CH301 GIBBS FREE ENERGY UNIT 4 Day 8 What are we going to learn - - PowerPoint PPT Presentation

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

SH2 = 131 J/K mol SO2 = 205 J/ K mol SH2O = 70 J/K mol ΔHf, 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 Questions

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

• f 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 1

Use the following tabulated data: ΔHf, 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 2

Use the following tabulated data: ΔHf, 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

QUIZ: iClicker Question 3

Free Energy, G – can predict spontaneity based

• nly on the system

DG = DH -TDS

Gibbs’ Free Energy (G)

POLL: iClicker Question

Given the following data, calculate the ΔGrxn for the combustion of 1 mole of hydrogen at 298 K. ΔHf, 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

• ΔGr = ΣnΔGf(products) – ΣnΔGf(reactants)
• Standard Gibbs’ Free Energy of formation, ΔGf

°, 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.

Reaction Gibbs’ Free Energy (ΔG)

Calculate the standard reaction free energy of 2Fe2O3(s) + C(s)  3CO2 (g) + 4Fe(s) from standard free energy of formation data. ΔGf (Fe2O3)s = -740 kJ/mol ΔGf (CO2)g = -394 kJ/mol

• A. 298 kJ/mol
• B. -298 kJ/mol
• C. 1660 kJ/mol
• D. -1660 kJ/mol

POLL: iClicker Question 6

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 non-

spontaneous—depending on the temperature.

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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.

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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

POLL: iClicker Question

DG = DH -TDS

DH DS DG Forward reaction spontaneity – + – – + + + –

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Equilibrium

• ΔGr° 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!

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DYNAMIC EQUILIBRIUM

DYNAMIC equilibrium is different from STATIC equilibrium.

Imagine 10,000 ants on a small see-saw: STATIC EQUILIBRIUM: DYNAMIC EQUILIBRIUM: 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!

side top side top

• At equilibrium, DStot = 0, therefore DG = 0.

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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? Br2 (l)  Br2 (g)

• A. 30 °C
• B. 60 °C
• C. 300 °C
• D. 330 °C
• E. 360 °C

POLL: iClicker Question

ΔHf Br2,g = 31 kJ/mol SBr2,l = 152.23 J/K mol S Br2,g = 245.46 J/ K mol

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

Summary

Calculate change in free energy for a chemical change from change in free energy

• f 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

Learning Outcomes

Important Information

HW due Monday, December 2nd, 8:45 am Thermochemistry Overview:

• Complete and bring to class Monday, December 2nd.
• 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