chemical thermodynamics blb 12 th chapter 19. chemical reactions 1. will the reaction occur, i.e. is...
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Chemical Thermodynamics
BLB 12th Chapter 19
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Chemical Reactions
1. Will the reaction occur, i.e. is it spontaneous? Ch. 5, 19
2. How fast will the reaction occur? Ch. 14
3. How far will the reaction proceed? Ch. 15
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Review terms
energy heat work pathway state function system surroundings
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Review terms, cont.
exothermic endothermic enthalpy enthalpy change standard state std. enthalpy of formation 1st Law of
Thermodynamics
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19.1 Spontaneous Processes Spontaneous – proceeds on its own without
any outside assistance product-favored; K > 1 not necessarily fast the direction a process will take if left alone and
given enough time. Nonspontaneous
opposite direction of spontaneous reactant-favored; K < 1 not necessarily slow
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Spontaneity and Energy
Examples of spontaneous systems: Brick falling Ball rolling downhill Hot objects cooling Combustion reactions
Are all spontaneous processes accompanied by a loss of heat, that is, exothermic?
Spontaneity is temperature-dependent.
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Reversible & Irreversible Systems
Reversible – a change in a system for which the system can restored by exactly reversing the change – a system at equilibriumex. melting ice at 0°C
Irreversible – a process that cannot be reversed to restore the system and surroundings to their original states – a spontaneous processex. melting ice at 25°C
See p. 788-790 (last paragraph of section)
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19.2 Entropy and the 2nd Law of Thermodynamics
Entropy, S – measure of randomness State function Temperature-dependent
A random (or dispersed) system is favored due to probability.
“Entropy Is Simple – If We Avoid the Briar Patches”
Frank Lambert, Occidental College, ret.
http://entropysimple.oxy.edu
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Entropy Change
ΔS = Sfinal − Sinitial (a state function)
(isothermal) as for phase changes.
ΔS > 0 is favorable
T) (constant T
qS rev
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Calculating ΔS for Changes of State
)(
:1
KinTT
H
T
qS
moleFor
Kmol
Jor
K
Jofunits
T
qS
syssyssys
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Problem 26 The freezing point of Ga is 29.8°C and the enthalpy of fusion is 5.59 kJ/mol.
a. Is ΔS + or − for Ga(l) → Ga(s) at the freezing point?
b. Calculate the value of ΔS when 60.0 g of Ga(l) solidifies.
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System & Surroundings
Dividing the universe:
1. System – dispersal of matter by reaction: reactants → products
2. Surroundings – dispersal of energy as heat
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2nd Law of Thermodynamics
The entropy of the universe increases for any spontaneous process. ΔSuniv > 0
ΔSuniv = ΔSsys + ΔSsurr
For a “reversible” process: ΔSuniv = 0. For an irreversible process:
Net entropy increase ►spontaneous Net entropy decrease ► nonspontaneous
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19.3 The Molecular Interpretation of Entropy
Molecules have degrees of freedom based upon their motion Translational Vibrational Rotational
Motion of water (Figure 19.8, p. 796) Lowering the temperature decreases
the entropy.
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Boltzman & Microstates S = k ln W
(W = # of microstates) If # microstates ↑,
then entropy ↑. Increasing volume,
temperature, # of molecules increases the # of microstates.
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Examples of systems that have increased entropy
Entropy increases for: Changes of state: solid → liquid → gas (T) Expansion of a gas (V) Dissolution: solid → solution (V) Production of more moles in a chemical
reaction (# of particles) Ionic solids: lower ionic charge
S° (J/mol·K)
Na2CO3 136
MgCO3 66
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Changes of State
H2O state S° (J/mol·K)
l 69.91
g 188.83
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# microstates ↑, system entropy ↑
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Dissolution
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Expansion of a Gas
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2 NO(g) + O2(g) → 2 NO2(g)
S° + or −?
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Problem 22 TNT (trinitrotoluene) Detonation
4 C3H5N3O9(l) → 6 N2(g) + 12 CO2(g) + 10 H2O(g) + O2(g)
a) Spontaneous?
b) Sign of q?
c) Can the sign of w be determined?
d) Can the sign of ΔE be determined?
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3rd Law of Thermodynamics
The entropy, S, of a pure crystalline substance at absolute zero (0 K) is zero.
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19.4 Entropy Changes in Chemical Reactions
Standard molar entropy values, S° (J/mol·K): increase in value as temperature increases
from 0 K have been determined for common
substances (Appendix C, pp. 1059-1061) increase with molar mass increase with # of atoms in molecule
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Calculating ΔS°sys
ΔS°sys = ∑nS°(products) - ∑mS°(reactants)
(where n and m are coefficients in the chemical equation)
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2 NO(g) + O2(g) → 2 NO2(g)
S° =
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Problem 54(c)
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Problem
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Entropy Changes in the Surroundings
T
H
T
qS syssyssurr
Heat flow affects surroundings. As T increases, ΔH becomes less important. As T decreases, ΔH becomes more important.
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Calculating ΔS°univ
ΔSuniv = ΔSsys + ΔSsurr
by obtaining ΔSsys and ΔSsurr
If ΔSuniv > 0, the reaction is spontaneous.
But there is a better way – one in which only the system is involved.
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19.5 Gibbs Free Energy
The spontaneity of a reaction involves both enthalpy (energy) and entropy (matter).
Gibbs Free Energy, ΔG makes use of ΔHsys and ΔSsys to predict spontaneity.
ΔGsys represents the total energy change for a system.
G = H – TS or ΔG = ΔH – TΔS or, under standard conditions:
ΔG° = ΔH° – TΔS°
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syssyssys
syssys
G
univ
syssysuniv
STHG
STHSTT
HSS
sys
)(
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Gibbs Free Energy
If: ΔG < 0, forward reaction is spontaneous ΔG = 0, reaction is at equilibrium ΔG > 0, forward reaction is nonspontaneous
In any spontaneous process at constant temperature and pressure, the free energy always decreases.
ΔG is a state function. ΔGf° of elements in their standard state is
zero.
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Calculating ΔG°sys
ΔG°sys = ΔH°sys − TΔS°sys
or
ΔG°sys = ∑nΔG°f (products) - ∑mΔG°f (reactants)
(where n and m are coefficients in the chemical equation)
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19.6 Free Energy and Temperature
ΔH ΔS −TΔS ΔG Reaction
− + − always −spontaneous at all TK > 1
+ − + always +nonspontaneous at all TK < 1
− − +− @ low T+ @high T
spontaneous at low Tnonspontaneous at high T
+ + −+ @ low T− @high T
nonspontaneous at low Tspontaneous at high T
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Problem 60(b) & 83
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Driving force of a reaction
For a reaction where ΔG < 0: Enthalpy-driven – if ΔH < 0 and ΔS < 0; at
low temp. Entropy-driven – if ΔH > 0 and ΔS > 0; at
high temp. “cross-over point” is where ΔG = 0
S
HTSTHSTH
0
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Problem
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19.7 Free Energy and K
If conditions are non-standard:
ΔG = ΔG° + RT lnQ R = 8.3145 J/mol·K If at equilibrium:
ΔG = ΔG° + RT lnQ = 0
ΔG° = −RT lnK
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Problem
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Problem 81
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