Sparks CH301 GIBBS FREE ENERGY UNIT 4 Day 8
What are we going to learn today? Quantify change in Gibbs Free Energy Predict Spontaneity at Specific Temperatures
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
QUIZ: iClicker Question 1 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. -200.1 J/Kmol B. -163.5 J/Kmol C. -115.8 J/Kmol D. -84.6 J/Kmol
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. 0.959 kJ/molK C. 1.223 kJ/molK D. 1.458 kJ/molK
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
Gibbs’ Free Energy (G) Free Energy, G – can predict spontaneity based only on the system D G = D H -T D S
POLL: iClicker Question Given the following data, calculate the Δ G 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. -237 kJ/mol D. 237 kJ/mol
Reaction Gibbs’ Free Energy (Δ G) • Δ G r = Σ n Δ G f(products) – Σ n Δ G f(reactants) • Standard Gibbs’ Free Energy of formation, Δ G 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.
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. -298 kJ/mol C. 1660 kJ/mol D. -1660 kJ/mol
The Temperature Dependence of Spontaneity • Spontaneity: – Exothermic reactions are favored for spontaneity. – Increasing entropy is favored for spontaneity. – From D G = D H -T D S we see that temperature also plays a role. • Some processes might be either spontaneous or non- spontaneous — depending on the temperature. 11
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
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
D G = D H -T D S D H D S D G Forward reaction spontaneity – + – – + + + – 14
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
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 ALL the ants start marching in a loop, evenly spaced, and keep marching. 5,000 ants stay on one side. Overall, there are 5,000 on each side. 5,000 ants stay on the other. Every individual ant is always NOBODY MOVES! changing sides !
• At equilibrium, D S tot = 0, therefore D G = 0. 17
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 ΔH f Br 2,g = 31 kJ/mol B. 60 °C S Br2,l = 152.23 J/K mol C. 300 °C D. 330 °C S Br2,g = 245.46 J/ K mol E. 360 °C
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
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
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
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