1. Chapter 14 Chemical Equilibrium
14.1 the concept of equilibrium and the equilibrium constant
14.4 writing equilibrium constant expression
14.4 what does the equilibrium constant tell us
14.5 factors that effect chemical equilibrium

2. Chemical equilibrium is achieved when:
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)
Chemical equilibrium
N2O4 (g)
H2O (g)
2NO2 (g)

3. The Equilibrium Constant
[A], [B], etc. are the equilibrium concentrations
K ≈ [products] / [reactants]
K >> 1 ; favors products >>> Lie to the right
K << 1 favors reactants >>> Lie to the left
K ≈ 1: roughly equal concentration of reactants and products
K is a constant at a given temperature
Solids drop out of the expression & water drops out when the solvent is water
K has no unit

4. 14.2 Writing Equilibrium Constant Expressions
A. Homogeneous Equilibria (all species are in the same phase)
N2O4 (g) NO2 (g)
Kc =
[NO2]2
[N2O4]
Kp =
NO2
P 2
N2O4 P
aA (g) + bB (g) cC (g) + dD (g)
Kp = Kc(RT) ∆n
In most cases
Kc  Kp
∆n = moles of gaseous products – moles of gaseous reactants
∆n = (c + d) products– (a + b)reactants
Kp = Kc , when ∆n= 0

5. Homogeneous Equilibria (all species are in the same phase)
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
Kc = ‘
[CH3COO-][H3O+]
[CH3COOH][H2O]
[H2O] = constant
[CH3COOH] =
Kc [H2O]
General practice not to include units for the equilibrium constant.

6. B. Heterogeneous Equilibria 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 = [CO2] =
Kc x
[CaO]
Kp = PCO 2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

7. Example 14-1 Kc = [H3O+ ][F-] [HF] Kc = [NO2]2 [NO]2[O2] Kc =
Write the equilibrium constant expression for the following reactions:
(a) HF (aq) + H2O (ℓ) ⇄ H3O+ (aq) + F- (aq)
(b) 2 NO (g) + O2 (g) ⇄ 2 NO2 (g)
(c) CH3COOH (aq) + C2H5COH (aq) ⇄ CH3COOC2H5+ H2O (ℓ)
Homogeneous
Kc =
[H3O+ ][F-]
[HF]
Kc =
[NO2]2
[NO]2[O2]
Kc =
[CH3COOC2H5]
[CH3COOH] [C2H5COH]
Kp =
P 2 NO2
P 2 NO P 2 O2

8. Example 14-2 The equilibrium concentrations
2 NO (g) + O2 (g) ⇄ 2 NO2 (g)
at 2300C are [NO] = M , [O2] = M, and [NO2] = M. Calculate the equilibrium constant Kc.
Homogeneous
Kc =
[NO2]2
[NO]2[O2]
Kc =
[15.5]2
[0.0542]2[0.127]
= 6.44 × 105
Dr.Laila Al-Harbi

9. Example 14-3 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) ⇄ 2NO (g) + O2 (g)
Homogeneous
Kp = 2
PNO PO
PNO PO 2
= Kp
PNO PO 2
= 158 x (0.400)2/(0.270)2
= 347 atm

10. Example 14-4 Methanol is manufactured industrially by the reaction
Kc = 10.5 at 220°C. What is the value of Kp at this temperature
Kp = Kc (RT)∆n
∆n = 1-3 = -2
Kp = Kc (0.0821× 493)-2 = 6.41 × 10-3
CO (g) + 2 H2 (g) ⇄ CH3OH(g)
Homogeneous

11. Example 14-5
Write the equilibrium constant expression for the following reactions:
(a) (NH4)2Se (s) ⇄ 2NH3 (g) + H2Se (g)
(b) AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)
(c) P4 (s) + 6Cl2(g) ⇄ 4PCl3(ℓ)
Heterogeneous
Kp =
PNH32 PH2Se
Kc =
[NH3]2 [H2Se]
Kc =
[Ag+ ]2 [Cl-]
Kc = 1
[Cl2]6

12. Example 14-6 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)
Heterogeneous
Kp = P
NH3
H2S P
= x =
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn
∆n = 2 – 0 = 2
T = 295 K
Kc = x ( x 295)-2 = 1.20 x 10-4
Dr.Laila Al-Harbi

13. 14.4 what does the equilibrium constant tell us
A) Predicting the direction of a reaction
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 right to left to reach equilibrium
Qc = Kc the system is at equilibrium
Qc < Kc system proceeds from left to right to reach equilibrium
B) Calculating equilibrium concentration
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. A) Predicting the direction of a reaction
Q has the same form as K, but uses existing concentrations
n-Butane iso-Butane
Kc =2.5
Q =
[iso]
[n] =
0.35
0.25
1.40
Q =
Since Q (1.4) < Kc (2.5) , the system at equilibrium
To reach equilibrium [iso-Butane] must increase
and [n-Butane] must decrease.

15. Example 14.8 0.611 = N2 (g) + 3H2(g) ⇄ 2NH3 (g) [NH3]2 Q c= [N2] [H2]3
0.249mol,N2 , 3.2×10-2 mol H2 , 6.42×10-4 mol NH3in A 3.50 L at 375°C , kc= 1.2, Decide whether the system is at equilibrium .if it is not predict which way the net reaction will proceed.
N2 (g) + 3H2(g) ⇄ 2NH3 (g)
Homogeneous
[NH3]2
[N2] [H2]3
Q c=
Since the molarity = number of moles /volume in L , M=n/Vin L
[N2] °= 0.249/3.50 L = M
[H2] ° = 3.2×10-2 /3.50 L = 9.17 ×10-3 M
[NH3] ° = 6.42×10-4 /3.50 L = 1.83 ×10-4 M =
[1.83 ×10-4 ]2
[0.0711][9.17 ×10-3]2
0.611
[NH3]2
[N2] [H2]3
Q c=
Since Q (0.611) < Kc (1.2) system , the system at equilibrium
To reach equilibrium [NH3] must increase and [N2], [H2] must decrease.
The net reaction will proceed from left to right untial equilibrium is reached .

16. B) Calculating equilibrium concentration
cis-stilbene trans-stilbene
Kc =24
Step 1 Define equilibrium condition in terms of initial condition and a change variable
[cis-stibene] [trans-stibene]
Initial
At equilibrium x x
Step 2 Put equilibrium Conic. into Kc .
Step 3. Solve for x (0.85-x) = x = 20.4 – 24x=x
25x= 20.4 >>>>>> x = M
[cis-stilbene] = 0.85 – = M
[trans-stilbene] = x = M
At equilibrium

17. Example 14.9 H2(g) + I2(g) 2 HI(g) Kc = 54.3
M= 0.5/1 =0.5 mol
Step 1 Define equilibrium condition in terms of initial condition and a change variable
[H2] [I2] [HI]
Initial
change x x x
At equilibrium x 0.5-x 2x
Step 2 Put equilibrium Conic. into Kc .
Step 3. Solve for x = (2x)2/(0.5-x)2
Square root of both sides & solve gives: = 2x/0.5-x
x =
At equilibrium
[H2] = [I2] = – = M
[HI] = 2x = M

18. Practice exercise 14.9
Consider the reaction in Ex.14.9 starting with [HI] =0.04 molar . Calculate [HI],[I2]& [H2] at equilibrium.
2 HI(g) ⇄ H2(g) + I2(g) Kc = 1/ 54.3 =
[HI] [I2] [H2]
Initial
At equilibrium x x x
[I2][H2]
[HI]2 x2
(0.04-2x)2
Kc = = =
Square root of both sides & solve gives: x = M
At equilibrium
[H2] = [I2] = = M
[HI] = x = M

19. 14.5 factors that effect chemical equilibrium
Le Chatelier’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.
factors that effect chemical equilibrium
Changes in Concentration
Changes in Volume and Pressure
Changes in Temperature
Adding a Catalyst

20. Changes in Concentration
N2 (g) + 3H2 (g) NH3 (g)
Add
NH3
Equilibrium shifts left to offset stress
Change
Shifts the Equilibrium
Increase concentration of product(s)
left
Decrease concentration of product(s)
right
Decrease concentration of reactant(s)
Increase concentration of reactant(s)

21. Example 14.11
N2 (g) + 3H2(g) ⇄ 2NH3 (g) >>>> Kc=2.37×10-3 , T=720°C
Equilibrium shifts left to offset stress
Add
NH3
3.65 M
[N2] = M
[H2] = M
[NH3] = 1.05M the concentration increase to 3.65 M
Calculate Qc compare it value with Kc
because Qc (2.86×10-2) > Kc (2.37×10-3) , The net reaction direction from right to left until Qc = Kc

22. Changes in Volume and Pressure
∆n = is the number of moles for substance in gaseous products &gaseous reactants
A (g) + B (g) C (g)
∆n=n products –n reactants
Note: Pressure and volume are inversely proportional.
Because the pressure of gases is related directly to the concentration by P = n/V, changing the pressure by increasing/decreasing the volume of a container will disturb an equilibrium system.
Change
Shifts the Equilibrium
Increase pressure (decrease volume)
Side with fewest moles of gas
Decrease pressure (Increase volume)
Side with most moles of gas

23. Example 14.12
Predict the net reaction direction (increasing P & decreasing V)
2 PbS (s) + 3O2 (g) ⇄ 2PbO (s) + 2HS2 (g)
∆n=n products –n reactants = 2-3 = -1
When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas (left to right toward product).
(b) PCl5 (g) ⇄ PCl3(g) + Cl2 (g)
∆n=n products –n reactants = 2-1 =1
When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas (right to left toward reactant).
(c) H2 (g) + CO2(g) ⇄ H2O(g) + CO(g)
∆n=n products –n reactants = 2-2 = 0
The change in P,v has no effect on the equilibrium .

24. Changes in Temperature
Exothermic Rx(∆H)=-
Increase temperature
K decreases
Decrease temperature
K increases
Endothermic Rx (∆H)= +
التفاعل من اليمين الى اليسار
التفاعل من اليسارالى اليمين
Adding a Catalyst
does not change K
does not shift the position of an equilibrium system
system will reach equilibrium sooner
uncatalyzed
catalyzed
Catalyst lowers Ea for both forward and reverse reactions
Catalyst does not change equilibrium constant or shift equilibrium.

25. 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

26. N2F4 (g) ⇄ 2 NF2 (g) >> ∆H= 38.5 kJ/mol
Example 14.13
Predict the net reaction direction a) if RXN heated at constant V, b) some N2F4 removed at constant T&V c) Decrease P?
d) catalyst is added
N2F4 (g) ⇄ 2 NF2 (g) >> ∆H= 38.5 kJ/mol
a) ∆H>0 >>endothermic reaction ,T increase, K decreases, a net change occurs in the direction is from left to right toward product).
b) Conc. of the reactant decrease the system shift to left( some NF2 combines to produce N2F4 )
c) P decrease the system shift to right .
d) if catalyst is added to reaction mixture ,the reaction will reach equilibrium faster but no change in the change equilibrium constant or shift equilibrium.

27. CuO(s) + H2(g) Cu(s) + H2O(g) is Kc = [H2]/[H2O] Kc = [H2]/[H2O][Cu]
Which is the correct equilibrium constant expression for the following reaction?
CuO(s) + H2(g)     Cu(s) + H2O(g) is
Kc = [H2]/[H2O]
Kc = [H2]/[H2O][Cu]
Kc = [Cu][H2O] /[Cu][H2]
Kc = [H2O]/[H2]
[23] Carbon tetrachloride reacts at high temperatures with oxygen to produce two gases,
phosgene and chlorine at 1000 K. The equilibrium constant Kp is 4.4  10-9 at the
same temperature, calculate Kc for the reaction .
  2COCl2(g) + 2Cl2(g).
2CCl4(g) + O2(g)  
4.5 x10-10
5.4 x1011
5.36 x10-11
4.01 x 10-5

28. [3] The equilibrium is achieved when ____ Q > K Q < K Q = K
[4]In which of these gas-phase equilibrium is the yield of products increased by increasing the
total pressure on the reaction mixture? 
CO(g) + H2O(g)
CO2(g) + H2(g)
PCl5(g) PCl3(g) + Cl2(g)
2SO3(g)
2SO2(g) + O2(g)
2NO(g) + Cl2(g)
2NOCl(g)

29. [5] The equilibrium constant, Kc for the reaction H2(g) + CO2(g) → H2O(g) + CO(g) is 4.2.
Initially 0.80 mole of H2 and 0.80 mole CO2 are placed in a 5.0 L flask. The concentration
of H2 at equilibrium is _____.
0.054 M
0.025 M
0.15 M
0.029 M
[6] Which of the following statements about chemical equilibrium is false?
Adding catalyst to the mixture do not affect the equilibrium constant.
Change in temperature can alter the equilibrium constant.
If an external stress is applied to a system at equilibrium, the system adjusts that the stress is
partially offset as the system reaches a new equilibrium position.
If Qc is greater than Kc; the system will be at equilibrium.

30. Kc = [Fe2O3] [H2]3/[Fe]2[H2O]3 Kc = [H2]/[H2O] Kc = [H2O]3/[H2]3
[7]Which is the correct equilibrium constant expression for the following reaction? Fe2O3(s) + 3H2(g)  
 2Fe(s) + 3H2O(g)  
Kc = [Fe2O3] [H2]3/[Fe]2[H2O]3
Kc = [H2]/[H2O]
Kc = [H2O]3/[H2]3
Kc = [Fe]2[H2O]3/[Fe2O3] [H2]3
[8] Carbon tetrachloride reacts at high temperatures with oxygen to produce two gases, phosgene
and chlorine at at 1000 K. The equilibrium constant Kp is 4.4  10-9 at the same temperature.
Calculate Kc for the reaction. 2CCl4(g) + O2(g)  →
2COCl2(g) + 2Cl2(g).
5.36 x10-11
7.6 x10-8
1.10 x10-10
2 x10-4

31. [9] When the substances in the equation below are at equilibrium, at pressure P and
temperature T, the equilibrium can be shifted to favor the products by         CuO(s) + H2(g)  
H2O(g) + Cu(s) Hºrxn = -2.0 kJ/mol
decreasing the temperature   
increasing the temperature.
increasing the pressure.
adding a catalyst.
10- The equilibrium constant, Kc for the reaction H2(g) + CO2(g) → 2O(g) + CO(g) is 4.2. Initially 0.80 mole of H2 and 0.80 mole CO2 are placed in a 5.0 L flask. The concentration of H2 at equilibrium is _____.
0.054 M
0.025 M
0.15 M
0.029 M

32. 11 Which of the following statements about chemical equilibrium is false?
Adding catalyst to the mixture do not affect the equilibrium constant.
Change in temperature can alter the equilibrium constant.
If an external stress is applied to a system at equilibrium, the system adjusts that the stress is
partially offset as the system reaches a new equilibrium position.
If Qc is greater than Kc; the system will be at equilibrium.
12] Carbon tetrachloride reacts at high temperatures with oxygen to produce two gases,
phosgene and chlorine at 500 K. The equilibrium constant Kp is 1.4  10-5 at same
Temperature.. Calculate Kc for the reaction CCl4(g) + O2(g)   ↔ 2COCl2(g) + 2Cl2(g).  
7.4 x10-7
3.4x10-7
5.2 x10-4
3 x10-11

33. position of the equilibrium? 1-a shift to produce more SO3
[13]Consider this reaction at equilibrium:         2SO2(g) + O2(g)   SO3(g) , Hºrxn = -198 kJ/mol
  If the volume of the system is compressed at constant temperature, what change will occur in the
position of the equilibrium?   a shift to produce more SO3
2-a shift to produce more O2
3-no change
4-a shift to produce more SO2
[14] When the following reaction is at equilibrium, which of these relationships is true
at equilibrium?           NOCl(g)   
2NO(g) + Cl2(g)
[NO]2[Cl2]2 = [NOCl]2
PNOCl / P NO PCl
[NO]2[Cl2] = Kc[NOCl]2
P2NO / P2 NOCl

34. 15- Kc is 3.8  10-5 at 1000K for the equilibrium I2(g) ↔ 2I(g)
Starting with moles of I2 in a 2.30 L flask, at 1000 K. The equilibrium concentration of I2 is: M M
0.12 M
0.081 M
[16] Carbon tetrachloride reacts at high temperatures with oxygen to produce two gases,
phosgene and chlorine at at 300 K. The equilibrium constant Kp is 1.7  10-3 at same
temperature. Calculate Kc for the reaction. 2CCl4(g) + O2(g)   ↔  2COCl2(g) + 2Cl2(g).  
7.7x10-7
1.06 x10-8
6.9 x10-5
1.10 x10-5

35. [17]Which is the correct equilibrium constant expression for the following reaction?
NH4HS(s)   NH3(g) + H2S(g)
A. Kc = [NH3] [H2S]
B. Kc = [NH3] [H2S]/ [NH4 HS]
C. Kc = [NH4 HS] / [NH3] [H2S]
D. Kc = [NH3] [H2S]2
[18] For the equilibrium reaction
2SO2(g) + O2(g)     ↔ SO3(g), Hºrxn = -198 kJ/mol.
Which one of these factors would cause the equilibrium constant to increase?
Add SO2 gas.
Remove O2 gas.
Decrease the temperature.
Increase the temperature.

36. [19] The reaction in which increased pressure has no affect on the equilibrium reaction is
N2(g)+3H2(g)↔2NH3(g)
2H2+CO(g)↔CH3OH(l)
CO(g)+H2O(g)↔CO2 (g)+H2(g)
CaCO3(s)↔CaO(s)+CO2 (g)
Consider the reaction
2NO(g)+O2(g)↔2NO2(g)
At 4300C,an equilibrium mixture consist of mole of O2,0.040 mole of NO,and 0.96 mole of NO2 .calculate Kp for the reaction,given that the total pressure is 0.20 atm.
A.1.5×10-5
B.150
C.1.32×103
D. 1.5×105

37. -For the following reaction at equilibrium in a reeaction vessel,which one of these stresses would cause the Br2
concentration to increase.
2NOBr(g)↔2NO(g)+Br2(g) ∆H0=70 KJ/mol
A-Lowering the temperature
B-Removing some NOBr
C-Increasing the temperature
D- Increasing the pressure of the system
-The substances in the equation below are at equilibrium.The equilibrium can be shifted to favor the products by:
CuO(s)+H2(g)↔H2O(g)+Cu(s) ∆H0= -2.0 KJ/mol
A-decreasing the temperature
B-Removing some CuO
C-Increasing the temperature
D- adding a catalyst

38. Consider the reaction
H2(g)+Cl2(g)↔2HCl(g)
At C,an equilibrium mixture consist of 0.57 mole of H2, 3.39 mole of HCl,and 0.1 mole of Cl2 .calculate Kp for the reaction,given that the total pressure is 2.00 atm.
A.203
B.2.03
C.20.3
D. 2.03×103