chapter 20 thermodynamics: entropy, free energy and the direction of chemical reactions
TRANSCRIPT
Chapter 20
Thermodynamics:
Entropy, Free Energy and the Direction of Chemical Reactions
Limitations of the First Law of Thermodynamics
E = q + w
Euniverse = Esystem + Esurroundings
Esystem = -Esurroundings
The total energy-mass of the universe is constant.
However, this does not tell us anything about the direction of change in the universe.
A spontaneous endothermic chemical reaction
water
Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l).
H0rxn = +62.3 kJ
The Concept of Entropy (S)
Entropy refers to the state of order.
A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.
solid liquid gasmore order less order
crystal + liquid ions in solution
more order less order
more order less order
crystal + crystal gases + ions in solution
The number of ways to arrange a deck of playing cards
1 atm evacuated
Spontaneous expansion of a gas
stopcock closed
stopcock opened
0.5 atm 0.5 atm
1877 Ludwig Boltzman S = k ln W
where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/NA (R = universal gas constant, NA = Avogadro’s number.
•A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy.
•A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy.
Suniverse = Ssystem + Ssurroundings > 0
This is the second law of thermodynamics.
Random motion in a crystal
The third law of thermodynamics.
A perfect crystal has zero entropy at a temperature of absolute zero.
Ssystem = 0 at 0 K
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Predicting Relative S0 Values of a System
1. Temperature changes
2. Physical states and phase changes
3. Dissolution of a solid or liquid
5. Atomic size or molecular complexity
4. Dissolution of a gas
S0 increases as the temperature rises.
S0 increases as a more ordered phase changes to a less ordered phase.
S0 of a dissolved solid or liquid is usually greater than the S0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.
A gas becomes more ordered when it dissolves in a liquid or solid.
In similar substances, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.
The increase in entropy from solid to liquid to gas
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The entropy change accompanying the dissolution of a salt
pure solid
pure liquid
solution
MIX
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Ethanol Water Solution of ethanol
and water
The small increase in entropy when ethanol dissolves in water
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The large decrease in entropy when a gas dissolves in a liquid
O2 gas
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NO
NO2
N2O4
Entropy and vibrational motion
Sample 1 Predicting Relative Entropy Values
PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]:
(a) 1mol of SO2(g) or 1mol of SO3(g)(b) 1mol of CO2(s) or 1mol of CO2(g)(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)(d) 1mol of KBr(s) or 1mol of KBr(aq)(e) Seawater in midwinter at 20C or in midsummer at 230C(f) 1mol of CF4(g) or 1mol of CCl4(g)
Sample 2 Calculating the Standard Entropy of Reaction, S0rxn
PROBLEM: Calculate S0rxn for the combustion of 1mol of propane at 250C.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Components of S0universe for spontaneous reactions
exothermic
system becomes more disordered
exothermic
system becomes more ordered
endothermic
system becomes more disordered
Sample 3 Determining Reaction Spontaneity
PROBLEM: At 298K, the formation of ammonia has a negative S0sys;
Calculate S0rxn, and state whether the reaction occurs
spontaneously at this temperature. H0fNH3 = -45.9 kJ/mol
N2(g) + 3H2(g) 2NH3(g) S0sys = -197J/K
G0system = H0
system - TS0system
G0rxn = mG0
products - nG0reactants
Sample 4 Calculating G0 from Enthalpy and Entropy Values
PROBLEM: Potassium chlorate, one of the common oxidizing agents in explosives, fireworks, and matchheads, undergoes a solid-state redox reaction when heated. In this reaction, note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation):
4KClO3(s) 3KClO4(s) + KCl(s)
+7 -1+5
Use H0f and S0 values to calculate G0
sys at 250C for this reaction.
Sample 5
PROBLEM:
Calculating G0rxn from G0
f Values
Use G0f values to calculate Grxn for the reaction:
4KClO3(s) 3KClO4(s) + KCl(s)
Sample 6
PROBLEM:
Determining the Effect of Temperature on G0
An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g):
2SO2(g) + O2(g) 2SO3(g)
At 298K, G0 = -141.6kJ; H0 = -198.4kJ; and S0 = -187.9J/K
(a) Use the data to decide if this reaction is spontaneous at 250C, and predict how G0 will change with increasing T.
(b) Assuming H0 and S0 are constant with increasing T, is the reaction spontaneous at 900.0C?
Sample 7
PROBLEM:
Determining the Effect of Temperature on G0
A reaction is nonspontaneous at room temperature but is spontaneous at -40 0C. What can you say about the signs and relative magnitudes of H0 S0 and -TS0
Reaction Spontaneity and the Signs of H0, S0, and G0
H0 S0 -TS0 G0 Description
- + - -
+ - + +
+ + - + or -
- - + + or -
Spontaneous at all T
Nonspontaneous at all T
Spontaneous at higher T;nonspontaneous at lower T
Spontaneous at lower T;nonspontaneous at higher T
The effect of temperature on reaction spontaneity
G and the Work a System Can Do
For a spontaneous process, G is the maximum work obtainable from the system as the process takes place: G = workmax
For a nonspontaneous process, G is the maximum work that must be done to the system as the process takes place: G = workmax
An example
The coupling of a nonspontaneous reaction to the hydrolysis of ATP.
The cycling of metabolic free enery through ATP
Free Energy, Equilibrium and Reaction Direction
•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0)
•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0)
•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (G = 0)
G = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and lnQ = 0 so
G0 = - RT lnK
FO
RW
AR
D R
EA
CT
ION
RE
VE
RS
E R
EA
CT
ION
The Relationship Between G0 and K at 250C
G0(kJ) K Significance
200
100
50
10
1
0
-1
-10
-50
-100
-200
9x10-36
3x10-18
2x10-9
2x10-2
7x10-1
1
1.5
5x101
6x108
3x1017
1x1035
Essentially no forward reaction; reverse reaction goes to completion
Forward and reverse reactions proceed to same extent
Forward reaction goes to completion; essentially no reverse reaction
Sample Problem 6
PROBLEM:
Calculating G at Nonstandard Conditions
The oxidation of SO2, which we considered in Sample Problem 6
2SO2(g) + O2(g) 2SO3(g)
is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature.
(a) Calculate K at 298K and at 973K. (G0298 = -141.6kJ/mol of reaction as
written using H0 and S0 values at 973K. G0973 = -12.12kJ/mol of reaction
as written.)(b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO2, 0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature?(c) Calculate G for the system in part (b) at each temperature.
Sample Problem 7
PROBLEM:
Calculating G at Nonstandard Conditions
At 298 K hypobromous acid (HBrO) dissociates in water with aKa of 2.3 x 10-9.
(a) Calculate G0298
(b) Calculate G if [H3O+] = 6.0 x 10-4M, [BrO-] = 0.1 M and [HBrO]=0.20 M
The relation between free energy and the extent of reaction
G0 < 0 K >1 G0 > 0
K <1