Thermodynamics 2 - Chemistry at Winthrop University

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THERMODYNAMICS 2
Dr. Harris
Suggested HW: Ch 23: 43, 59, 77, 81
Recap: Equilibrium Constants and Reaction Quotients
• For any equilibrium reaction:
 + 
 + 
• The equilibrium constant, K, is equal to the ratio of the concentrations/
pressures of products and reactants to their respective orders at
equilibrium.
[] []
1
=
=   
[] [] −1
• The reaction quotient, Q, has the same form as K, but does not
use equilibrium concentrations/pressures. Using the previous
reaction:
[] []
   =
[] []
• The subscript ‘0’ denotes initial concentrations before shifting
toward equilibrium. Unlike K, Q is not constant.
Recap: Direction of Spontaneity
• The direction of spontaneity is always toward equilibrium.
• The value of Q/Kc tells us the direction in which a system not at
equilibrium will proceed to reach equilibrium.
Determining the Direction of A Reaction Toward Equilibrium
Too much product
K
Q
Too much reactant
Q
Q
Q
Equilibrium
Recap: Entropy and the 2nd Law of Thermodynamics
• Entropy is a measure of the disorder of a system. Increasing disorder
means that the change in entropy is positive.
• 2nd Law of Thermodynamics:
Entropy is not conserved. The Entropy of the universe is continually
Increasing.
∆ = ∆ + ∆
∆ ≥ 0
• The universe can never become more ordered after a process.
Therefore, if a particular system becomes more ordered (ΔSsys<0), the
surroundings must become even more disordered (ΔSsurr >0)
Recap: Thermodynamics of Equilibrium
• When a system reaches equilibrium, the entropy is at a maximum, so
the change in entropy is 0 (ΔSsys = 0 at equilibrium)
Recap: Spontaneity Depends on Enthalpy AND
Entropy
∆ = ∆ − ∆
Dictates if a process is
energetically favored
Dictates if a process
is entropically
favored
Minimizing ΔG
• In general, a system will change spontaneously in such a way that its
Gibbs free energy is minimized.
• The enthalpy term is independent of concentration and pressure.
Entropy is not.
• During a reaction, the composition of the system changes, which
changes concentrations and pressures, leading to changes in the
TΔS term.
• As the system approaches an entropically unfavorable composition,
the back reaction occurs to prevent ΔG from becoming more positive.
This is the basis of equilibrium.
• Once equilibrium is reached, the free energy no longer changes
ΔG < 0
Reactants
Equilibrium
ΔG > 0
Products
K = 0.01
K=25
K = 1000
When ΔG is Negative, the Value Tells Us the Maximum Portion
of ΔU That Can Be Used to do Work
ΔG = wmax
Gasoline with
internal
energy U
qmin
Work not accounted for by
change in free energy
must be lost as heat
Maximum possible
portion of U converted
to work = ΔG
Relating the Equilibrium Constant, Reaction Quotient, and
ΔGorxn
• Keep in mind that the standard free energy change, ΔGo, is not
the same as the nonstandard free energy change, ΔG. ΔGo is
determined under standard conditions. Those conditions are
listed below.
State of Matter
Standard State
Pure element in most stable
state
ΔGo is defined as ZERO
Gas
1 atm pressure, 25oC
Solids and Liquids
Pure state, 25oC
Solutions
1M concentration
Relating K, Q, and ΔGorxn
• For many elements, ΔGorxn can be obtained from a table of values.
∆ =
∆  −
∆ 
• In terms of the equilibrium constant of a particular reaction, the
driving force to approach equilibrium under standard conditions is
given by:
∆ = −  
• When the reaction conditions are not standard, you must use the
reaction quotient, Q.
• The free energy change of a reaction (or the driving force to approach
equilibrium) under non-standard conditions, ΔG, is given by:
∆ = ∆ +   
∆ = ∆ +   
Example #1 (No K value given)
2  + 32 
23 ()
• Calculate ΔG at 298oK for a reaction mixture that consists of 1.0
atm N2, 3.0 atm H2, and 0.50 atm NH3. Which direction must the
reaction shift to reach equilibrium?
• We are finding the free energy change under non-standard
conditions (ΔG). We must first Q.
(0.50)2
=
= .0277
(1.0)(3.0)3
• Now determine the standard free energy, ΔGo. If K is not
given, you can calculate it from the standard table.
• From appendix D in the back of the book:
∆  = 
∆  = 
∆  = −. 


∆


=  −. 
= −. 


• Solve for ΔG
 +  ln 
∆ = ∆
∆


= −32800
+ 8.314

 
∆ = −41684.6


298  ln(.0277)
Reaction moves to
the right to reach
equilibrium.
Example #2 (Value of K given)
 
 +  ln 
∆ = ∆
  +  
• At 298oK, the initial partial pressures of H2, F2 and HF are 0.150 bar,
.0425 bar, and 0.500 bar, respectively. Given that Kp = .0108,
determine ΔG. Which direction will the reaction proceed to reach
equilibrium?
• Find Q
.150 (.0425)
=
= 0.0255
(.500)2
• We have K, so we can determine ΔGrxn without using the standard
table.
∆

= −   +  ln  = RT ln

∆ = 4.27 kJ/mol
Reaction moves
left to reach
equilibrium.
Deriving The van’t Hoff Equation
• We know that rate constants vary with temperature.
• Considering that equilibrium constants are ratios of rate constants of
the forward and back reaction, we would also expect equilibrium
constants to vary with temperature.
• Using our relationship of the standard free energy with standard
enthalpy and entropy:
o
o
o
∆Grxn
= ∆Hrxn
− T∆Srxn
• And relating this expression to the equilibrium constant, K, we obtain:
−RT ln K = ∆H o − T∆S o
o
o
∆Hrxn
∆Srxn
ln K = −
+
RT
R
Deriving The van’t Hoff Equation
o
o
∆Hrxn
∆Srxn
ln K = −
+
RT
R
• As we see in this expression, as we increase temperature, the
enthalpy term becomes very small. The entropy term then becomes
more important in determining K as T increases.
• Thus, entropy is the dominant factor in determining equilibrium
distributions at high temperatures, and enthalpy is the dominant
factor at low temperatures.
• A plot of ln K vs. 1/T will yield a linear plot with a slope of (–ΔHorxn)/R
Expanded Form of The van’t Hoff Equation
• If you run the same reaction at different temperatures, T1 and T2:
o
o
∆Hrxn
∆Srxn
ln K1 = −
+
RT1
R
o
o
∆Hrxn
∆Srxn
ln K 2 = −
+
RT2
R
• Then subtraction yields:
o
∆Hrxn
1
1
ln K 2 − ln K1 =
−
R
T1 T2
• Which equals:
o
K 2 ∆Hrxn
1
1
ln
=
−
K1
R
T1 T2
Expanded van’t
Hoff equation
• So if you know the equilibrium constant at any temperature, and the
standard enthalpy of reaction, you can determine what K would be at
any other temperature.

2 ∆
2 − 1

=
1

1 2
Example
• CO(g) + 2H2(g)
ΔHorxn= -90.5 kJ/mol
CH3OH(g)
The equilibrium constant for the reaction above is 25000 at 25oC.
Calculate K at 325oC. Which direction is the reaction favored at T2? Is
this in line with LeChatlier’s Principle.
J
−90500
K2
mol
ln
=
J
25000
8.314
mol K
use ex to cancel ln
term
1
1
−
298K 598K
e
K2
ln25000
K2
= 1.1 x 10−8
25000
ln
K2
= −18.32
25000
= e−18.32
  = .   −
Left. This is expected for an exothermic reaction at increased temperature.

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