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Entropy and the Driving Force. Each substance has its own entropy value. This is an absolute scale, because a perfectly ordered substance at 0 K has an entropy of 0. Compare this to enthalpy, where only changes, Δ H, can be measured. Some entropy values. Temperature effects on entropy. - PowerPoint PPT Presentation
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Entropy and the Driving Force
• Each substance has its own entropy value. This is an absolute scale, because a perfectly ordered substance at 0 K has an entropy of 0.
• Compare this to enthalpy, where only changes, ΔH, can be measured.
Some entropy values
Substance (1 mole) So J/mol.K
C (diamond) 2.44
B (s) 5.86
Hg (l) 76.0
O2 (g) 205
Temperature effects on entropy
• S of water • (J/mol.K):
• Increasing temperature increases entropy because random motion of molecules has increased
0o(solid)
43.2
100o(liquid)
86.8
0o(liquid)
65.2
100o(gas)
196.9
Why isn’t everything a gas?
• Gases have more entropy than the equivalent solids, but energy factors – such as intermolecular forces – can make it favorable to exist as a solid
• Other things being equal, reactions that produce gases are favored
Entropy change of reactions
• ∆S can be calculated just like ΔH:
• ∆So = ΣSo(prod) - ΣSo(react)
• For: ½N2(g) + O2 (g) → NO2(g)
• So: 191.6 205.1 240.1 J/mol.K
• ΔSo = So(NO2) – [½So(N2) + So(O2)]
• = 240.1 - [½(191.6) + 205.1)]
• = -60.8 J J/mol.K
½N2(g) + O2 (g) → NO2(g)
• ΔSo = -60.8 J/mol.K
• Entropy decreases in this process, consistent with the fewer moles of gas in the products than in the reactants
Entropy, enthalpy and spontaneity:
“The Universe”
Surroundings
SystemΔH
An increase in temperature also causes an increase in entropy
The “Second Law” of Thermodynamics
• In a spontaneous process, the total entropy change of the universe must be positive (i.e., ∆S > 0)
• The S of the system can decrease, but only if S of the surroundings increases to a greater extent.
• Exothermic reactions always raise the entropy of the surroundings, which is why they are intrinsically more favorable.
The four cases:
• ΔH < 0; ΔSsys > 0: reaction is spontaneous
• ΔH < 0; ΔSsys < 0: depends on values
• ΔH > 0; ΔSsys > 0: depends on values
• ΔH > 0; ΔSsys < 0: reaction not spontaneous
• There ought to be just one function that tells us whether the reaction is spontaneous [“will tend to occur”] or not!
A quick derivation
• ΔSsurroundings = - ΔH/T• Negative sign, because if reaction is
exothermic, ΔH is negative, but if it makes the temperature of the surroundings increase, the entropy of the surroundings should increase (positive ΔS)
• 1/T because effect of increased random molecular motion is more pronounced at lower temperatures than higher
A quick derivation
S S S
SH
TT
T S T S H
define T S G
then G H T S
to tal s y s tem surround ings
sy s tem
to tal s y s tem
to tal
. . .
:
G H T S
G
Reactants
Products
For a spontaneous process, ΔG is always negative
F ind G o for: 1
2N 2 + O 2 N O 2 at 25 o C
W e found S o = - 60 .8 J / m ol K
H = H fo (prod) - H f
o (react)
= H fo (N O 2 ) - [
1
2H f
o (N 2 ) H fo (O 2 )]
(from tab les): 33 .2 k J / m ol 0
(recall H fo for elem ents is 0 )
So G o = 33 .2 k J / m ol - (298 K )(-60 .8 J / m ol K )
A fter converting - 60 .8 J to - 0 .0608 k J,
G o = 51 .4 k J
o
[ ( ) ]1
20
/ m ol
G = 51. 4 k J / m o lo
• The positive sign indicates that this reaction is not spontaneous at 25oC
• The o indicates standard conditions (P = 1 atm; concentrations = 1 M)
• ΔGo = 33.2 kJ/mol – T(-60.8 J/mol.K)
• Because ΔH is positive and ΔS is negative, this reaction can never be spontaneous.
Another way to calculate ΔGo
• ΔGfo values are also tabulated
• ΔGfo = Gibbs free energy change when
one mole of a substance is formed from its elements in their standard states
• ΔGo = ΣΔGfo(prod) – ΣΔGf
o(reactants)
So is ΔGo related to Keq?And what can ΔG tell us about systems
not at equilibrium?
• Consider: -ln(K/Q)
• Recall: ln(1) = 0
• ln (>1) = positive number
• ln (<1) = negative number
• If Q < K, reaction goes to right, -ln(K/Q) is -
• If Q> K, reaction goes to left, -ln(K/Q) is +
If Q < K, reaction goes to right, -ln(K/Q) is -If Q> K, reaction goes to left, -ln(K/Q) is +
• So we can say ∆G α -ln(K/Q)
• Specifically: ∆G = -RTln(K/Q)
• At standard conditions, Q = 1 (!) . . . so . . .
• ∆Go = -RT lnK a marvelous equation!
• R = 8.31 J/mol.K
• This allows us to calculate an equilibrium constant from basic thermodynamic information