10 Nov 97 Entropy & Free Energy (Ch 20) - lect. 2 1 Entropy and Free Energy (Kotz Ch 20) - Lecture #2 • Spontaneous vs. non-spontaneous • thermodynamics vs. kinetics • entropy = randomness (S o ) Gibbs free energy (G o ) G o for reactions - predicting spontaneous direct thermodynamics of coupled reactions G rxn versus G o rxn predicting equilibrium constants from G o rxn
Entropy and Free Energy (Kotz Ch 20) - Lecture #2. Spontaneous vs. non-spontaneous thermodynamics vs. kinetics entropy = randomness (S o ). Gibbs free energy ( G o ) G o for reactions - predicting spontaneous direction thermodynamics of coupled reactions G rxn versus G o rxn - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 21
Entropy and Free Energy (Kotz Ch 20) - Lecture #2
• Spontaneous vs. non-spontaneous• thermodynamics vs. kinetics• entropy = randomness (So)
• Gibbs free energy (Go)• Go for reactions - predicting spontaneous direction
• thermodynamics of coupled reactions• Grxn versus Go
rxn
• predicting equilibrium constants from Gorxn
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 22
Entropy and Free Energy ( Kotz Ch 20 )
• How can we predict if a reaction can occur, given enough time?
• Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur.
• To predict if a reaction can occur at a reasonable rate, one needs to consider:
• some processes are spontaneous; others never occur. WHY ?
THERMODYNAMICS
KINETICS
9-paper.mov20m02vd1.mov
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 23
Product-Favored Reactions
E.g. thermite reaction
Fe2O3(s) + 2 Al(s)
2 Fe(s) + Al2O3(s)
H = - 848 kJ
In general, product-favored reactions are exothermic.
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 24
Non-exothermic spontaneous reactions
But many spontaneous reactions or processes are endothermic . . .
NH4NO3(s) + heat NH4+ (aq) + NO3
- (aq)Hsol = +25.7 kJ/mol
or have H = 0 . . .
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 25
PROBABILITY - predictor of most stable state
WHY DO PROCESSES with H = 0 occur ?
Consider expansion of gases to equal pressure:
This is spontaneous because the final state,with equal # molecules in each flask, is much more probable than the initial state,with all molecules in flask 1, none in flask 2
SYSTEM CHANGES to state of HIGHER PROBABILITYFor entropy-driven reactions - the more RANDOM state.
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 26
Standard Entropies, So
• Every substance at a given temperature and in a specific phase has a well-defined Entropy• At 298o the entropy of a substance is called
So - with UNITS of J.K-1.mol-1
• The larger the value of So, the greater the degree of disorder or randomness
e.g. So (in J.K-1mol-1) : Br2 (liq) = 152.2
Br2 (gas) = 245.5
For any process: So = So(final) - So(initial)
So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 27
S (gases) > S (liquids) > S (solids)
So (J/K•mol)
H2O(gas) 188.8
H2O(liq) 69.9
H2O (s) 47.9
Ice Water
Vapour
Entropy and Phase
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 28
The entropy of a substance increases with temperature.
Molecular motions of heptane at different temps.
Entropy and Temperature
9_heptane.mov20m04an2.mov
Higher T means :• more randomness• larger S
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 29
Entropy and complexity
Increase in molecular complexity generally leads to increase in S.
9_alkmot.mov20m04an3.mov
SSoo (J/K•mol) (J/K•mol)
CHCH44 248.2248.2
CC22HH66 336.1 336.1
CC33HH88 419.4419.4
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 210
• Ionic Solids : Entropy depends on extent of motion of ions. This depends on the strength of coulombic attraction.
Entropy of Ionic Substances
• Entropy increases when a pure liquid or solid dissolves in a solvent.
Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.
Calculating S for a Reaction
So = So (products) - So (reactants)
If S DECREASES, why is this a SPONTANEOUS REACTION??
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 213
2nd Law of Thermodynamics
Suniverse = Ssystem + Ssurroundings
Suniverse > 0 for product-favored process
First calc. entropy created by matter dispersal (Ssystem)
Next, calc. entropy created by energy dispersal (Ssurround)
A reaction is spontaneous (product-favored) if S for the universe is positive.
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 214
Calculating S(universe)
2 H2(g) + O2(g) 2 H2O(liq)
Sosystem = -326.9 J/K
T
H- =
T
q =
systemgssurroundingssurroundin
ooS
K 298.15
J/kJ) kJ)(1000 (-571.7 - = gssurroundin
oS
Sosurroundings = +1917 J/K
Can calculate that Horxn = Ho
system = -571.7 kJ
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 215
Calculating S(universe) (2)
2 H2(g) + O2(g) 2 H2O(liq)
Sosystem = -326.9 J/K (less matter dispersal)
Sosurroundings = +1917 J/K (more energy dispersal)
The entropy of the universe increases so the reaction is spontaneous.
Souniverse = +1590 J/K
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 216
E = q + w
The Laws of Thermodynamics
0. Two bodies in thermal equilibrium are at same T
1. Energy can never be created or destroyed.
2. The total entropy of the UNIVERSE ( = system plus surroundings) MUST INCREASE in every spontaneous process.
STOTAL = Ssystem + Ssurroundings > 0
3. The entropy (S) of a pure, perfectly crystalline compound at T = 0 K is ZERO. (no disorder)
ST=0 = 0 (perfect xll)
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 217
Gibbs Free Energy, G
Suniv = Ssurr + Ssys
Suniv = Hsys
T + Ssys
Go = Ho - TSo
Multiply through by -T-TSuniv = Hsys - TSsys
-TSuniv = change in Gibbs free energy
for the system = Gsystem
Under standard conditions —The GibbsEquation
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 218
Standard Gibbs Free Energies, Gof
• Every substance in a specific state has a Gibbs Free Energy, G = H - TS• recall: only H can be measured. Therefore: there is no absolute scale for G• only G values can be determined
• Gof the Gibbs Free Energy of formation (from
elements) is used as the “standard value”
• We set the scale of G to be consistent with that
for H - Gof for elements in standard states = 0
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 219
Go < 0 for all SPONTANEOUS processes
Sign of G for Spontaneous processes
STOTAL = Ssystem + Ssurroundings > 0
2nd LAW requirement for SPONTANEITY is :
Multiply by T TSsystem + TSsurroundings > 0
and Ssurroundings = Hosystem/T
Thus TSsystem - Hosystem > 0
Multiply by -1 (->reverse > to <), drop subscript “system”
Ho -TS < 0 and Go = Ho
-TS
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 220
Go = Ho - TSo
• change in Gibbs free energy =
(total free energy change for system - free energy lost in disordering the system)
• If reaction is exothermic (Ho is -ve) and entropy increases (So is +ve), then
Go must be -ve and reaction CAN proceed.• If reaction is endothermic (Ho is +ve), and entropy decreases (So is -ve), then Go must be +ve; reaction CANNOT proceed.
Sign of Gibbs Free Energy, G
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 221
Gibbs Free Energy changes for reactions
Ho So Go Reaction
exo (-) increase(+) - Product-favored
endo(+) decrease(-) + Reactant-favored
exo (-) decrease(-) ? T dependent
endo(+) increase(+) ? T dependent
Spontaneous in last 2 cases only ifTemperature is such that Go < 0
Go = Ho - TSo
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 222
Methods of calculating G
Two methods of calculating Go
GGoorxnrxn = = GGff
oo (products) - (products) - G Gffoo (reactants) (reactants)
a) Determine Horxn and So
rxn and use Gibbs
equation.
b) Use tabulated values of free energies of
formation, Gfo.
Go = Ho - TSo
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 223
Calculating Gorxn
EXAMPLE: Combustion of acetylene
C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g)
From standard enthalpies of formation: Horxn = -1238 kJ
From standard molar entropies: Sorxn = - 0.0974 kJ/K
Calculate Gorxnfrom Go = Ho - TSo
Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)
= -1209 kJ
Reaction is product-favored in spite of negative So
rxn. Reaction is “enthalpy driven”
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 224
Is the dissolution of ammonium nitrate product-favored?
If so, is it enthalpy- or entropy-driven?
EXAMPLE 2:
NH4NO3(s) NH4NO3(aq)
Calculating Gorxn for NH4NO3(s)
9_amnit.mov20 m07vd1.mov
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 225
Gorxn for NH4NO3(s) NH4NO3(aq)
From tables of thermodynamic data we find
Horxn = +25.7 kJ
Sorxn = +108.7 J/K or +0.1087 kJ/K
Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)
= -6.7 kJ
Reaction is product-favored
. . . in spite of positive Horxn.
Reaction is “entropy driven”
10 Nov 97 Entropy & Free Energy (Ch 20) -
lect. 226
Calculating Gorxn
EXAMPLE 3: Combustion of carbon
C(graphite) + O2(g) CO2(g)
Gorxn = Gf
o(CO2) - [Gfo(graph) + Gf
o(O2)]
Gorxn = -394.4 kJ - [ 0 + 0]
Note that free energy of formation of an element in its standard state is 0.