Sparks CH301 GIBBS FREE ENERGY. UNIT 4 Day 8
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1 Sparks CH301 GIBBS FREE ENERGY UNIT 4 Day 8
2 What are we going to learn today? Quantify change in Gibbs Free Energy Predict Spontaneity at Specific Temperatures
3 QUIZ: iclicker Questions S H2 = 131 J/K mol S O2 = 205 J/ K mol S H2O = 70 J/K mol ΔH f, H2O = -286 kj/mol Determine for the combusion of 1 mole of hydrogen under standard conditions: 1. the change in entropy of the system 2. the change in entropy for the surroundings 3. whether or not this process is spontaneous
4 Use the following tabulated data: S H2 = 131 J/K mol S O2 = 205 J/ K mol S H2O = 70 J/K mol to determine the change in entropy for the combustion of 1 mole of hydrogen. A J/Kmol B J/Kmol C J/Kmol D J/Kmol QUIZ: iclicker Question 1
5 QUIZ: iclicker Question 2 Use the following tabulated data: ΔH f, H2O = -286 kj/mol to determine the change in entropy for the surroundings for the combustion of 1 mole of hydrogen under standard conditions. A. 0.0 kj/molk B kj/molk C kj/molk D kj/molk
6 QUIZ: iclicker Question 3 Use the following tabulated data: ΔH f, H2O = -286 kj/mol Is the combustion of 1 mole of hydrogen spontaneous under standard conditions? A. Yes B. No C. No way to tell
7 Gibbs Free Energy (G) Free Energy, G can predict spontaneity based only on the system DG = DH -TDS
8 POLL: iclicker Question Given the following data, calculate the ΔG rxn for the combustion of 1 mole of hydrogen at 298 K. ΔH f, H2O = -286 kj/mol ΔS H2 = 131 J/mol ΔS O2 = 205 J/mol ΔS H2O = 70 J/mol A. -48,284 kj/mol B. 48,284 kj/mol C kj/mol D. 237 kj/mol
9 Reaction Gibbs Free Energy (ΔG) ΔG r = ΣnΔG f(products) ΣnΔG f(reactants) Standard Gibbs Free Energy of formation, ΔG f, of a substance is the standard reaction Gibbs Free Energy per mole for the formation of a compound from its elements in their most stable form.
10 POLL: iclicker Question 6 Calculate the standard reaction free energy of 2Fe 2 O 3 (s) + C(s) 3CO 2 (g) + 4Fe(s) from standard free energy of formation data. ΔG f (Fe 2 O 3 )s = -740 kj/mol ΔG f (CO 2 )g = -394 kj/mol A. 298 kj/mol B kj/mol C kj/mol D kj/mol
11 The Temperature Dependence of Spontaneity Spontaneity: Exothermic reactions are favored for spontaneity. Increasing entropy is favored for spontaneity. From DG = DH -TDS we see that temperature also plays a role. Some processes might be either spontaneous or nonspontaneous depending on the temperature. 11
12 POLL: iclicker Question Consider an exothermic reaction for which the entropy of the system is increasing. This reaction will be: A. Spontaneous at all temperatures B. Spontaneous only at relatively high temperatures. C. Spontaneous only at relatively low temperatures. D. Nonspontaneous at all temperatures. 12
13 POLL: iclicker Question If ΔH is positive and ΔS is positive, but the reaction is nonspontaneous at a given temperature you can: a) Not force it to change, once a reaction is nonspontaneous it is always nonspontaneous b) Increase the temperature and move to a spontaneous change c) Decrease the temperature and move to a spontaneous change d) This makes absolutely no sense, please explain again
14 DG = DH -TDS DH DS DG Forward reaction spontaneity
15 Equilibrium ΔG r is a comparison between ALL Products and ALL Reactants Therefore, we assume that the reactions will go to completion, in which all reactants are converted to the desired product. However, some reactions stop in the middle because they are at equilibrium. A system at equilibrium does not change in either the forward or reverse direction. DYNAMIC EQUILIBRIUM: Forward and Backward reactions still occur: BUT net result is no change in the amounts of products or reactants! 15
16 DYNAMIC EQUILIBRIUM DYNAMIC equilibrium is different from STATIC equilibrium. Imagine 10,000 ants on a small see-saw: STATIC EQUILIBRIUM: DYNAMIC EQUILIBRIUM: side side top top 5,000 ants stay on one side. 5,000 ants stay on the other. NOBODY MOVES! ALL the ants start marching in a loop, evenly spaced, and keep marching. Overall, there are 5,000 on each side. Every individual ant is always changing sides!
17 At equilibrium, DS tot = 0, therefore DG = 0. 17
18 POLL: iclicker Question The vaporization of liquid bromine will be spontaneous at temperatures for which ΔG is negative. At what temperature is this process first spontaneous at 1 atm? Br 2 (l) Br 2 (g) A. 30 C B. 60 C C. 300 C D. 330 C E. 360 C ΔH f Br 2,g = 31 kj/mol S Br2,l = J/K mol S Br2,g = J/ K mol
19 Summary There are two important questions in Thermodynamics 1. How much energy is released or absorbed with change? We use the change in Enthalpy to answer this question 2. Will the change happen (in isolation)? We use Gibbs Free Energy to answer this question
20 Learning Outcomes Calculate change in free energy for a chemical change from change in free energy of formation of the products and the reactants Understand the concept of the change in free energy Calculate change in free energy for a chemical change from change in enthalpy and change in entropy values Recognize based on the sign and magnitude of change in enthalpy and change in entropy how changing the temperature affects the spontaneity of a physical or chemical change Calculate the temperature at which a particular change (physical or chemical) will be spontaneous
21 Important Information HW due Monday, December 2 nd, 8:45 am Thermochemistry Overview: Complete and bring to class Monday, December 2 nd. ONLY predict signs; you do not (and should not) calculate values. AFTER you predict the signs, you can calculate values if you want extra practice. UNIVERSITY COURSE INSTRUCTOR SURVEY!! MOST IMPORTANT! PLEASE COMPLETE! IT s LIKE VOTING, IT IS YOUR DUTY TO YOUR UNIVERSITY
Vanden Bout/LaBrake. Important Information. HW11 Due T DECEMBER 4 th 9AM. End of semester attitude survey closes next Monday
UNIT4DAY6-LaB Page 1 UNIT4DAY6-LaB Thursday, November 29, 2012 8:13 AM Vanden Bout/LaBrake CH301 The 2 nd Law of Thermodynamics GIBBS FREE ENERGY UNIT 4 Day 6 Important Information HW11 Due T DECEMBER
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