entropy, free energy, and equilibrium

Entropy, Free Energy, and Equilibrium

Chapter 16

Spontaneous Physical and Chemical Processes

• A waterfall runs downhill

• A lump of sugar dissolves in a cup of coffee

• At 1 atm., water freezes below 00C and ice melts above 00C

• Heat flows from a hotter object to a colder object

• The expansion of a gas in an evacuated bulb

• Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

18.2

spontaneous

nonspontaneous

18.2

Does a decrease in enthalpy mean a reaction proceeds spontaneously?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H0 = -890.4 kJ

H+ (aq) + OH- (aq) H2O (l) H0 = -56.2 kJ

H2O (s) H2O (l) H0 = 6.01 kJ

NH4NO3 (s) NH4+(aq) + NO3

- (aq) H0 = 25 kJH2O

Spontaneous reactions

18.2

Spontaneous reactions could be either exothermic or endothermic.

Therefore, to predict the spontaneity of a reaction, another thermodynamic entity must be considered, i.e., entropy.

Entropy (S) is a measure of the randomness or disorder of a system.

order SdisorderS

S = Sf - Si

If the change from initial to final results in an increase in randomness

Sf > Si S > 0

For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state

Ssolid < Sliquid << Sgas

H2O (s) H2O (l) S > 018.2

Thermodynamics

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy, enthalpy, pressure, volume, temperature

6.7

, entropy

At higher temperatures,• molecular motion increases• randomness increases.• entropy increases.

• 2 possible arrangements

• 50 % chance of finding the left empty

Microstates and Entropy

4 possible arrangements

25% chance of finding the left empty

50 % chance of them being evenly dispersed

4 possible arrangements

8% chance of finding the left empty

50 % chance of them being evenly dispersed

Processes that lead to an increase in

entropy (S > 0)

18.2

How does the entropy of a system change for each of the following processes?

(a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 600C to 800C

(d) Subliming dry ice

18.2

Dec

Dec

Inc

Inc

How does the entropy of a system change for each of the following processes?

(a) A solid melts

(b) A liquid freezes

(c) Sugar dissolves in water

(d) A vapor condenses to a liquid

18.2

Inc

Dec

Inc

Dec

First Law of Thermodynamics

Energy can be converted from one form to another but energy cannot be created or destroyed.

Second Law of Thermodynamics

The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.

Suniv = Ssys + Ssurr > 0Spontaneous process:

Suniv = Ssys + Ssurr = 0Equilibrium process:

18.3

Entropy Changes in the System (Ssys)

aA + bB cC + dD

S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

S0rxn nS0(products)= mS0(reactants)-

The standard entropy of reaction (S0 ) is the entropy change for a reaction carried out at 1 atm and 250C.

rxn

18.3

What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)

S0(CO) = 197.9 J/K•molS0(O2) = 205.0 J/K•mol

S0(CO2) = 213.6 J/K•mol

S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

S0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

Entropy Changes in the System (Ssys)

18.3

When gases are produced (or consumed)

• If a reaction produces more gas molecules than it consumes, S0 > 0.

• If the total number of gas molecules diminishes, S0 < 0.

• If there is no net change in the total number of gas molecules, then S0 may be positive or negative BUT S0 will be a small number.

What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s)

The total number of gas molecules goes down, and a gas is converted to a solid, S is negative.

18.3

Predict the sign of the entropy change for the following reactions?

a) 3 O2 (g) 2 O3 (g)

b) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g)

c) N2 (g) + O2 (g) 2 NO (g)

d) I2 (s) 2 I (g)

e) 2 Hg (l) + O2 (g) 2 HgO (s)

-

+

-

+

-

Entropy Changes in the Surroundings (Ssurr)

Exothermic ProcessSsurr > 0

Endothermic ProcessSsurr < 0

18.3

ssurr is proportional to - Hsys

Ssurr =Hsys

T

Suniv = Ssys + Ssurr > 0Spontaneous process:

Is the following reaction spontaneous at 25°C?N2(g) + 3H2(g) 2NH3(g); H° = -92.6 kJ

Ssys = -199 J/K

Hsys = -92.6 kJ

Ssurr =Hsys

T=

- (-92600 J)

298 K= 311 J/K

Suniv = Ssys + Ssurr = -199 J/K + 311 J/K = 112 J/K > 0

Process is spontaneous, but may be very slow.

= [2 x S0(NH3) ] – [S0(N2) + 3 x S0(H2)]

Third Law of Thermodynamics

The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

18.3

Suniv = Ssys + Ssurr > 0Spontaneous process:

Gibbs Free Energy

For a constant-temperature process:

G = Hsys -TSsysGibbs free energy (G)

18.4

Suniv = Ssys - > 0Hsys

T

Suniv = Hsys - TSsys < 0

G < 0 The reaction is spontaneous in the forward direction.

G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.

G = 0 The reaction is at equilibrium.

18.4

aA + bB cC + dD

G0rxn dG0 (D)fcG0 (C)f= [ + ] - bG0 (B)faG0 (A)f[ + ]

G0rxn nG0 (products)f= mG0 (reactants)f-

The standard free-energy of reaction (G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.

rxn

Standard free energy of formation (G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

f

G0 of any element in its stable form is zero.

f

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

G0rxn nG0 (products)f= mG0 (reactants)f-

What is the standard free-energy change for the following reaction at 25 0C?

G0rxn 6G0 (H2O)f12G0 (CO2)f= [ + ] - 2G0 (C6H6)f[ ]

G0rxn = [12(–394.4) + 6(–237.2)] – [2(124.5)] = -6405 kJ

Is the reaction spontaneous at 25 0C?

G0 = -6405 kJ < 0

spontaneous

18.4

a) CaCO3 (s) CaO (s) + CO2 (g)

Calculate the standard free-energy change for the following reactions at 25 0C and determine whether the reactions are spontaneous.

18.4

b) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g)

c) 2HgO (s) 2Hg (l) + O2 (g)

d) 2C2H4 (g) + 6O2 (g) 4CO2 (g) + 4H2O (l)

a) [(-603.5 + -394.4) ] – [( -1128.8) ] =

Calculate the standard free-energy change for the following reactions at 25 0C and determine whether the reactions are spontaneous.

18.4

b) [-1044.4+ -237.2+ -394.4] - [ 2 NaHCO3 (s) ] =

c) [2 (0) + 0] - [2 (-58.49)] =

d) [4 (-394.4) + 4 (237.2) ] - [2C2H4 (g) + 0) ] =

G = H - TS

18.4

CaCO3 (s) CaO (s) + CO2 (g)

H0 = 177.8 kJ

S0 = 160.5 J/K

G0 = H0 – TS0

At 25 0C, G0 = 130.0 kJ

Tat which G0 = 0 is when the reaction starts to become spontaneous

18.4

Temperature and Spontaneity of Chemical Reactions

nonspontaneous

G0 = 0 at 835 0C

MgCO3 (s) MgO (s) + CO2 (g)

To what temperature must MgCO3 must be heated to decompose it? Is this higher or lower than the temperature to decompose CaCO3?

18.4

G0 = H0 – TS0 = 0

T = H0 S0

Gibbs Free Energy and Phase Transitions

H2O (l) H2O (g)

G0 = 0 = H0 – TS0

Svap = T

Hvap =

40.79 kJ373 K

= 109 J/K

18.5

At phase transitions, the system is at equilibrium (G0 = 0)

Ar (l) Ar (g)

The molar heat of vaporization of argon is 6.3 kJ/mol, and argon’s boiling point is -186°C. Calculate the entropy change of vaporization.

18.5

Svap = H vap

T

Svap = 6300 J 87 K

= 72.4 J/K

Gibbs Free Energy and Chemical Equilibrium

G = G0 + RT lnQ

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

18.5

G0 < 0 G0 > 0

18.5

G0 = RT lnK

18.5

Using thermodynamic data, calculate the equilibrium constants for the following reactions at 25°C.

18.5

a) CaCO3 (s) CaO (s) + CO2 (g)

b) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g)

c) 2HgO (s) 2Hg (l) + O2 (g)

d) 2C2H4 (g) + 6O2 (g) 4CO2 (g) + 4H2O (l)

G0 = RT lnK

BaF2 (s) Ba+2 (aq) + 2 F- (aq)

Calculate G0 for the following process at 25°C. The Ksp of BaF2 is 1.7 x 10-6.

18.5

Calculate G0 for the following process at 25C. The Ksp of CaCO3 is 8.7 x 10-9.

CaCO3 (s) Ca+2 (aq) + CO3-2 (aq)

18.5

Calculate the Ksp at 25°C for the following reaction, using standard free energy of formation (G0

f ) values.

CaCO3 (s) Ca+2 (aq) + CO3-2 (aq)