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Second Law of Thermodynamics

Second Law of Thermodynamics

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Second Law of Thermodynamics. ENTROPY (S). Entropy is a measure of disorder Low entropy (S) = low disorder High entropy (S) = greater disorder Operates at the level of atoms and molecules. Law of Disorder the disorder (or entropy) of a system tends to increase. - PowerPoint PPT Presentation

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Page 1: Second Law of Thermodynamics

Second Law of Thermodynamics

Page 2: Second Law of Thermodynamics

Law of Disorder

the disorder (or entropy) of a system tends to increase

ENTROPY (S)

•Entropy is a measure of disorder

• Low entropy (S) = low disorder

•High entropy (S) = greater disorder

•Operates at the level of atoms and molecules

• hot metal block tends to cool

• gas spreads out as much as possible

Page 3: Second Law of Thermodynamics

Factors affecting Entropy

A. Entropy increase as matter moves from a solid to a liquid to a gas

Increasing Entropy

B. Entropy increases when a substance is divided into parts

Increasing Entropy

Page 4: Second Law of Thermodynamics

C. Entropy tends to increase in reactions in which the number of molecules increases

Increasing Entropy

D. Entropy increase with an increase in temperature

Page 5: Second Law of Thermodynamics

Entropy Changes in the System (∆Ssys)

aA + bB cC + dD

DS0

rxndS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

DS0

rxnnS0(products)= S mS0(reactants)S-

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

rxn

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

S0(CO) = 197.9 J/K•mol

S0(O2) = 205.0 J/K•mol

S0(CO2) = 213.6 J/K•mol

DS0

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

DS0

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

Page 6: Second Law of Thermodynamics

Entropy Changes in the System (∆Ssys)

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, ∆ S is negative.

Page 7: Second Law of Thermodynamics

DSuniv = DSsys + DSsurr > 0Spontaneous process:

DSuniv = DSsys + DSsurr = 0Equilibrium process:

Gibbs Free Energy

For a constant-temperature process:

DG = DHsys -TDSsys

Gibbs free energy (G)

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

DG > 0 reaction is spontaneous in the reverse direction. The reaction is non-spontaneous as written. The

DG = 0 The reaction is at equilibrium.

Page 8: Second Law of Thermodynamics

DG = DH - TDS

Page 9: Second Law of Thermodynamics

aA + bB cC + dD

DG0

rxndDG0 (D)

fcDG0 (C)

f= [ + ] - bDG0 (B)f

aDG0 (A)f

[ + ]

DG0

rxnnDG0 (products)

f= S mDG0 (reactants)f

S-

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

DG0 of any element in its stable form is zero.

f

f

Page 10: Second Law of Thermodynamics

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

DG0

rxnnDG0 (products)

f= S mDG0 (reactants)f

S-

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

DG0

rxn6DG0 (H2O)

f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f

[ ]

DG0

rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ

Is the reaction spontaneous at 25 0C?

DG0 = -6405 kJ < 0

spontaneous

Page 11: Second Law of Thermodynamics

Recap: Signs of Thermodynamic Values

Negative PositiveEnthalpy (ΔH)

Exothermic Endothermic

Entropy (ΔS)

Less disorder

More disorder

Gibbs Free Energy (ΔG)

Spontaneous

Not spontaneous

Page 12: Second Law of Thermodynamics

Gibbs Free Energy and Chemical Equilibrium

DG = DG0 + 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

DG = 0 Q = K

0 = DG0 + RT lnK

DG0 = - RT lnK

Page 13: Second Law of Thermodynamics

DG0 = - RT lnK