Thermodynamics: Entropy, Free Energy Direction of Chemical Reactions Chapter 20 Entropy and Free Energy - Spontaneity of Reaction 1. The Second Law of.
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Thermodynamics: Entropy,
Free Energy
Direction of Chemical Reactions
Chapter 20 Entropy and Free Energy - Spontaneity of Reaction
1. The Second Law of Thermodynamics: Predicting Spontaneous Change
First Law and enthalpy, Second Law and entropy, Standard Entropy and Third Law
2. Calculating the Change in Entropy of a ReactionThe standard entropy of reaction, entropy change and equilibrium
3. Entropy, Free Energy, and WorkFree energy change and spontaneity, Standard free energy changes
4. Free Energy, Equilibrium, and Reaction Direction
Important criterions
Spontaneous reaction, in chemical reaction terms, is one which occurs
with the system releasing releasing free energyfree energy in some form (often, but not
always, heat) and moving to a lower energylower energy, hence more
thermodynamically stablethermodynamically stable, state.
Enthalpy, H, is the sum of the sum of the internal energyinternal energy and the product of product of
the pressure and volumethe pressure and volume of a thermodynamic system.
Entropy, S, is a state function which measures the degree of disorderdegree of disorder
(randomness)(randomness) of a system. The larger the disorder, the greater the
value of S in units of J/K.
Free Energy is the measure of the spontaneityspontaneity of a process and of the
useful energy available from it.
The First Law of Thermodynamics indicates that to conserve energyconserve energy,
the change in energy must equal the heat transferred plus the work
performed, E = q + wE = q + w.
The Second Law of Thermodynamics states that for a spontaneousspontaneous
process, ∆∆SSuniverseuniverse = ∆Ssystem + ∆Ssurroundings > 0 > 0 ((∆∆SSuniverseuniverse positive))
Figure 1 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
Spontaneous reaction and exothermic /endothermic
Common mistake – exothermic is spontaneous.
The Concept of Entropy (S)
Entropy refers to the state of order, measures the degree of disorder of a system. .
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
Figure 2
1 atm evacuated
Spontaneous expansion of a gas
stopcock closed
stopcock opened
0.5 atm 0.5 atm
In thermodynamic terms, a change in order is the number of ways of arranging the particles.
The change in order is likely the key factor to determine the direction of a spontaneous reaction.
The number of possible arrangement is increased.
The number is of ways of arranging a system is related to the entropy.
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.
Figure 4 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
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.
Thermodynamics:
Entropy, Free Energy, and the Direction of Chemical Reactions
1 Three Laws of Thermodynamics: First Law : ∆Universe = 0, Second Law: ∆S>0, G<0, Third law :
Ssolid=0
2 Calculating the Change in Entropy of a Reaction
∆S0 = mS0products - nS0
reactants
Combine standard molar entropies to find the standard entropy for a reaction
3 Entropy, Free Energy, and WorkG = H-TS, ∆ G= ∆H- ∆TS, ∆G= ∆H- T∆S, ∆G0= ∆H0 - T∆S0,
4 Free Energy, Equilibrium, and Reaction Direction
Lecture 2
Summary
Spontaneous reaction happens at a specific condition without continuous input of energy.
Neither first law of thermodynamic or the sign of ∆H predicts the direction of reaction.
The spontaneous reaction involves a change in the universe from more to less order.
Entropy is the measure of disorder and directly related to the number of ways of arrangement for components in the system.
The second law of thermodynamic states that the entropy of universe increases in a spontaneous process.
Entropy value are absolute since the pure single crystal has a zero entropy at absolute zero degree.
Standard molar entropy is dependant on the temperature, phase change, dissolution, atomic size, and molecular complexity.
Sample Problem
SOLUTION:
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)
PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature.
(a) 1mol of SO3(g) - more atoms
(b) 1mol of CO2(g) - gas > solid
(c) 3mol of O2(g) - larger #mols
(d) 1mol of KBr(aq) - solution > solid
(e) 230C - higher temperature
(f) CCl4 - larger mass
Important criterions
Spontaneous reaction, in chemical reaction terms, is one
which occurs with the system releasing free energy in some
form (often, but not always, heat) and moving to a lower
energy, hence more thermodynamically stable, state.
Entropy, S, measures the degree of disorder of a system.
The larger the disorder, the greater the value of S in units of
J/K.
Free Energy, G, is the measure of the spontaneity of a
process and of the useful energy available from it.
Sample Problem Calculating the Standard Entropy of Reaction, S0rxn
PROBLEM: Calculate S0rxn for the combustion of 1mol of propane at 25 0C.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
PLAN: Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas.
SOLUTION: Find standard entropy values in the Appendix or other table.
S = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)]
S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)]
S = - 374 J/K
When a reaction occurs, chemists expect to learn how to predict and calculate the change in entropy using the following equation:
∆S0 = mS0products - nS0
reactants
Lecture 2
In many spontaneous reaction, the system become more ordered at a given condition, given condition, ∆∆SS00
rxnrxn<0<0.
The Second Law of Thermodynamics indicated that the decrease in entropy of a system only can occur when the increase the entropy in surrounding overweigh them.
The entropy change of the surrounding is related to an opposite change in the heat of the system. ∆S0
surr - qsys
The entropy change of the surrounding is also reversely proportional to the temperature of the heat transferred. ∆S0
surr 1/T
Combining these changes, we obtain the equation:
∆S0surr = -
∆S0surr = -
Determining Reaction Spontaneity
qsys
T When a pressure is constant, the heat is ∆H. ( ∆H = q + ∆ PV)∆ Hsys
T
Lecture 2
Sample Problem
SOLUTION:
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.
N2(g) + 3H2(g) 2NH3(g) S0sys = -197 J/K
PLAN: S0universe must be > 0 in order for this reaction to be spontaneous, so
S0surroundings must be > 197 J/K. To find S0
surr, first find Hsys; Hsys = Hrxn which can be calculated using H0
f values from tables.
S0universe = S0
surr + S0sys.
H0rnx = [(2 mol)(H0
fNH3)] - [(1 mol)(H0fN2) + (3 mol)(H0
fH2)]
H0rnx = -91.8 kJ
S0surr = -H0
sys/T = -(-91.8x103J/298K) = 308 J/K
S0universe = S0
surr + S0sys = 308 J/K + (-197 J/K) = 111 J/K
S0universe > 0 so the reaction is spontaneous.
Lecture 2
G0system = H0
system - TS0system
G0rxn = mG0
products - nG0reactants
Gibbs Free Energy (G)
G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process at equilibrium
Lecture 2
Sample Problem
SOLUTION:
Calculating G0 from Enthalpy and Entropy Values
PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated.
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 (G0rxn) at 250C for this reaction.
PLAN: Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve.
H0rxn
= mH0products - nH0
reactants
H0rxn
= (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) -
(4 mol)(-397.7 kJ/mol)
H0rxn
= -144 kJ
Lecture 2
Sample Problem Calculating G0 from Enthalpy and Entropy Values
continued
S0rxn
= mS0products - nS0
reactants
S0rxn
= (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) -
(4 mol)(143.1 J/mol*K)
S0rxn
= -36.8 J/K
G0rxn
= H0rxn - T S0
rxn
G0rxn
= -144 kJ - (298K)(-36.8 J/K)(kJ/103 J)
G0rxn
= -133 kJ
Lecture 2
Entropy, Free Energy, Work
Free Gibbs energy is the measure of the spontaneity of a process. • ∆G0
sys = ∆H0
sys - T∆S0sys
• T∆S0univ = - (∆H0
sys - T∆S0sys)
• The sign for ∆G0sys
and ∆S0univ can tell us the spontaneity of a reaction.
• ∆G0sys < 0 for a spontaneous reaction;
• ∆G0sys = 0 for a reaction at equilibrium;
• ∆G0sys > 0 for a nonspontaneous reaction;
• ∆S0univ > 0 for a spontaneous reaction;
• ∆S0univ = 0 for a reaction at equilibrium;
• ∆S0univ < 0 for a nonspontaneous reaction.
Lecture 2
The spontaneity of reaction and the sign of ∆H, ∆G, ∆S.
∆G0 = ∆H0 - T∆S0
∆H0 ∆S0 -T∆S0 ∆G0 Description
- + - - Spontaneous at all T
+ - + + Non-spontaneous at all T
+ + - + or - Spontaneous at high TNon-spontaneous at low T
- - + + or - Spontaneous at low TNon-spontaneous at high T
PROBLEM:
Determining the effect of temperature on G
An important reaction in production of sulfuric acid is the oxidation of SO2 (g) to SO3(g),
2SO2(g) + O2(g) 2SO3(g)
At 298K, ∆G0 = -141.6 kJ/mol, ∆H0 = -198.4 kJ/mol, ∆S0 = -187.9kJ/mol,
(a). Use the data to predict if this reaction is spontaneous reaction at 25 C.
(b). Predict how ∆G0 changes when temperature increases.
(c). Assume ∆H0 and ∆S0 are constant with increasing T, is the reaction spontaneous at 900 C?
PLAN: Note the sign of ∆G0, ∆H0, and T∆S0
Use the equations ∆G0 = ∆H0 - T∆S0
SOLUTION:
(a) Noting G = -141.6 kJ, is negative, so the reaction at room temperature is spontaneous.
Determining the effect of temperature on G
(b) Noting G = H – TS
= -198.4 - (- 0.1879T) = -198.4 + 0.1879T
If T increases, the G is less negative. Therefore, the reaction is less spontaneous.
(c) Calculating G = H – TS at 900 C
= -198.4 - (- 0.1879T)
= -198.4 + 0.1879(900+273.15)
= 22.0 kJ
At 900 C, G is positive. Therefore, the reaction is non- spontaneous.
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 ln Q = 0 so
G0 = - RT lnK
FO
RW
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D R
EA
CT
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RE
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RS
E R
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Table 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
PROBLEM:
Calculating G at Nonstandard Conditions
The oxidation of SO2, 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.
PLAN: Use the equations and conditions. G0 = -RTlnK
SOLUTION:
Calculating G at Nonstandard Conditions
(a) Calculating K at the two temperatures:
G0 = -RTlnK so
K e (G 0 /RT)
At 298, the exponent is -G0/RT = -(-141.6kJ/mol)(103J/kJ)
(8.314J/mol*K)(298K)= 57.2
K e (G 0 /RT) = e57.2 = 7x1024
At 973, the exponent is -G0/RT(-12.12kJ/mol)(103J/kJ)
(8.314J/mol*K)(973K)= 1.50
K e (G 0 /RT) = e1.50 = 4.5
Calculating G at Nonstandard Conditions
(b) The value of Q =pSO3
2
(pSO2)2(pO2)=
(0.100)2
(0.500)2(0.0100)= 4.00
Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight.
(c) The nonstandard G is calculated using G = G0 + RTLnQ
G298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(Ln4.00)
G298 = -138.2kJ/mol
G973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(Ln4.00)
G298 = -0.9kJ/mol
2SO2(g) + O2(g) 2SO3(g)
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