che1102, chapter 18 learn, 1 chapter 18 thermodynamics

CHE1102, Chapter 18Learn, 1

Chapter 18Thermodynamics

• Spontaneous process – any process, once started, that proceeds without the external input of energy

• Nonspontaneous process – any process which requires the continual, external input of energy to keep the process going

Any process that is spontaneous in one direction will be nonspontaneous in the reverse direction !

Spontaneous vs Nonspontaneous Change

Nonspontaneous Change • Occurs only with outside assistance• Never occurs by itself:

– Room gets straightened up– Pile of bricks turns into a brick wall– Decomposition of H2O by electrolysis

• Continues only as long as outside assistance occurs:– Person does work to clean up room– Bricklayer layers mortar and bricks– Electric current passed through H2O

Spontaneous Change• What factors influence spontaneity?

• Spontaneous Change– Occurs by itself– Without outside assistance until finished

• e.g.– Water flowing over waterfall– Melting of ice cubes in glass on warm day

Spontaneous Processes

• Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.

• Above 0 C, it is spontaneous for ice to melt.• Below 0 C, the reverse process is spontaneous.

Spontaneous processes tend to favor:

1. Decrease in Energy

2. Increase in Disorder

Direction of Spontaneous Change• Many reactions which occur spontaneously are

exothermic:– Iron rusting– Fuel burning

• H and Esystem are negative – Heat given off– Energy leaving system

• Thus, H is one factor that influences spontaneity

Direction of Spontaneous Change• Some endothermic reactions occur

spontaneously:– Ice melting– Evaporation of water– Expansion of CO2 gas into vacuum– The operation of a chemical “cold pack”

• H and E are positive – Heat absorbed– Energy entering system

• Clearly other factors influence spontaneity

Entropy (Symbol S )• Thermodynamic quantity• Describes number of equivalent ways that energy

can be distributed• Can be thought of as a measure of the randomness

or disorder of a system• The greater the statistical probability of a particular

state the greater the entropy!– Larger S, means more possible ways to distribute

energy and that it is a more probable result

Disorder Driven by Statistical Probability

Statistical probability – refers to the number of possible arrangements of a system

A disordered state is more probable because it can be achieved in more statistical ways

the larger the number of different possible combinations, the greater the probability of getting a disordered state

Examples of Spontaneity• Spontaneous reactions

– Things get rusty spontaneously• Don't get shiny again

– Sugar dissolves in coffee • Stir more—it doesn't undissolve

– Ice liquid water at RT• Opposite does NOT occur

– Fire burns wood, smoke goes up chimney• Can't regenerate wood

• Common factor in all of these:– Increase in randomness and disorder of system– Something that brings about randomness more

likely to occur than something that brings order

There are 2,598,960 possible five-card poker hands

Entropy, S

less disorderS = smaller #

more disorderS = larger #

Ludwig Boltzmann1844 – 1906

order disorder is favored to occur spontaneously

order disorder ΔS = + #

2nd Law of Thermodynamics – the entropy of the universe is increasing over time

disorder order ΔS = - #

Entropy, S• State function• Independent of path • S = Change in entropy

• For chemical reactions or physical processes

• Factors that affect entropy – volume, temperature, physical state, number of particles

Effect of Volume on Entropy

• For gases, entropy increases as volume increasesa) Gas separated from vacuum by partition b) Partition removed, more ways to distribute energy c) Gas expands to achieve more probable particle

distribution• More random, higher probability, more positive S

3rd Law of Thermodynamics – entropy decreases as temperature decreases (and vice-versa)

Effect of Temperature on Entropy

Effect of Physical State on Entropy• Crystalline solid very low entropy• Liquid higher entropy, molecules can move freely

– More ways to distribute KE among them• Gas highest entropy, particles randomly distributed

throughout container – Many, many ways to distribute KE

Entropy Affected by Number of Particles• Adding particles to system• Increase number of ways energy can be distributed

in system• So all other things being equal• Reaction that produces more particles will have

positive S

Learning Check

Which represents an increase in entropy?

A. Water vapor condensing to liquidB. Carbon dioxide sublimingC. Liquefying helium gasD. Proteins forming from amino acids

Entropy Changes in Chemical Reactions

• Reactions without gases - Calculate number of mole molecules

n = nproducts – nreactants

If n is positive, entropy increases More molecules, means more disorder

• Reactions involving gases- Calculate change in # of moles of gas, ngas

If ngas is positive , S is positive ngas is more important than nmolecules

Predict Sign of S for Following Reactions

CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O(l) + CO2(g)

– ngas = 1 mol – 0 mol = 1 mol– since ngas is positive, S is positive

2 N2O5(g) 4 NO2(g) + O2(g) – ngas = 4 mol + 1 mol – 2 mol = 3 mol– since ngas is positive, S is positive

OH–(aq) + H+(aq) H2O(l) – ngas = 0 mol– n = 1 mol – 2 mol = –1 mol– since n is negative, S is negative

Learning Check

Which of the following has the most entropy at standard conditions?

A. H2O(l )B. NaCl(aq)C. AlCl3(s)D. Can’t tell from the information

Standard entropy of reaction – the (entropy, or disorder) that accompanies ANY reaction under standard conditions (units are J/K)

= [(sum S products) – (sum S reactants)]

1. write and balance the reaction

2. use table of thermodynamic data of S to calculate

Standard entropy, S – disorder of a substance at standard conditions(units are J/mole·K)

N2O (g) + O2 (g) NO2 (g)3 4

Determine the standard entropy of reaction,

= – 93.0 J/KrxnS

= – #, suggests the reaction is nonspontaneous

ΔHrxn only slightly varies with temperature

ΔSrxn is highly dependent on temperature

Recap: 2 forces in nature which drive processes to occur spontaneously

1. Decrease in energy ΔHrxn = – # (exo)

2. Increase in disorder ΔSrxn = + # (disorder)

Willard Gibbs1839 – 1903

ΔGrxn = ΔHrxn – T ΔSrxn ΔHrxn is the heat of reaction

ΔSrxn is the entropy of reaction

T is the temperature in Kelvin

ΔGrxn is the Gibbs Free Energy

ΔGrxn or Gibbs Free Energy, is the ultimate, final deciding factor as to whether a reaction will occur spontaneously, anywhere in the universe

Bottom Line:

When ΔGrxn = – #, the reaction is spontaneous.

When ΔGrxn = + #, the reaction is nonspontaneous.

When ΔGrxn = 0, the reaction is at equilibrium (has no tendency to go one way or the other)

Is the following reaction spontaneous at 175 °C ?

ΔGrxn = + #, the reaction is nonspontaneous ΔGrxn = +13.7 kJ

42 N2O (g) + O2 (g) NO2 (g)3

= – 93.0 J/KrxnS favors nonspontaneous

favors spontaneous rxnH = – 28.0 kJ

Who wins ??

ΔGrxn = ΔHrxn – T ΔSrxn

Learning Check

For the reaction 2NO(g) + O2(g) 2NO2(g), H° = -113.1 kJ/mol and S° = -145.3 J/K mol.

Which of these statements is true? A. The reaction is spontaneous at all temperatures.B. The reaction is only spontaneous at low temperatures.C. The reaction is only spontaneous at high temperatures.D. The reaction is at equilibrium at 25°C under standard conditions.

Learning CheckAt what temperature (K) will a reaction become nonspontaneous when H = –50.20 kJ mol–1 and S = +20.50 J K–1 mol–1?A. 298 KB. 1200 KC. 2448 KD. The reaction cannot become non-spontaneous at any temperature

We are looking for when ΔG > 0 So ΔH – TΔS > 0 or ΔH > TΔS ΔH/ΔS > T(-50.20 kJ mol-1)/(0.02050 kJ K-1 mol-1) > T-2448 K > T so not possible

Calculate S° for reduction of aluminum oxide by hydrogen gas

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g) Substance S ° (J/ K mol)

Al(s) 28.3Al2O3(s) 51.00

H2(g) 130.6H2O(g) 188.7

Calculate ΔS°

S ° = 56.5 J/K + 566.1 J/K – (51.00 J/K + 391.8 J/K )

S ° = 179.9 J/K

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

= + #, suggests the reaction is spontaneous

Standard Free Energy Changes • Standard Free Energy Change, G °

– G measured at 25 °C (298 K) and 1 atm• Two ways to calculate, depending on what data is

available• Method 1:• G ° = H ° – TS ° • Method 2:

Calculate G ° Method 1

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

Step 1: Calculate H° for reaction using heats of formation below

Calculate G ° for reduction of aluminum oxide by hydrogen gas

H ° = 0.0 kJ – 725.4 kJ – 0.00 kJ – (– 1669.8 kJ) H ° = 944.4 kJ

Calculate G ° Method 1 Step 2: Calculate S ° see Example 3

S ° = 179.9 J/KStep 3: Calculate G ° = H ° – (298.15 K)S ° G ° = 944.4 kJ – (298 K)(179.9 J/K)(1 kJ/1000 J)G ° = 944.4 kJ – 53.6 kJ = 890.8 kJ

– G ° is positive– Indicates that the reaction is not spontaneous

Calculate G ° Method 2• Use Standard Free Energies of Formation

• Energy to form 1 mole of substance from its elements in their standard states at 1 atm and 25 °C

Calculate G ° Method 2Calculate G ° for reduction of aluminum oxide by hydrogen gas.

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

Substance (kJ/mol)Al(s) 0.0Al2O3(s) –1576.4 H2(g) 0.0H2O(g) –228.6

G° = 0.0 kJ – 685.8 kJ – 0.00 kJ – (– 1576.4 kJ) G° = 890.6 kJ

Both methods same within experimental error

G ° = Maximum Possible Work

• G ° is maximum amount of energy produced during a reaction that can theoretically be harnessed as work– Energy that need not be lost to

surroundings as heat– Energy that is “free” or available to do work

How K is related to G°

G ° = –RT lnK

K = e–G°/RT

G° and Position of Equilibrium• When G ° > 0 (positive)

– Position of equilibrium lies close to reactants– Little reaction occurs by the time equilibrium is

reached– Reaction appears nonspontaneous

• When G ° < 0 (negative)– Position of equilibrium lies close to products– Mainly products exist by the time equilibrium is

reached– Reaction appears spontaneous

G ° and Position of Equilibrium• When G ° = 0

– Position of equilibrium lies ~ halfway between products and reactants

– Significant amount of both reactants and products present at time equilibrium is reached

– Reaction appears spontaneous, whether start with reactants or products

• Can Use G ° to Determine Reaction Outcome– G ° large and positive

• No observable reaction occurs– G ° large and negative

• Reaction goes to completion

Calculating G ° from K• Ksp for AgCl(s) at 25 °C is 1.8 10–10 Determine G °

for the process• Ag+(aq) + Cl–(aq) AgCl(s) • Reverse of Ksp equation, so

G ° = –RT lnK = –(8.3145 J/K mol)(298 K) × ln(5.6 109) × (1 kJ/1000

J)• G ° = –56 kJ/mol • Negative G ° indicates precipitation will occur

Determining Effect of Temp on Spontaneity

• Calculate G ° at 25 °C and 500 °C for the Haber process

N2(g) + 3H2(g) 2NH3(g)

• Assume that H ° and S ° do not change with temperature– H ° = – 92.38 kJ– S ° = – 198.4 J/K

Step 1. Calculate G ° for the reaction at 25 °C using H ° and S ° N2(g) + 3H2(g) 2NH3(g) – H ° = –92.38 kJ– S ° = –198.4 J/K

• G ° = H ° – TS° • G ° = –92.38 kJ – (298 K)(–198.4 J/K)• G ° = –92.38 kJ + 59.1 kJ = –33.3 kJ• So the reaction is spontaneous at 25 °C

Step 2. Calculate G ° for the reaction at 500 °C using H ° and S °.– T = 500 °C + 273 = 773 K– H ° = – 92.38 kJ– S ° = – 198.4 J/K

• G ° = H ° – TS ° • G ° = –92.38 kJ – (773 K)(–198.4 J/K)• G ° = –92.38 kJ + 153 kJ = 61 kJ• So the reaction is NOT spontaneous at 500 °C

che1102, chapter 18 learn, 1 chapter 18 thermodynamics

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