1. Chemical Equilibrium Chapter 6 E-mail: benzene4president@gmail.com
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2. Chemical Equilibrium – Ch. 6
1. Consider the following reaction:
N2 (g) H2 (g) ⇌ 2 NH3 (g)
a. Write the equilibrium expression (K and Kp)
b. What is the value of Kp at 25°C if the equilibrium pressures for N2, H2 and NH3 were 1.8 atm, 0.5 atm and atm respectively.
c. What is the value of K at 25°C? Kp = K(RT)Δn
d. Does the reaction favor reactants or products?

3. Chemical Equilibrium – Ch. 6
Equilibrium Expression
jA + kB ⇌ lC + mD
K = 𝐶 𝑙 𝐷 𝑚 𝐴 𝑗 𝐵 𝑘
Pure solids and liquids
are given the value of 1 in the
expression
Where does the equilibrium lie? K > 1 ⇒ The products are favored K < 1 ⇒ The reactants are favored

4. Chemical Equilibrium – Ch. 6

5. Chemical Equilibrium – Ch. 6
2. Which reaction does the K = Kp?
a. P4 (s) + 6 Cl2 (g) ⇌ 4 PCl3 (g)
b. H2O (l) ⇌ H2O (g)
c. H2 (g) + Cl2 (g) ⇌ 2 HCl (g)
d. C3H8 (g) + 5 O2 (g) ⇌ 3 CO2 (g) + 4 H2O (g)
Kp = K(RT)Δn

6. Chemical Equilibrium – Ch. 6
3. Given the following information:
H2(g) + I2(g) ⇌ 2 HI(g) K = 54.0
N2(g) + 3 H2(g) ⇌ 2 NH3(g) K = 96.0
Determine the value of the equilibrium constant for the following reaction:
2 NH3(g) + 3 I2(g) ⇌ 6 HI(g) + N2(g)

7. Chemical Equilibrium – Ch. 6
Since reversible reactions can be written in either direction or balanced with any multiple of coefficients you must consider how this affects the equilibrium constant – K
1. If you flip a reaction take the reciprocal of the constant
ex: if K=5 for 2A ⇌ B then K= for B ⇌ 2A
2. If you multiply the reaction coefficients raise the constant to the same multiple
ex: if K=5 for 2A ⇌ B then K=5½ for A ⇌ ½B
3. If you add two or more reactions take the product of the constants
ex: if K=5 for 2A ⇌ B and if K= 8 for B ⇌ 3C then
K = (5)(8) = 40 for 2A ⇌ 3C

8. Chemical Equilibrium – Ch. 6
4. Consider the following reaction:
UO2 (s) + 4HF (g) ⇌ UF4 (g) + 2H2O (g) Kp = 11
If you fill a 5 L container with 15 g of UO2, 1.1 atm of HF, 0.75 atm of UF4 and 5.9 atm of H2O, what will be the mass of the solid UO2 at equilibrium?
a. 15 g
b. > 15 g
c. < 15 g
d. not enough information

9. Chemical Equilibrium – Ch. 6
Reactions favor a state of equilibrium - so if a reaction mixture is not at equilibrium
it will automatically shift one way or the other in order to achieve equilibrium status -
The equation for Q looks identical to the equation for K - however you can apply Q
at any stage of a reaction whereas you can only apply K at equilibrium
Q = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑐𝑜𝑒𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑐𝑜𝑒𝑓 K Q
K > Q
System has too
much reactant
K = Q
System is at
equilibrium
K < Q
much product
Shift reaction
to the right
to the left
causes Q to
increase
decrease

10. Chemical Equilibrium – Ch. 6
5. Consider the following reaction:
N2 (g)  +  O2 (g) ⇌ 2  NO (g)    Kp = ?
Initially the partial pressures of N2 and O2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the Kp for this reaction?

11. Chemical Equilibrium – Ch. 6
6. Pure PCl5 is introduced into an empty rigid 5-L flask at 24 °C with a pressure of 0.45 atm.  The PCl5 is heated to 90 °C upon which it decomposes into solid P and gaseous Cl2. 
2 PCl5 (g) ⇌ 2 P (s) + 5 Cl2 (g)
At equilibrium the total pressure is measured to be 0.93 atm.  What is the equilibrium constant (Kp) at 90 °C? What percentage of the PCl5 decomposed? How many grams of P are in the flask at equilibrium?

12. Chemical Equilibrium – Ch. 6
7. At a certain temperature the partial pressures of an equilibrium mixture of N2O4 (g) and NO2 (g) are 0.34 atm and 1.2 atm respectively.
N2O4 (g) ⇌ 2 NO2 (g)
The volume of the container is doubled. Find the new partial pressures when the equilibrium is re-established.

13. Chemical Equilibrium – Ch. 6
8. Consider the following reaction: 
2 NOBr (g) ⇌ 2 NO (g) + Br2 (g)     Kp = 34
An unknown pressure of NOBr is put into a rigid container at 40 °C, equilibrium is reached when 86 % of the original partial pressure of NOBr has decomposed.  What was the initial pressure of the NOBr?  What is the total pressure in the flask at equilibrium?

14. Chemical Equilibrium – Ch. 6
9. At a particular temperature Kp = 5.7x10-8 for the following reaction: 
2 C (s) + 3 H2 (g) ⇌ C2H6 (g)
What is the pressure of C2H6 at equilibrium if initially there's 5 g of C and 3 atm of H2?

15. Chemical Equilibrium – Ch. 6
10. Consider the following endothermic reaction at equilibrium: 
CCl4 (g) ⇌ C (s) + 2 Cl2 (g)
a. Is heat absorbed or released as the reaction goes forward?
b. Which way will the reaction shift if the temperature is increased?
c. What happens to K as the temperature is increased?
d. Which way will the reaction shift if the partial pressure of CCl4 is increased?
e. Which way will the reaction shift if you add more C?
f. Which way will the reaction shift if you remove Cl2?
g. Which way will the reaction shift if you compress the system?
h. Which way will the reaction shift if you add a catalyst?

16. Answer Key – Ch. 6
You have completed ch. 6

17. K = 𝑁𝐶𝑙 3 2 𝑁 2 𝐶𝑙 2 3 and Kp = 𝑃 𝑁𝐶𝑙3 2 𝑃 𝑁2 𝑃 𝐶𝑙2 3
Answer Key – Ch. 6
1. Consider the following reaction:
N2 (g) Cl2 (g) ⇌ 2 NCl3 (g)
a. Write the equilibrium expression (K and Kp)
K = 𝑁𝐶𝑙 𝑁 2 𝐶𝑙 and Kp = 𝑃 𝑁𝐶𝑙 𝑃 𝑁2 𝑃 𝐶𝑙2 3
b. What is the value of Kp at 25°C if the equilibrium pressures for N2, Cl2 and NCl3 were 1.8 atm, 0.50 atm and atm respectively.
Kp = =
c. What is the value of K at 25°C? K = Kp (RT)Δn ⇒
K = (0.0013) ( 𝑥298) (2−4) = 0.77

18. Answer Key – Ch. 6 d. Does the reaction favor reactants or products?
1. …continued
d. Does the reaction favor reactants or products?
Since Kp < 1 and/or K < 1 the reaction favors the reactants
2. Which reaction does the K = Kp?
a. P4 (s) + 6Cl2 (g) ⇌ 4PCl3 (g)
b. H2O (l) ⇌ H2O (g)
c. H2 (g) + Cl2 (g) ⇌ 2HCl (g)
d. C3H8 (g) + 5O2 (g) ⇌ 3CO2 (g) + 4H2O (g)
If Δn=0 ⇒ Kp=K
Δn=0 when there are the same number of moles
of gaseous reactants as gaseous products

19. Answer Key – Ch. 6 3. Given the following information:
H2(g) + I2(g) ⇌ 2 HI(g) K = 54.0
N2(g) + 3 H2(g) ⇌ 2 NH3(g) K = 96.0
Determine the value of the equilibrium constant for the following reaction:
2 NH3(g) + 3 I2(g) ⇌ 6 HI(g) + N2(g)
Rxn 2
Flipped
Rxn 1
x 3
Rxn 1
x 3
Rxn 2
Flipped
3 H2(g) + 3 I2(g) ⇌ 6 HI(g) K = 54.03
2 NH3(g) ⇌ N2(g) + 3 H2(g) K =
2 NH3(g) + 3 I2(g) ⇌ 6 HI(g) + N2(g) K = (54.03)(96.0-1) = 1640

20. Answer Key – Ch. 6 4. Consider the following reaction:
UO2 (s) + 4HF (g) ⇌ UF4 (g) + 2H2O (g) Kp = 11
If you fill a 5 L container with 15 g of UO2, 1.1 atm of HF, 0.75 atm of UF4 and 5.9 atm of H2O, what will be the mass of the solid UO2 at equilibrium?
a. 15 g
b. > 15 g
c. < 15 g
d. not enough information
Qp = PUF4 PH2O 2 PHF = = 17.8
Since K < Q the reaction will go backwards thereby
increasing the mass of the UO2

21. Answer Key – Ch. 6 5. Consider the following reaction:
N2 (g)  +  O2 (g) ⇌ 2  NO (g)    Kp = ?
Initially the partial pressures of N2 and O2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the Kp for this reaction?
Since the P of NO at equilibrium is 1.5
2x = 1.5 => x = 0.75
So at equilibrium the
PN2 and PO2 are 0.25 atm and 2.25 atm respectively
Kp= PNO 2 PN2 PO2 = (1.5)2 (0.25)(2.25) = 4 N2 O2 ⇌
2 NO I
1 atm
3 atm Δ
- x
+ 2x Eq
1 – x
3 – x 2x

22. Answer Key – Ch. 6
6. Pure PCl5 is introduced into an empty rigid flask at 24 °C with a pressure of 0.45 atm.  The PCl5 is heated to 90 °C upon which it decomposes into solid P and gaseous Cl2. 
2 PCl5 (g) ⇌ 2 P (s) + 5 Cl2 (g)
At equilibrium the total pressure is measured to be 0.93 atm.  What is the equilibrium constant (Kp) at 90 °C? What percentage of the PCl5 decomposed?
Using the combined gas law
𝑃1𝑉1 𝑛2𝑇2 = 𝑃2𝑉2 𝑛2𝑇2
Where n and V are constant
0.45𝑎𝑡𝑚 𝐾 = 𝑃 𝐾
P2 = 0.55 atm
2 PCl5 (g) ⇌
2 P (s)
+ 5 Cl2 (g) I
0.55 atm
N/A Δ
- 2x
+ 5x Eq
0.55 – 2x 5x
Continue to next slide…

23. % decomposed = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑠𝑒𝑑 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 x100
Answer Key – Ch. 6
6. …continued
Ptotal = PPCl5 + PCl2
0.93 = (0.55 – 2x) + (5x)
x = 0.13
PPCl5 = 0.55 – 2(0.13) = 0.29
PCl2 = 5(0.13) = 0.65
Kp = (0.65)5 (0.29)2
Kp = 1.38
% decomposed = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑠𝑒𝑑 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 x100
% decomposed = 2(0.13) x100 = 47%

24. Answer Key – Ch. 6
7. At a certain temperature the partial pressures of an equilibrium mixture of N2O4 (g) and NO2 (g) are 0.34 atm and 1.2 atm respectively. The volume of the container is doubled. Find the new partial pressures when the equilibrium is reestablished.
N2O4 (g) ⇌ 2 NO2 (g)
Since the equilibrium pressures are given we can find Kp⇒
Kp = PNO2 2 PN2O4 = (1.2)2 (0.34) = 4.24
According to P1V1 = P2V2 ⇒ if the volume is doubled the pressures will be halved ⇒ PN2O4 = 0.17 atm and PNO2 = 0.6
Calculate Q to know the direction the reaction proceeds
Q = PNO2 2 PN2O4 = (0.6)2 (0.17) = 2.12 since K > Q the reaction goes forward in
order to re-establish equilibrium
…continue to next slide

25. Answer Key – Ch. 6 7. …continued Use K to solve for x
4.24 = ( x)2 (0.17 − x)
Manipulate to polynomial format
0 = 4x x –
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
x = or -0.71
So the new equilibrium pressures ⇒
PN2O4 = 0.17 – 0.053
PNO2 = (0.053)
N2O4 (g) ⇌
2 NO2 (g) I
0.17
0.6 Δ
- x
+ 2x Eq x x
PN2O4 = 0.12 atm
PNO2 = 0.71 atm

26. Answer Key – Ch. 6 8. Consider the following reaction:
2 NOBr (g) ⇌ 2 NO (g) + Br2 (g)     Kp = 34
An unknown pressure of NOBr is put into a rigid container at 40 °C, equilibrium is reached when 86 % of the original partial pressure of NOBr has decomposed.  What was the initial pressure of the NOBr?  What is the total pressure in the flask at equilibrium?
2 NOBr (g) ⇌
2 NO (g)
+ Br2 (g) I x Δ
-0.86x
+ 0.86x
+0.43x Eq
0.14x
0.86x
0.43x
…continue to
next slide

27. Answer Key – Ch. 6
8. …continued Use K to solve for x 34 = (0.86x)2(0.43x) (0.14x)2 x = 2.2 atm ⇒ the initial pressure of NOBr is 2.2 atm The total pressure in the flask at equilibrium Ptotal = 0.14x x x = 3.1 atm

28. Negligible b/c Kp is so small
Answer Key – Ch. 6
9. At a particular temperature Kp = 5.7x10-8 for the following reaction: 
2 C (s) + 3 H2 (g) ⇌ C2H6 (g)
What is the pressure of C2H6 at equilibrium if initially there's 5 g of C and 3 atm of H2?
Since Kp is so small we can make the assumption that x is negligible with respect to a non-zero value
Kp = PC2H6 PH2 3
5.7x10-8 = x (3)3
x = 1.5x10-6 atm
3 H2 ⇌
C2H6 I 3 Δ
- 3x
+ x Eq
3 - 3x x
Negligible b/c Kp is so small

29. Answer Key – Ch. 6
10. Consider the following endothermic reaction at equilibrium: 
CCl4 (g)  ⇌ C (s) + 2 Cl2 (g)
a. Is heat absorbed or released as the reaction goes forward?
Since the reaction is endothermic heat is absorbed if the reaction goes forward and vice versa
b. Which way will the reaction shift if the temperature is increased? As the temperature is increased the system will have more heat causing the reaction to shift right
c. What happens to K as the temperature is increased? As the temperature is increased heat is being added causing the reaction to shift right producing more products thereby causing an increase in K
d. Which way will the reaction shift if the partial pressure of CCl4 is increased? More reactant pushes the reaction to the right
… continue to next slide

30. Answer Key – Ch. 6
e. Which way will the reaction shift if you add more C? Adding more solid will have no effect on the equilibrium
f. Which way will the reaction shift if you remove Cl2? Removing a product causes the reaction to shift to the right
g. Which way will the reaction shift if you compress the system? Decreasing the volume of the system causes the reaction to shift to the side with fewer moles of gas – in this case the left
h. Which way will the reaction shift if you add a catalyst? Catalysts will speed up the forward and reverse direction therefore having no effect on the equilibrium

31. Answer Key – Ch. 6