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Molarity of A ' [ A ] '^ moles of A volume in liters
units : mole L
' M
A ( g ) % B ( g )! C ( g ) % D ( g )
A (g) + B (g) C (g) + D (g)
indicates equilibrium
IV. Chemical Equilibrium
A) Law of Chemical Equilibrium Concentration expressed as molarity
Consider the hypothetical reaction
Compound A is allowed to react with B and the concentrations of the reactants and products are plotted as a function of time.
Note: After eleven seconds the concentrations do not change with time.
Chemical equilibrium is the condition that exists when the concentrations of the reactants and products no longer change with time. Chemical equilibrium is a dynamic state. It occurs when the rate of the forward reaction (A + B 6 C + D) equals the rate of the reverse reacion (A + B 7 C + D).
[ C ] eq [ D ] eq [ A ] eq [ B ] eq
' 0.167 ' constant ' Kc
[ C ] peq [ D ] qeq ã [ A ] meq [ B ] neq ã
' Kc
Kc '
[ C ] (^2) eq [ A ] eq [ B ] (^3) eq
units : ï^
mole L (^)
2
ï
mole L (^)
4 ' ð
L mole ë
2
At equilibrium there is a specific relationship between the concentrations of the reactants and products.
Consider three experiments for the reaction
A (g) + B (g) C (g) + D (g)
in which the initial concentrations of A and B vary from experiment to experiment.
Table of Equilibrium Concentrations Exp. # [A] (^) eq [B] (^) eq [C] (^) eq [D]eq 1 3.00 2.00 1.00 1. 2 9.60 10.0 4.00 4. 3 0.500 3.00 0.500 0.
The following relationship fits the data in each experiment.
For the general reaction
m A (g) + n B (g) + ã p C (g) + q D (g) + ã
K is definedc
The magnitude of the constant K depends on the nature of the reaction and the temperature.c
The units of K depend on the specific reaction.c
Example: A + 3 B 2 C
Example: A + B C + D
Equilibrium moles of Cl 2 = (initial moles Cl 2 ) + (change in moles Cl 2 )
Negative sign indicates moles of Cl 2 are consumed
PCl 3 + Cl 2 PCl 5
initial: 0.2660 0.3340 0.0 mole
change: -0.0660 -0.0660 0.0660 mole
equil: 0.2000 0.2680 0.0660 mole
Change in moles of PCl 5 = (equilibrim moles PCl 5 ) - (initial moles of PCl 5 )
Change in moles of Cl 2 = (change in moles PCl 5 )(1 mole Cl 2 /1 mole PCl 5 )
Kc '
[ PCl 5 ] eq [ PCl 3 ] eq [ Cl 2 ] eq
' ï^
0.0660 mole 2.00 L (^)
ï
0.200 mole 2.00 L (^) ï
0.268 mole 2.00 L (^)
L
mole
- Initial moles of a substance + change in moles of substance = equilibrium moles of substance
Example: initial moles of PCl + change in moles of PCl 5 5 = equilibrium moles of PCl 5
0.0 moles of PCl 5 + 0.0660 moles of PCl 5 = 0.0660 moles of PCl 5
- The changes in number of moles of reactants and products are related by mole ratios.
These observations can be utilized in an Equilibrium Table to solve equilibrium problems.
Equilibrium Table
Substituting the equilibrium moles into the equilibrium expression
Example #2: Initially, 3.00 moles of PCl and 3.00 moles of Cl are placed in a 5.00 L container 3 2 at 300EC. Calculate the equilibrium concentration of PCl , [PCl ]. K is 2.46 L/mole. 5 5 eq c
Unknown: [PCl ]5 eq , Let x = moles of PCl at equilibrium. 5 Knowns: initial moles of PCl = 3.00 mole, initial moles of Cl = 3.00 mole, volume of container = 3 2
Kc '
[ PCl 5 ] eq [ PCl 3 ] eq [ Cl 2 ] eq
' ï^
x 5.00 L (^)
ï
3.00 & x 5.00 L (^) ï
3.00 & x 5.00 L (^)
' 2.46 L mole
2.46 x^2 & 19.76 x % 22.14 ' 0
x ' &^ b^ ±^ b^
(^2) & 4 ac 2 a
x ' &^ b^ ±^ b^
(^2) & 4 ac 2 a
' &^ (&19.76 ) ±^ (&^ 19.76)
(^2) & 4(2.46) (22.14) 2 (2.46)
x ' 1.35, 6.
5.00L, K = 2.46 L/molec Concepts: chemical equilibrium, molarity
Substitute known values into equilibrium table.
PCl 3 + Cl 2 PCl 5 initial 3.00 3.00 0 mole change mole equil x mole
Use the concepts presented above to fill in the table.
PCl 3 + Cl 2 PCl 5 initial 3.00 3.00 0 mole change - x - x x mole equil 3.00 - x 3.00 - x x mole
Substituting equilibrium moles into the equilibrium expression
For the quadratic equation
ax 2 + bx + c = 0
the solutions are
Substituting
Although both roots (1.35 and 6.67) are mathematically correct, only one root satisfies the physical constraints of the problem and that root is 1.35. If root 6.67 is used, then the moles of PCl and Cl 3 2 at equilibrium would be - 3.67 (3.00 - 6.67) and the equilibrium concentrations of PCl and Cl would be 3 2 - 0.734 M. A negative or zero equilibrium concentration for reactants or products is physically impossible.
Kc '
[ A ] (^2) eq [ B ] eq
' ï^
2 x 1.00 L (^)
2
ï
3.00 & x 1.00 L (^)
' 0.100 moleL
4.00 x^2 % 0.100 x & 0.300 ' 0
x ' 0.261, &0.
'^ change in moles of B initial moles of B
'^ x 3.00 mole
'^ 0.261 mole 3.00 mole
' 8.70@ 10 &^2
Q '
[ C ] cin [ D ] din [ A ] ain [ B ] bin
x = 0.261 mole = change in moles of B
C) Factors Affecting Equilibrium Concentrations at Constant Temperature
Le Châtelier’s principle - when a system is in equilibrium , a change in any one of the factors upon which the equilibrium depends will cause the equilibrium to shift in such a way as to diminish the effect of the change.
To determine if the reaction
a A (g) + b B (g) c C (g) + d D (g)
will go to the left or to the right to reestablish the equilibrium, Q is defined as
If Q < K (^) c , then the reaction will go to the right to reestablish the equilibrium. Moles of reactants A and B will be consumed and moles of products C and D will be produced.
If Q > K (^) c , then the reaction will go to the left to reestablish the equilibrium. Moles of products C and D will be consumed and moles of reactants A and B will be produced.
- Addition or Removal of Moles of Reactants or Products
Example #4: If 1.00 mole of PCl is added to the equilibrium in Example #2, what will be 5 the NEW equilibrium concentration of PCl? 5
Unknown: new equilibrium molarity of PCl after the addition of 1.00 mole of PCl 5 5 Knowns: equilibrium moles of PCl , Cl , and PCl 3 2 5 (1.65, 1.65, and 1.35 mole) before the addition of 1.00 mole of PCl (see Example #2); K = 2.46 L/mole; volume of the 5 c container = 5.00 L
Q '
[ PCl 5 ] in [ PCl 3 ] in [ Cl 2 ] in
' ï^
(2.35 mole ) 5.00 L (^)
ï
(1.65 mole ) 5.00 L (^) ï
(1.65 mole ) 5.00 L (^)
' 4.31 L mole
Q ' 4.31 L mole
2.46 L mole
' Kc
Kc '
[ PCl 5 ] eq [ PCl 3 ] eq [ Cl 2 ] eq
' ï^
2.35 & x 5.00 L (^)
ï
1.65 % x 5.00 L (^) ï
1.65 % x 5.00 L (^)
' 2.46 L mole
0.492 x^2 & 2.624 x % 1.010 ' 0
Concepts: chemical equilibrium, Q, molarity
Step #1: Evaluate Q and determine whether the reaction will go to the left or right to reestablish the equilibrium.
PCl 3 + Cl 2 PCl 5 initial 1.65 1.65 1.35 + 1.00 mole
The moles in the table are the equilibrium values from Example #2 and the 1.00 mole of PCl added. 5
Since Q > K , the reaction will go to the left to reestablish the equilibrium. Moles ofc PCl 5 will be consumed and moles of PCl and Cl will be produced. 3 2
Step #2: Solve the equilibrium problem.
Let x = change in moles of PCl. 5 Since the reaction goes to the left to reestablish the equilibrium, moles of PCl will be consumed and the entry for PCl in the row labeled 5 5 “change” of the equilibrium table will be -x.
PCl 3 + Cl 2 PCl 5 initial 1.65 1.65 2.35 mole change x x - x mole equil 1.65 + x 1.65 + x 2.35 - x mole
x = 0.361 mole = change in moles of PCl 5
y ' ( slope ) x % ( intercept )
ln( Kp ) ' ð
& H
R ë ð^
T ë^
% C
Equation for a straight line is
where C is a constant.
Consider K at T and K at T for a specific reaction, then 2 2 1 1
ln( Kp ) 2 ' ð
& H
R ëð^
T 2 ë^
% C
& ð
ln( Kp ) 1 ' ð
& H
R ë ð^
T 1 ë^
% C
ë
ln( Kp ) 2 & ln( Kp ) 1 ' ð
& H
R ëð^
T 2
&^1
T 1 ë
ln( Kp ) 2 ' ð
H
R ë ð^
T 2 & T 1
T 2 T 1 ë^
% ln( Kp ) (^1)
ln( Kp ) 2 ' ð
H R ë ð^
T 2 & T 1 T 2 T 1 ë^
% ln( Kp ) (^1)
H ' R ln ð
( Kp ) (^2) ( Kp ) 1 ëð^
T 2 T 1 T 2 & T 1 ë^
' ( 8.314 (^) K J ) ln ð
(0.4554) (0.09823)ë ð^
(313 K )(293 K 20 K ë^
' 5.85 @ 104 J
Rearranging
When H > 0 , heat is absorbed and the reaction is said to be endothermic. When H < 0 , heat is evolved and the reaction is said to be exothermic.
Example: If K = 0.09823 at 20p EC and 0.4554 at 40EC, calculate H for the reaction.
N O 2 4 (g) 2 NO 2 (g)
Rearranging
E) Entropy and Free Energy
& T Suniv ' H & T S (4)
G / H & T S
G ' H & T S (5)
G ' & T Suniv
G E ' H E & T S E ' 436 kJ & (298 K ) (98.5 (^) K J ) ð
1 kJ 1000 J (^) ë '^407 kJ
)G = )Go^ + RT ln
P C
c P D
d
P A
a P B
b
Partial Pressure of Gas C Partial Pressure of Gas D
Partial Pressure of Gas A
Reaction FreeEnergy Partial Pressure of Gas B
Standard Reaction Free Energy
Ideal Gas Constant
Absolute Temp
Gibbs free energy, G, is defined as
For a constant pressure and constant temperature process
Comparing eqs 4 and 5
A constant pressure and constant temperature process will be spontaneous when S (^) univ> 0 or G < 0.
Example: Calculate G E at 25EC for the following reaction?
H 2 (g) 6 2 H (g) HE = 436 kJ, SE = 98.5 J/ K
F) Reaction Free Energy
For the generic reaction
a A (g) + b B (g) 6 c C (g) + d D (g)
the reaction free energy, G, is defined as
A plot of G for the above reaction versus the extent of reaction yields
G ' 0 ' G E % R T ln ð
( PC ) ceq ( PD ) deq ( PA ) aeq ( PB ) beq ë^
The degree of reaction is a measure of the extent to which the reaction has advanced from the left to right.
The slope of the above curve at any point is equal to the reaction free energy, G. When the degree of reaction is 0, only the reactants A and B are present, G < 0 , and the reaction will proceed spontaneously to the right. When the degree of reaction is 1.0, only the products are present, G > 0 , and the reaction will proceed spontaneously to the left. When the degree of reaction is 0.16, G = 0 and the reaction is at equilibrium.
At equilibrium, G = 0 and
K is defined asp
G E reaction ' j (^) product G E f ( product ) & j (^) reactant G E f ( reactant )
S E reaction ' j (^) product S E ( product ) & j (^) reactant S E ( reactant )
The free energy change for the formation of a compound in its standard state from its elements (most stable form) in their standard states is call the standard free energy of formation, GE (^) f.
Example:
N 2 (g) + 3 H 2 (g) 6 2 NH 3 (g) GE (^) f = - 33.2 kJ
By definition GE (^) f for elements in their most stable form is zero.
A table of GE (^) fis found in the text. See example below.
To calculate the standard reaction free energy, GE, from the standard free energies of formation
where , called the mole number , is the stoichiometric coefficient in the thermochemical equation.
- SE
The third law of thermodynamics states that the entropy for all pure, perfectly ordered, crystalline substances is zero at absolute zero(0 K ). Consequently the absolute entropies, SE, of all compounds and elements in their most stable form can be determined. Some of these entropies are list in a table of thermodynamic data found in the text.
Example: Table of Selected Thermodynamic Data
Substance G E (^) f(kJ/mole) S E (J/mole-K) CO (g) 2 - 394.4 213. H O (l) 2 - 237.2 70. H (g) 2 0.0 130. H (g) 203.3 115
To calculate the change in the standard entropy for a reaction
