (don’t hurt those delicate lips). Do you experience a warming or cooling sensation? Carefully release the rubber band and experience the sensation. Is stretching a spontaneous or a non-spontaneous process? What are the correct signs for ΔG, ΔH, and ΔS when you allow the rubber band to relax? Δ G Δ H Δ S a) – + + b) – – + c) + + – d) + – – Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 14

Hess’s Law / 3 rd Law of Thermo Hess’s Law: A reaction enthalpy/free energy/entropy is the sum of the enthalpies/free energies/entropies of any sequence of reactions (at the same temperature and pressure) into which the overall reaction can be divided. Things to remember:  If you add reactions together, add ΔH/ΔG/ΔS.  If you flip a reaction, flip the sign of ΔH/ΔG/ΔS.  If you multiply a reaction by a constant, multiply ΔH/ΔG/ΔS by the same constant . Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 15

Hess’s Law / 3 rd Law of Thermo Student Question What is ΔG ° for SO 2 (g) + ½O 2 (g)  SO 3 (g) given the following information? 2S(s) + 3O 2 (g)  2SO 3 (g) ΔG ° = -742 𝑙𝐾 𝑛𝑝𝑚 S(s) + O 2 (g)  SO 2 (g) ΔG ° = -300. 𝑙𝐾 𝑛𝑝𝑚 a) -71 𝑙𝐾 𝑛𝑝𝑚 b) -442 𝑙𝐾 𝑛𝑝𝑚 c) -671 𝑙𝐾 𝑛𝑝𝑚 d) -1042 𝑙𝐾 𝑛𝑝𝑚 e) None of the above Chapt pter er 10: Sponta ntaneity eity, Entropy, and d Free e Energy 16