1. Chemical Equilibrium Chapter 14
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2. Last chapter, the rate of a chemical reaction was described as:Dt
When A B
What happens when the reaction nears completion? A B
rate = -
D[A] Dt
D[B] =
= 0
Equilibrium is a state in which there are no observable changes as time goes by.

3. Physical equilibrium Chemical equilibrium
Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when:
the rates of the forward and reverse reactions are equal and
the concentrations of the reactants and products remain constant
Physical equilibrium
H2O (l)
H2O (g)
Chemical equilibrium
N2O4 (g)
2NO2 (g)
14.1

4. forward reaction rate = reverse reaction rate
On a molecular level, reactants never stop reacting.
So why does the rate of reactant disappearance drop to zero?
Rate of products reacting is the same as the rate of reactants reacting.
forward reaction rate = reverse reaction rate
Consider the following reaction:
H2+ I2
2 HI
Rate forward = kf [H2][I2] Rate reverse = kr[HI]2
At equilibrium: kf [H2]eq[I2] eq = kr[HI]2 eq
Rearranging: kf = [HI]2 eq  K Equilibrium constant
kr [H2]eq[I2] eq

5. N2O4 (g) 2NO2 (g) equilibrium kinetics kinetics kinetics equilibrium
Start with NO2
Start with N2O4
Start with NO2 & N2O4
It does not matter where you start, the ratios at equilibrium are always the same (at the same temperature).

6. constant
14.1

7. Equilibrium Will N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3
aA + bB cC + dD
Products
K =
[C]c[D]d
[A]a[B]b
Law of Mass Action
Equilibrium Will
Reactants
K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants
When at equilibrium, the ‘eq’ subscript is assumed.
No reaction mechanism is assumed !!
14.1

8. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) NO2 (g)
Kp =
NO2 P2
N2O4 P
Kc =
[NO2]2
[N2O4]
In most cases
Kc  Kp
Concentration units for K’s:

9. P C P D P A P B aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn
Remember: PV = nRT n V
PA = RT = [A] RT RT 1
[A] = P C
P C
P D D
[C]C [D]D
Kc =
Kp =
[A]A [B]B
P A
P B A B
Dn = moles of gaseous products – moles of gaseous reactants
= (c + d) – (a + b)

10. Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
[CH3COO-][H3O+]
[CH3COOH][H2O]
Kc = ‘
[H2O] = constant
[CH3COO-][H3O+]
[CH3COOH] =
Kc [H2O] ‘
Kc =
General practice not to include units for the equilibrium constant.
14.2

11. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
[COCl2]
[CO][Cl2] =
0.14
0.012 x 0.054
Kc =
= 220
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R =
T = = 347 K
Kp = 220 x ( x 347)-1 = 7.7
14.2

12. The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = atm and PNO = atm?
2NO2 (g) NO (g) + O2 (g) 2
Kp = 2
PNO PO
PNO PO 2
= Kp
PNO PO 2
= 158 x (0.400)2/(0.270)2
= 347 atm
14.2

13. They must just be present.
Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
Kc = ‘
[CaO][CO2]
[CaCO3]
[CaCO3] = constant
[CaO] = constant
Kc x ‘
[CaCO3]
[CaO]
Kc = [CO2] =
Kp = PCO 2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
The equilibrium state is not affected by the quantity of solids and liquids, if they are pure.
They must just be present.
14.2

14. does not depend on the amount of CaCO3 or CaO
CaCO3 (s) CaO (s) + CO2 (g)
PCO 2
= Kp
PCO 2
does not depend on the amount of CaCO3 or CaO
14.2

15. Consider the following equilibrium at 295 K:
The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction?
NH4HS (s) NH3 (g) + H2S (g)
Kp = P
NH3
H2S P
= x =
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
Dn = 2 – 0 = 2
T = 295 K
Kc = x ( x 295)-2 = 1.20 x 10-4
14.2

16. ‘ ‘ ‘ ‘ ‘ Kc = [C][D] [A][B] Kc = [E][F] [C][D] Kc A + B C + D Kc
C + D E + F
[E][F]
[A][B]
Kc =
A + B E + F Kc
Kc = Kc ‘ x
If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
14.2

17. Add two reactions => multiply K’s
[NO2][O2] K1
O3 + NO NO2 + O2
K1 =
[O3][NO]
NO2 + O NO + O2 K2
[NO][O2]
K2 =
O3 + O O2
Knet
[NO2][O .] K1
Knet = x K2
[NO2][O2]
[NO][O2]
Knet =
[O3][NO]
[NO2][O .]
[O2]2
Knet =
[O3][O .]
Add two reactions => multiply K’s
14.2

18. ‘ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4]
= 216
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
14.2

19. Writing Equilibrium Constant Expressions
The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.
The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.
The equilibrium constant is a dimensionless quantity.
In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.
If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
14.2

20. Chemical Kinetics and Chemical Equilibrium
ratef = kf [A][B]2
A + 2B AB2 kf kr
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2] kf kr
[AB2]
[A][B]2 =
Kc =
14.3

21. Qc < Kc system proceeds from left to right to reach equilibrium
The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF
Qc < Kc system proceeds from left to right to reach equilibrium
Qc = Kc the system is at equilibrium
Qc > Kc system proceeds from right to left to reach equilibrium
14.4

22. For the reaction NH3 (g) N2 (g) + 3 H2 (g)
Kc = 2.6 x 10-5 at 1270C. The initial concentrations are [N2 ]0 = 0.50 mol/L, [H2]0 = 1.5 mol/L, and [NH3]0 = mol/L. Are we at equilibrium? If not, which way will the reaction proceed?
Qc =
[N2] [H2]3
[NH3] 2 =
(0.50)(1.5)3
0.0202
= 4.2 x 103
Is Qc = Kc ? x  2.6 x 10-5
Compare Qc to Kc, if Qc > Kc then [Products] are too high and the [Reactants] are too low. Thus reaction must cnvert products to reactants.
4.2 x > 2.6 x 10-5

23. Calculating Equilibrium Concentrations
Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.
Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.
Having solved for x, calculate the equilibrium concentrations of all species.
14.4

24. ICE At 12800C the equilibrium constant (Kc) for the reaction
Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.
Br2 (g) Br (g)
Let x be the change in concentration of Br2
Br2 (g) Br (g)
Initial (M)
0.063
0.012
ICE
Change (M) -x
+2x
Equilibrium (M) x x
[Br]2
[Br2]
Kc =
Kc =
( x)2 x
= 1.1 x 10-3
Solve for x
14.4

25. Kc =
( x)2 x
= 1.1 x 10-3
4x x = – x
4x x = 0 
-b ± b2 – 4ac
ax2 + bx + c =0
x = 2a
x =
x =
Br2 (g) Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063
0.012 -x
+2x x x
At equilibrium, [Br] = x = M
or M
At equilibrium, [Br2] = – x = M
14.4

26. Assume zero if not given
The equilibrium constant (Kc) for the reaction
is 4.00 x If the initial concentration of [HCN] = 0.39 mol/L, calculate [C2N2] at equilibrium.
2 HCN (g) H2 (g) + C2N2(g)
Let x be the change in concentration of H2 and C2N2
2 HCN (g) H2 (g) + C2N2(g)
Assume zero if not given
Initial (M)
0.39 x
Change (M)
-2x x x x
Equilibrium (M) x
[H2][C2N2]
(x)(x)
( x)2
= 4.00 x 10-4
Solve for x
Kc =
Kc =
[HCN] 2
14.4

27. x = 0.0075 mol/L = [C2N2] at equilibrium
(x)(x)
( x)2
= x 10-4   x x
= x 10-2
=>
x = ( x)
x = x
1.04 x =
x = mol/L = [C2N2] at equilibrium
Use the definition of x to help you answer the question given in the problem.

28. Assume zero if not given
When solid NH4HS is stored is a sealed container, ir decomposed to some extent to NH3 and H2S gases. Given:
and Kc = 1.2 x 10-4 at 21.8°C. Determine [NH3] and [H2S] at equilibrium.
NH4HS (s) NH3 (g) + H2S (g)
Let x be the change in concentration of NH3 and H2S
NH4HS (s) NH3 (g) + H2S (g)
Assume zero if not given
Initial (M)  x
Change (M) -x x x x
Equilibrium (M)
( - x)
[NH3] [H2S]
(x)(x)
= 1.2 x 10-4
Solve for x
Kc =
Kc =
“1”
x = mol/L = [NH3] = [H2S ] at equilibrium

29. Given some information about the equilibrium state, we can calculate the remainder of the information.
What is the partial pressure of NH3 in equilibrium with atm H2 and atm N2, given:
N2 (g) + 3 H2 (g) NH3 (g)
and KP = x 10-5 at 25°C. Determine [NH3] and [H2S] at equilibrium.
N2 (g) H2 (g) NH3 (g)
Equilibrium
0.527 atm
0.733 atm x x2
PNH3 2
KP = =
= x 10-5
PN2 PH23
(0.527)(0.733)3
Solve for x
x = 1.73 x 10-3 atm = PNH3 at equilibrium

30. Assume zero if not given Initial (M) 1.000 atm x x Change (M) -x
At 250°C, PCl5 is introduced into a vessel at an initial pressure of atm. When equilibrium is established for the following reaction, the total pressure increses to atm. What is KP?
PCl5 (g) PCl3 (g) + Cl2 (g)
Assume zero if not given
Initial (M)
1.000 atm x x
Change (M) -x
Equilibrium x x x
PPCl3 PCl2
Dalton’s Law PTOT = PPCl3 + PCl2 + PPCl5
KP =
PPCl5
1.714 atm = ( x) + x + x
1.714 atm = x
(x) (x)
KP =
( x)
Solve for x x = atm
KP = (0.714 atm)(0.714 atm) = 1.78 = KP
( ) atm
PPCL5 = ( ) atm = atm at equilibrium

31. Must increase PCl5 and decrease PCl3
Add atm Cl2. What will happen?
PCl5 (g) PCl3 (g) + Cl2 (g)
0.714 atm
( ) atm
Initial (M)
0.268 atm
- x
- x
Change (M) x
Must increase PCl5 and decrease PCl3
Equilibrium x x x
( x)( x)
PPCl3 PCl2
KP = =
= 1.78
PPCl5
( x)
x x + x2 = x
x x = 0 use quadratic equation
x =  (4)(1)(1.179) = 11.24
(2)(1)
0.0527
PPCl3 = = atm PCl2 = = atm
3.451 does not make sense
PPCl5 = = atm
KP = (0.661 atm)(0.911 atm) = 1.78
0.339 atm

32. Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Changes in Concentration
N2 (g) + H2 (g) NH3 (g)
Add
NH3
Equilibrium shifts left to offset stress
14.5

33. Le Châtelier’s Principle
Changes in Concentration continued
Add
Add
aA + bB cC + dD
Remove
Remove
Change
Shifts the Equilibrium
Increase concentration of product(s)
left
Decrease concentration of product(s)
right
Increase concentration of reactant(s)
right
Decrease concentration of reactant(s)
left
14.5

34. Le Châtelier’s Principle
Changes in Volume and Pressure
A (g) + B (g) C (g)
Change
Shifts the Equilibrium
Increase pressure
Side with fewest moles of gas
Decrease pressure
Side with most moles of gas
Increase volume
Side with most moles of gas
Decrease volume
Side with fewest moles of gas
14.5

35. Le Châtelier’s Principle
Changes in Temperature
Change
Exothermic Rx
Endothermic Rx
Increase temperature
K decreases
K increases
Decrease temperature
K increases
K decreases
Adding a Catalyst
does not change K
does not shift the position of an equilibrium system
system will reach equilibrium sooner
14.5

36. Catalyst lowers Ea for both forward and reverse reactions.
uncatalyzed
catalyzed
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
14.5

37. Le Châtelier’s Principle
Change Equilibrium
Constant
Change
Shift Equilibrium
Concentration
yes no
Pressure
yes no
Volume
yes no
Temperature
yes
yes
Catalyst no no
14.5