chapter 19: chemical thermodynamics

52
Joe DiLosa Hannah Robinson Justin Whetzel Emily Penn Casey White

Upload: morwen

Post on 05-Jan-2016

253 views

Category:

Documents


17 download

DESCRIPTION

Chapter 19: Chemical Thermodynamics. Joe DiLosa Hannah Robinson Justin Whetzel Emily Penn Casey White. 19.1 – Spontaneous Processes Vocabulary. Spontaneous Process – a process that proceeds on its own without any outside assistance - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Chapter 19:  Chemical Thermodynamics

Joe DiLosaHannah Robinson

Justin WhetzelEmily Penn

Casey White

Page 2: Chapter 19:  Chemical Thermodynamics

Spontaneous Process – a process that proceeds on its own without any outside assistance

Reversible Process – A process that can go back and forth between states along exactly the same path

Irreversible Process – A process that is not reversible

Isothermal Process – A process that occurs at a constant temperature

Page 3: Chapter 19:  Chemical Thermodynamics

Occur in a definite direction Ex: When a brick is dropped and it falls

down (gravity is always doing work on the brick)

Opposite direction (reverse) would be nonspontaneous (takes work to go against gravity)

If the surroundings must do work on the system to return it to its original state, the process is irreversible

Any spontaneous process is irreversible

Page 4: Chapter 19:  Chemical Thermodynamics

http://wps.prenhall.com/wps/media/objects/3312/3392504/blb1901.html

Page 5: Chapter 19:  Chemical Thermodynamics

A spontaneous reaction does not mean the reaction occurs very fast

Iron rusting is spontaneous Experimental conditions are important in

determining if a reaction is spontaneous (temperature and pressure)

To determine spontaneity, distinguish between the reversible and irreversible paths between states

Page 6: Chapter 19:  Chemical Thermodynamics

Which of the following processes are spontaneous and which are nonspontaneous: (a) the melting of ice cubes at -5°C and 1 atm pressure, (b) dissolution of sugar in a cup of hot coffee, (c) the reaction of nitrogen atoms to form N2 molecules at 25°C (d) alignment of iron filings in a magnetic field, (e) formation of CH4 and O2 molecules from CO2 and H2O at room temperature and 1 atm of pressure?

Page 7: Chapter 19:  Chemical Thermodynamics

(a) is nonspontaneous because the ice will not change state

(b) is spontaneous because to get the sugar back, the temperature of the system must be altered.

(c) is spontaneous because work must be done to form N2 and it is an irreversible process

(d) is spontaneous because the system must do work against the magnetic field to go back to its original state

(e) is nonspontaneous because a change in temperature and work must be done to break the products up and form the original reactants

Page 8: Chapter 19:  Chemical Thermodynamics

Practice Problems - #7, 8, and 9

Page 9: Chapter 19:  Chemical Thermodynamics

Entropy – A thermodynamic function associated with the number of different equivalent energy states or spatial arrangements in which a system may be found

Second Law of Thermodynamics – in general, any irreversible process results in an overall increase in entropy whereas a reversible process results in no overall change in entropy

Infinitesimal difference - an extremely small difference

Page 10: Chapter 19:  Chemical Thermodynamics

Entropy = q/T = S A reversible change produces the

maximum amount of work that can be achieved by the system on the surroundings (wrev = wmax)

Reversible processes are those that reverse direction whenever an infinitesimal change is made in some property of the system

Page 11: Chapter 19:  Chemical Thermodynamics

Example of Reversible Process◦ Flow of heat between a system and its

surroundings Example of Irreversible Process

◦ Gas that has been limited to part of a container expands to fill the entire container when the barrier is removed

Page 12: Chapter 19:  Chemical Thermodynamics

∆S = Sfinal – Sinitial

∆S = qrev/T (where T = constant, isothermal process)

qrev = ∆Hfusion when calculating ∆Sfusion for a phase change

Page 13: Chapter 19:  Chemical Thermodynamics

∆Ssystem = qrev/T ∆Ssurroundings= -qrev/T ∆Stotal = ∆Suniv = ∆Ssystem + ∆Ssurroundings

For a reversible process ∆Stotal = 0 For an irreversible process ∆Stotal > 0

Page 14: Chapter 19:  Chemical Thermodynamics

The normal boiling point of methanol (CH3OH) is 64.7˚C and its molar enthalpy of vaporization is ∆Hvap=71.8 kJ/mol. When CH3OH boils at its normal boiling point does its entropy increase or decrease? Calculate the value of ∆S when 1.00 mol CH3OH (l) is vaporized at 64.7 ˚C

Increases, the process is endothermic∆S = qrev/T = ∆Hvap/T

64.7 ˚C + 273 = 337.7 K(1 mol)(71.8 kJ/mol)/(337.7 K)= 0.213 kJ/K = 213 J/Kentropy increases

Page 15: Chapter 19:  Chemical Thermodynamics

Practice Problems- # 20 and 21

http://ffden-2.phys.uaf.edu/212_fall2003.web.dir/Anca_Bertus/page_7.htm

Page 16: Chapter 19:  Chemical Thermodynamics

Translational Motion – movement in which an entire molecule moves in a definite direction

Vibrational Motion – the atoms in the molecule move periodically toward and away from one another

Rotational Motion – the molecules are spinning like a top Microstate – the state of a system at a particular

instant, one of many possible states of the system Third Law of Thermodynamics – the entropy of a

pure crystalline substance at absolute zero is zero

Page 17: Chapter 19:  Chemical Thermodynamics

Vibration Vibration

Vibration Rotation

Img src = wps.prenhall.com

Page 18: Chapter 19:  Chemical Thermodynamics

S = k ln W (W = number of microstates) k = Boltzmann’s constant = 1.38 x 10-23

J/K Entropy is a measure of how many

microstates are associated with a particular macroscopic state

∆S= k ln Wfinal - k ln Winitial = k ln Wfinal/Wintial

Entropy increase with the number of microstates of the system

Page 19: Chapter 19:  Chemical Thermodynamics

The # of microstates available to a system increases with1. Increase in volume2. Increase in temperature3. Increase in the number of molecules Because any of these changes increases the

possible positions and energies of the molecules of the system

Page 20: Chapter 19:  Chemical Thermodynamics

Generally expect the entropy of a system to increase for processes in which1. Gases are formed from either solids or

liquids2. Liquids or solutions are formed from

solids3. The # of gas molecules increases during

a chemical reaction

Page 21: Chapter 19:  Chemical Thermodynamics

At absolute zero there is no thermal motion so there is only one microstate

As the temperature increases the molecules gain energy in the form of vibrational motion

When the molecules increase their motion they have a greater number of microstates

Page 22: Chapter 19:  Chemical Thermodynamics

http://wpscms.pearsoncmg.com/au_hss_brown_chemistry_1/57/14649/3750314.cw/content/index.html

Page 23: Chapter 19:  Chemical Thermodynamics

Practice Problems- # 28, 30, 38, 40

Page 24: Chapter 19:  Chemical Thermodynamics

Standard Molar Entropy – The molar entropy value of substances in their standard states

Page 25: Chapter 19:  Chemical Thermodynamics

The standard molar entropies of elements are not 0 at 298K

The standard molar entropies of gases are greater than those of liquids and solids

standard molar entropies generally increase with increasing molar mass

Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance

Page 26: Chapter 19:  Chemical Thermodynamics

Pictures taken from p820 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay, and Bursten

Page 27: Chapter 19:  Chemical Thermodynamics

∆S˚= ΣnS˚(products) - ΣmS˚(reactants) ∆Ssurr= -qsys/T

For a reaction at constant pressure qsys= ∆H

Page 28: Chapter 19:  Chemical Thermodynamics

Calculate ∆S˚ values for the following reaction:

C2H4 (g) + H2 (g) C2H6 (g)

∆S˚= ΣnS˚(products) - ΣmS˚(reactants) = 229.5 J/molK – (219.4 J/molK + 130.58 J/molK)

= -120.5 J/molK

Page 29: Chapter 19:  Chemical Thermodynamics

Practice Problems - # 48

Page 30: Chapter 19:  Chemical Thermodynamics

Gibbs free energy- a thermodynamic state function that combines entropy and enthalpy

Standard free energies of formation- the change in free energy associated with the formation of a substance from its elements under standard conditions

Page 31: Chapter 19:  Chemical Thermodynamics

Free energy (G) is defined by:G = H-TS

Where T is absolute temperature, S is entropy, and H is enthalpy

Page 32: Chapter 19:  Chemical Thermodynamics

When temperature is constant the equation for the change in free energy for the system is

ΔG=ΔH-TΔS If T and P are constant then the sign of ΔG and

spontaneity of a reaction are related.1) If ΔG < 0, reaction proceeds forward2) If ΔG = 0, reaction is at equilibrium3) If ΔG > 0, the forward reaction is not spontaneous

because work must be supplied from the surroundings but the reverse reaction is be spontaneous.

Page 33: Chapter 19:  Chemical Thermodynamics

In any spontaneous process at constant temperature and pressure the free energy always decreases

Page 34: Chapter 19:  Chemical Thermodynamics

Conditions of Standard free energies of formation (implied)

1 atm pressure for gases pure solids pure liquids 1M solutions 25˚ C (normally)

The free energy of elements is 0 in their standard states.

Page 35: Chapter 19:  Chemical Thermodynamics

ΔGf˚ = standard free energies of formation Calculated the same way as ΔH

ΔG˚ = ΣnΔGf˚ (products) - ΣmΔGf˚ (reactants)

ΔG = -wma

x

Page 36: Chapter 19:  Chemical Thermodynamics

http://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Gibbs_free_energy_graph

Page 37: Chapter 19:  Chemical Thermodynamics

Practice Problems - #49, 51, 53, 54, 56, 58, 63, 66

Page 38: Chapter 19:  Chemical Thermodynamics

For a certain reaction ΔH = -35.4 kJ and ΔS = -85.5 J/K. Calculate ΔGo for the reaction at 298 K. Is the reaction spontaneous at 298 K?

ΔG = ΔH – ΔS*T ΔG = -35.4KJ – (-85.5 J/K * 298K) ΔG = -9920 J

Page 39: Chapter 19:  Chemical Thermodynamics

No new vocabulary

Page 40: Chapter 19:  Chemical Thermodynamics

When ΔH and –TΔS are negative, ΔG is negative and the process is spontaneous

When ΔH and –TΔS are positive, ΔG is positive and the process is non-spontaneous

When ΔH and –TΔS are opposite signs, ΔG will depend on the magnitudes

http://www.chem1.com/acad/webtext/thermeq/TE4.html

Page 41: Chapter 19:  Chemical Thermodynamics

Picture taken from p828 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay, and Bursten

Page 42: Chapter 19:  Chemical Thermodynamics

For a certain reaction ΔH = -35.4 kJ and ΔS = -85.5 J/K. is the reaction exothermic or endothermic? Does the reaction lead to an increase or decrease of disorder in the system? Calculate ΔGo for the reaction at 298 K. Is the reaction spontaneous at 298 K?

Page 43: Chapter 19:  Chemical Thermodynamics

A) The reaction is exothermic because ΔH is negative

B) It leads to a decrease in the disorder of the system (increase in the order of the system) because ΔS is negative

C) ΔG = ΔH – ΔS*T ΔG = -35.4KJ – (-85.5 J/K * 298K) ΔG = -9920 J D) Yes, if all reactants are in their

standards states because ΔG is negative

Page 44: Chapter 19:  Chemical Thermodynamics

Practice problems (same as 19.5) - #49, 51, 53, 54, 56, 58, 63, 66

Page 45: Chapter 19:  Chemical Thermodynamics

Nonstandard conditions – conditions other than the standard conditions

Page 46: Chapter 19:  Chemical Thermodynamics

ΔG = ΔGo + RT ln Q Where R = 8.314 J/mol-K, T = absolute temperature, Q = reaction quotient

ΔGo = ΔG under standard conditionsB

ecause Q = 1 and lnQ = 0

Page 47: Chapter 19:  Chemical Thermodynamics

ΔGo = -RT ln KA

t equilibrium, ΔG = 0

K = e –(delta G standard)/RT

Smaller (more negative ΔG0), the larger the K value

Page 48: Chapter 19:  Chemical Thermodynamics

Direction of Reaction Keq ΔG

Toward forming more products >1 negative

Toward forming more reactants <1 positive Products and reactants equal 1 0

http://www.wiley.com/legacy/college/boyer/0470003790/reviews/thermo/thermo_keq.htm

Page 49: Chapter 19:  Chemical Thermodynamics

#71 – Explain Quantitatively how ΔG changes for each of the following reactions as the partial pressure of O2 is increased:

A) 2 CO(g) + O2(g) → 2 CO2 (g)

Page 50: Chapter 19:  Chemical Thermodynamics

2 CO(g) + O2(g) → 2 CO2 (g)

-137.2 kJ/mol 0 kJ/mol -394.4 kJ/mol

ΔG = products – reactantsΔG = -394.4 kJ/mol + 137.2 kJ/molΔG = -257.2 kJ/mol

Page 51: Chapter 19:  Chemical Thermodynamics

Practice Problems: #71,76,79

Page 52: Chapter 19:  Chemical Thermodynamics

#82, 87, 88, 89, 90, 92