chapter 19: chemical thermodynamics

Joe DiLosaHannah Robinson

Justin WhetzelEmily Penn

Casey White

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

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

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

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

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?

(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

Practice Problems - #7, 8, and 9

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

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

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

∆S = Sfinal – Sinitial

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

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

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

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

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

Practice Problems- # 20 and 21

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

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

Vibration Vibration

Vibration Rotation

Img src = wps.prenhall.com

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

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

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

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

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

Practice Problems- # 28, 30, 38, 40

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

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

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

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

For a reaction at constant pressure qsys= ∆H

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

Practice Problems - # 48

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

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

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

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.

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

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.

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

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

ΔG = -wma

x

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

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

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

No new vocabulary

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

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

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?

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

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

Nonstandard conditions – conditions other than the standard conditions

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

ΔGo = -RT ln KA

t equilibrium, ΔG = 0

K = e –(delta G standard)/RT

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

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

#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)

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

Practice Problems: #71,76,79

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