Second Law of Thermodynamics
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Transcript of Second Law of Thermodynamics
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Second Law of Thermodynamics
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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
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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
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C. Entropy tends to increase in reactions in which the number of molecules increases
Increasing Entropy
D. Entropy increase with an increase in temperature
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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
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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.
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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.
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DG = DH - TDS
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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
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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
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Recap: Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH)
Exothermic Endothermic
Entropy (ΔS)
Less disorder
More disorder
Gibbs Free Energy (ΔG)
Spontaneous
Not spontaneous
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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
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DG0 = - RT lnK