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Chemistry 100 Chapter 19 Spontaneity of Chemical and Physical Processes: Thermodynamics
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Chemistry 100 Chapter 19

Feb 24, 2016

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Chemistry 100 Chapter 19. Spontaneity of Chemical and Physical Processes: Thermodynamics. What Is Thermodynamics?. Study of the energy changes that accompany chemical and physical processes. Based on a set of laws. - PowerPoint PPT Presentation
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Page 1: Chemistry 100 Chapter 19

Chemistry 100 Chapter 19

Spontaneity of Chemical and Physical Processes: Thermodynamics

Page 2: Chemistry 100 Chapter 19

What Is Thermodynamics?

Study of the energy changes that accompany chemical and physical processes.

Based on a set of laws. In chemistry, a primary application

of thermodynamics is as a tool to predict the spontaneous directions of a chemical reaction.

Page 3: Chemistry 100 Chapter 19

What Is Spontaneity?

Spontaneity refers to the ability of a process to occur on its own!

Can the Niagara Falls suddenly reverse?

“Ice will melt, water will boil,” Neil Finn, Tim Finn of Crowded House/Plant ‘It’s Only Natural’.

Water spontaneously freezes on a cold winter day!

Page 4: Chemistry 100 Chapter 19

The First Law of Thermodynamics

The First Law deals with the conservation of energy changes.

E = q + w The First Law tells us nothing

about the spontaneous direction of a process.

Page 5: Chemistry 100 Chapter 19

Entropy and Spontaneity

Need to examine the entropy change of the process as well

as its enthalpy change (heat flow). Entropy – the degree of randomness

of a system. Solids – highly ordered low entropy. Gases – very disordered high entropy. Liquids – entropy is variable between that

of a solid and a gas.

Page 6: Chemistry 100 Chapter 19

Entropy Is a State Variable

Changes in entropy are state functions

S = Sf – Si Sf = the entropy of the final stateSi = the entropy of the initial state

Page 7: Chemistry 100 Chapter 19

Entropy Changes for Different Processes

S > 0 entropy increases (melting ice or making steam)

S < 0 entropy decreases (examples freezing water or condensing

steam)

Page 8: Chemistry 100 Chapter 19

The Solution Process For the dissolution of NaCl (s) in water

NaCl (s) Na+(aq) + Cl-(aq)

Highly ordered – low entropy

Disordered or random state – high entropy

The formation of a solution is always accompanied by an increase in the

entropy of the system!

Page 9: Chemistry 100 Chapter 19

The Entropy Change in a Chemical Reaction

Burning ethane! C2H6 (g) + 7/2O2 (g) 2CO2 (g) + 3H2O (l)

The entropy change rS np S (products) - nr S (reactants)

np and nr represent the number of moles of products and reactants, respectively.

Page 10: Chemistry 100 Chapter 19

Finding S Values

Appendix C in your textbook has entropy values for a wide variety of species.

Units for entropy values J / (K mole)

Temperature and pressure for the tabulated values are 298.2 K and 1.00 atm.

Page 11: Chemistry 100 Chapter 19

Finding S Values

Note – entropy values are absolute!

Note – the elements have NON-ZERO entropy values!

e.g., for H2 (g) fH = 0 kJ/mole (by def’n)

S = 130.58 J/(K mole)

Page 12: Chemistry 100 Chapter 19

Some Generalizations

For any gaseous reaction (or a reaction involving gases).

ng > 0, rS > 0 J/(K mole).ng < 0, rS < 0 J/(K mole).ng = 0, rS 0 J/(K mole).

For reactions involving only solids and liquids – depends on the entropy values of the substances.

Page 13: Chemistry 100 Chapter 19

The Second Law of Thermodynamics

The entropy of the universe (univS) increases in a spontaneous process. univS unchanged in an equilibrium

process

Page 14: Chemistry 100 Chapter 19

What is univS?

univS = sysS + surrSsysS = the entropy change of the

system.surrS = the entropy change of the

surroundings.

Page 15: Chemistry 100 Chapter 19

How Do We Obtain univS?

We need to obtain estimates for both the sysS and the surrS.

Look at the following chemical reaction.

C(s) + 2H2 (g) CH4(g) The entropy change for the systems is

the reaction entropy change, rS. How do we calculate surrS?

Page 16: Chemistry 100 Chapter 19

Calculating surrS Note that for an exothermic process,

an amount of thermal energy is released to the surroundings!

Heat

Insulation

surroundings System

Page 17: Chemistry 100 Chapter 19

Calculating surrS

Note that for an endothermic process, thermal energy is absorbed from the surroundings!

Heat

surroundings System

Page 18: Chemistry 100 Chapter 19

Connecting surrS to sysH

For a constant pressure process qp = H

surrS surrH = -sysH surrS = -sysH / T

For a chemical reactionsysH = rH

surrS = -rH/ T

Page 19: Chemistry 100 Chapter 19

The Use of univS to Determine Spontaneity

Calculation of TunivS two system parameters rS rH

Define a system parameter that determines if a given process will be spontaneous?

Page 20: Chemistry 100 Chapter 19

The Definition of the Gibbs Energy

The Gibbs energy of the systemG = H – TS

For a spontaneous processsysG = Gf – G i

Gf = the Gibbs energy of the final stateGi = the Gibbs energy of the initial state

Page 21: Chemistry 100 Chapter 19

Gibbs Energy and Spontaneity

sysG < 0 - spontaneous processsysG > 0 - non-spontaneous process

(note that this process would be spontaneous in the reverse

direction)sysG = 0 - system is in equilibrium

Note that these are the Gibbs energies of the system under non-

standard conditions

Page 22: Chemistry 100 Chapter 19

Standard Gibbs Energy Changes

The Gibbs energy change for a chemical reaction?

Combustion of methane. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

Define rG = np fG (products) - nr fG

(reactants) fG = the formation Gibbs energy of the

substance

Page 23: Chemistry 100 Chapter 19

Gibbs Energy Changes

fG (elements) = 0 kJ / mole. Use tabulated values of the Gibbs

formation energies to calculate the Gibbs energy changes for chemical reactions.

Page 24: Chemistry 100 Chapter 19

The Third Law of Thermodynamics

Entropy is related to the degree of randomness of a substance.

Entropy is directly proportional to the absolute temperature.

Cooling the system decreases the disorder.

Page 25: Chemistry 100 Chapter 19

The Third Law of Thermodynamics

The Third Law - the entropy of any perfect crystal is 0 J /(K mole) at 0 K (absolute 0!)

Due to the Third Law, we are able to calculate absolute entropy values.

Page 26: Chemistry 100 Chapter 19

At a very low temperature, the disorder decreases to 0 (i.e., 0 J/(K mole) value for S).

The most ordered arrangement of any substance is a perfect crystal!

Page 27: Chemistry 100 Chapter 19

Applications of the Gibbs Energy

The Gibbs energy is used to determine the spontaneous direction of a process.

Two contributions to the Gibbs energy change (G) Entropy (S) Enthalpy (H)

G = H - TS

Page 28: Chemistry 100 Chapter 19

Spontaneity and Temperature

H S G

+ + < 0 at high temperatures

+ - > 0 at all temperatures

- + < 0 at all temperatures

- - < 0 at low temperatures

Page 29: Chemistry 100 Chapter 19

Gibbs Energies and Equilibrium Constants

rG < 0 - spontaneous under standard conditions

rG > 0 - non-spontaneous under standard conditions

Page 30: Chemistry 100 Chapter 19

The Reaction Quotient

Relationship between QJ and Keq

Q < Keq

- reaction moves in the forward directionQ > Keq

- reaction moves in the reverse directionQ = Keq

- reaction is at equilibrium

Page 31: Chemistry 100 Chapter 19

rG° refers to standard conditions only!

For non-standard conditions - rG rG < 0 - reaction moves in the

forward directionrG > 0 - reaction moves in the

reverse directionrG = 0 - reaction is at equilibrium

Page 32: Chemistry 100 Chapter 19

Relating Keq to rG

rG = rG +RT ln QrG = 0 system is at equilibrium

rG = -RT ln Qeq

rG = -RT ln Keq

Page 33: Chemistry 100 Chapter 19

Phase Equilibria

At the transition (phase-change) temperature only - trG = 0 kJ

tr = transition type (melting, vapourization, etc.)

trS = trH / Ttr