Download Chapter 17 Thermodynamics: Directionality of Chemical Reactions

Reactant- & Product-Favored Processes
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
Why are equilibria product- or
reactant- favored?
Why do some reactions occur
spontaneously?
http://academic.cengage.com/chemistry/moore
Why do others require help (heat,
spark…)?
Chapter 17
Thermodynamics: Directionality of
Chemical Reactions
Stephen C. Foster • Mississippi State University
Exothermic reactions are often
product favored (but not always).
2 Al(s) + 3 Br2(ℓ) → 2 AlBr3(ℓ)
Product favored; spontaneous
Chemical Reactions & Dispersal of Energy
Probability and Dispersal of Energy
“Energy will spread out (disperse) unless it is hindered
from doing so”.
What are the possible outcomes if 2 excited atoms
make contact with 2 more atoms?
Most exothermic reactions are product favored.
• E is transferred to the surroundings.
• Bond (potential) E dispersed into many more atoms
and molecules.
Dispersal of energy is probability driven:
• E is more likely to spread over many atoms than
over just a few.
Excited atoms
(A, B)
A** B
A* B
A B*
A B
C
C
C
C
A*B*
+
C D
Unexcited
atoms
E packet
D
D*
D*
D**
A* B* C D
A B** C D
A B C** D
A* B C* D
A B* C* D
A B C* D*
E stays with original atoms in only 3 (of 10) cases:
• 70% chance that E will disperse.
• More atoms: E concentration is much less likely.
Dispersal of Matter
Measuring Dispersal of E: Entropy
E is dispersed when a system expands.
• Gases expand to fill a container.
In summary:
summary
Processes are product favored if E is
dispersed from a few levels into many
levels.
• Quantum theory: E-levels
for gas motion get closer
together when volume
expands.
• More E-levels become
accessible at a given T.
• E will disperse over more
levels.
vacuum
removable
barrier
This typically occurs during:
• exothermic reactions (bond E
dispersal).
• processes in which matter is
dispersed.
Br2(g)
In thermodynamics, nanoscale E dispersal is measured
by the Entropy (S) of a system.
1
Measuring Dispersal of E: Entropy
At constant T:
ΔS = Sfinal – Sinitial =
qrev
T
Absolute T
“rev”: Only applies to reversible changes.
A reversible process:
• can be reversed by a slight change in conditions.
• e.g. ice melting at 0.0°C and 1 atm is reversible.
 small decrease in T will cause it to refreeze.
Measuring Dispersal of E: Entropy
Calculate ΔS when 25.0 g of Al(s) melts at its normal
melting point (660.3 °C; 1 atm). ΔH°fus(Al)=10.7 kJ mol-1
qrev
ΔHfus
qp = ΔH
ΔS =
=
Reversible if
T
T
normal melting
T = 660.3 °C = 933.5 K
nAl =
ΔS =
25.0 g
= 0.9266 mol
26.98 g mol-1
0.9266 mol(10.7 kJ mol-1) = 0.01062 kJ K-1
933.5 K
ΔS = 10.6 J K-1
Absolute Entropy Values
Absolute Entropy Values
A perfect crystal, at absolute zero (0 K) has:
• Minimum molecular motion (min. E dispersal).
Standard molar entropies (S°), are measured by
heating 1 mol of substance from 0 K to:
A specified T (often 25°C = 298.15 K)
At constant P = 1 bar
S=0
(perfect crystal at 0 K)
(The 3rd law of thermodynamics)
Measure the heat to change from 0 K  room-T
(reversibly). Gives ΔS and the absolute S at room T.
ΔS = Sfinal – Sinitial =
S° has J K-1mol-1 units.
S° is always positive, for all materials.
qrev
T
Sroom T = ΔS – Sinitial = ΔS
Standard Molar Entropies (298.15 K)
Compound
C (graphite)
C(g)
CH4(g)
C2H6(g)
C3H8(g)
C2H4(g)
CH3OH(ℓ)
CO(g)
CO2(g)
F2(g)
S° (J K-1 mol-1)
5.740
158.096
186.264
229.60
269.9
219.4
126.8
197.674
213.74
202.78
Compound
Cl2(g)
Br2(ℓ)
I2(s)
HCl(g)
H2O(g)
H2O(ℓ)
NaCl(s)
KOH(s)
NaCl(aq)
KOH(aq)
S° (J K-1 mol-1)
223.066
152.231
116.135
186.908
188.825
69.91
72.13
78.9
115.5
91.6
Qualitative Guidelines for Entropy
Sgas >> Sliquid > Ssolid
Solid molecules vibrate about fixed lattice points.
Liquid molecules can slide past each other.
Gas molecules have few restrictions on motion.
More motion = higher S
S°(I2)
S°(Br2)
S°(Cl2)
gas
261
245
223
liquid
152
-
solid
116
-
(J K-1 mol-1)
2
Qualitative Guidelines for Entropy
Larger, more complex, molecules have larger S:
S°complex molecule > S°simple molecule
Species
CH4
C2 H6
C3 H8
type S°(J K-1 mol-1) Species
alkane
186
Ar
alkane
230
CO2
alkane
270
C3 H8
mass S°(J K-1 mol-1)
40
155
44
214
44
270
Qualitative Guidelines for Entropy
Ionic solids
Weaker ionic forces = larger S
(Less tightly bound = easier motion)
S°(J K-1mol-1)
Comment
MgO
27
+2 and -2 charges
NaF
51
+1 and -1 charges
NaCl
72
Cl- much larger than F-, less attraction.
Larger molecules have more ways to distribute E
(more bonds to hold potential and kinetic E).
Qualitative Guidelines for Entropy
Qualitative Guidelines for Entropy
Solid or liquid dissolving
Larger matter dispersal. S usually* increases:
Gas dissolving
Gas-molecule motion becomes restricted. ΔS < 0.
Substance
CH3COOH(ℓ)
NH4NO3(s)
S° (pure)
160
151
S° (aq)
179
260
J K-1 mol-1 units
*Strong solvation may increase order. ΔS < 0 is
possible.
Substance S° (gas) S° (aq. soln)
CO2
214
118
H2
131
58
CH3OH
240
133
J K-1 mol-1 units
Predicting Entropy Changes
Predicting Entropy Changes
CaCO3(s) → CaO(s) + CO2(g) ΔS = 161 J K-1mol-1
Large entropy increase:
• Solid → gas, ΔS > 0.
• 1 molecule → 2 molecules, ΔS > 0.
Predict whether S will increase or decrease for:
2 CO(g) + O2(g) → 2 CO2(g)
Decrease
3 gas molecules → 2 gas molecules.
NaCl(s) → Na+(aq) + Cl-(aq)
H2(g) + F2(g) → 2 HF(g)
ΔS = 14 J K-1mol-1
• All gases, S ≈ constant.
• 2 reactant gases, 2 product gases, S ≈ constant.
Actual ΔS shows a small increase.
Increase
Ions locked in a crystal → ions dispersed
in water.
H2O(ℓ) → H2O(s)
Decrease Liquid molecules less restricted than solid
molecules.
3
Calculating Entropy Changes
Calculating Entropy Changes
Entropy is a state function (like H). For a reaction:
Example:: ΔS° at 298 K for: 2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ)
ΔS° = (nproductS°product) – S(nreactantS°reactant)
Example
Methanol is a common fuel additive. Calculate ΔS°
at 298 K for its combustion:
2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ)
K-1
ΔS° = (nproductS°product) – (nreactantS°reactant)
= (2S°CO2 + 4S°H2O) – (2S°CH3OH + 3S°O2)
= [2(213.74) + 4(69.91)] – [2(126.8) + 3(205.14)]
= 707.12 – 869.02
= –161.90 J K-1 mol-1
mol-1):
Look up S° values (J
CH3OH(ℓ) = 126.8
CO2(g)
= 213.74
O2(g) = 205.138
H2O(ℓ) = 69.91
S decreases as expected: 3 gas + 2 ℓ → 2 gas + 4 ℓ
Entropy & the 2nd Law of Thermodynamics
Entropy & the 2nd Law of Thermodynamics
E dispersal is accompanied by an increase in disorder
of a system:
The entropy of a system can increase or decrease.
Increased disorder = increased S
• Suniverse always goes up.
• Suniverse = Ssystem + Ssurr
(No subscript? S = Ssystem ).
• If Ssystem falls, Ssurr must increase by a larger amount.
Second Law of Thermodynamics
“The total entropy of the universe is continuously
increasing.”
• The universe is slowly becoming more disordered.
ΔS°surr =
qrev
T
=
ΔHsurr
T
=
-ΔHsys
T
Entropy & the 2nd Law of Thermodynamics
Gibbs Free Energy
Calculate ΔSuniv
° for the combustion of octane in air:
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(ℓ)
For the reaction ΔH° = -11,024 kJ and ΔS° = -1383.9
J/K at 298.15 K.
Neither entropy (S), nor enthalpy (H ), alone can predict
whether a reaction is product favored.
Spontaneous (product favored) reactions can:
° = ΔSsys
° + ΔSsurr
°
ΔSuniv
• Increase or decrease the Ssystem.
• Be exothermic or endothermic.
-ΔHsys / T
given
ΔSuniv = -1383.9 J/K + (+
+11,024 x 103 J)/298.15 K
The Gibbs free energy (G
(G), combines H and S.
ΔG does predict if a reaction is product favored or not.
= -1383.9 J/K + 36.975 kJ/K
= +35,591 J/K
4
Gibbs Free Energy
Gibbs Free Energy
For a constant T change, ΔG is equal to:
ΔG° can be calculated from ΔH° and ΔS° values…
… or from Gibbs free energies of formation (ΔGf°).
ΔG = ΔH - TΔS
If G:
Decreases (ΔG < 0) a reaction is product favored
Increases (ΔG > 0) a reaction is reactant favored
(at constant P and T).
ΔG° = (nproductΔGf°product) – (nreactantΔGf°reactant)
ΔGf° equals:
• ΔGreaction to make 1 mol of compound from its
elements.
• 0 for an element in its most stable state (like ΔHf° )
Gibbs Free Energy
Effect of T on Reaction Direction
Calculate ΔG° for the following reaction at 25°C:
CO(g) + 2 H2(g) → CH3OH(ℓ)
ΔGf° values: CO(g) = -137.2 kJ/mol
CH3OH(ℓ) = -166.3 kJ/mol
Reactions go from reactant to product-favored when:
ΔG = 0 = ΔH − TΔS
ΔG° = ΔGf°CH3OH – (2ΔGf°H2 + 1ΔGf°CO)
or when: ΔH = TΔS
or T = ΔH
ΔS
element in its
standard state
= -166.3 kJ/mol – [2(0) + 1(-137.2 kJ/mol)]
= -29.1 kJ/mol
Product favored
Effect of T on Reaction Direction
Effect of T on Reaction Direction
Since ΔG = ΔH – TΔS
At what T will the following reaction be product
favored?
CH4(g) + H2O(g)
CO(g) + 3 H2(g)
Sign of ΔH
Sign of ΔS
Product Favored?
Negative (exothermic)
Positive
Yes
Negative (exothermic)
Negative
Yes at low T; No at high T
Positive (endothermic) Positive
No at low T; Yes at high T
Positive (endothermic) Negative
No
ΔHf° (kJ/mol)
−74.81
S° (J K-1 mol-1) 186.26
−241.82
188.83
−110.53
197.67
0
130.68
ΔS°=[(197.67) +3(130.68)] − [186.26 + 188.83]J K-1mol-1
= 214.62 J K-1 mol-1
ΔH° = [(-110.53) + 3(0)] − [(-74.81) + (-241.82)] kJ mol-1
= 206.10 kJ mol-1
It will switch direction at:
2.0610 x 105 J mol-1
ΔH
T=
=
= 960.30 K
214.62 J K-1 mol-1
ΔS
5
Gibbs Free Energy and Equilibrium Constants
Gibbs Free Energy and Equilibrium Constants
Gibbs free energy G and the equilibrium constant K
both show if a reaction is product-favored and are
related.
ΔG° = −RT
−RT ln K°
Evaluate K° (298 K) for the reaction:
2 NO2(g)
N2O4(g)
ΔG° = ΔGf°(N2O4) - 2ΔGf°(NO2)
= 97.89 – 2(51.31) kJ/mol = -4.73 kJ/mol
K° must be unitless.
Each quantity in the K° expression is divided by its
standard state value:
ln K° = –
ΔG°
-4.73 x 103 J/mol
=–
= 1.91
RT
(8.314 J K-1 mol-1)(298 K)
• concentrations by 1 mol L-1 (solids and liquids).
K° = 6.75
• pressures by 1 bar (gas reactions).
Gibbs Free Energy & Maximum Work
Gibbs Free Energy & Maximum Work
ΔG = maximum useful work that can be done by a
reaction on its surroundings (constant T and P).
ΔG also equals the minimum work required to cause a
reactant-favored process to occur.
ΔG = wsystem = - wmax
2 H2O(g)
“free” energy = available energy
For:
2 H2(g) + O2(g)
2 H2O(g)
2 H2(g) + O2(g)
ΔG°= 474.3 kJ
A minimum of 474.3 kJ of work must be used to
produce 2 mol of H2 from liquid water.
ΔG°= -474.3 kJ
Every 2 moles of H2 consumed can do up to 474.3 kJ
of work.
Thermodynamic and Kinetic Stability
Thermodynamically stable
Reaction is not product-favored.
• Thermodynamically stable:
Pd + O2 → PdO2
ΔG° = +326 kJ/mol
• Thermodynamically unstable:
4 Al + 3 O2 → 2 Al2O3
ΔG° = -3164.6 kJ/mol
Kinetically stable
Product-favored, but too slow to be important.
• Diamond is thermodynamically unstable
Cdiamond → Cgraphite ΔG° = -2.9 kJ/mol
• Kinetically stable, too slow to be important.
• Ea is too large
6