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Equilibrium Exercises
Show all your work and use the R.I.C.E. method when appropriate.
1. 2 CO2(g)  2 CO(g) + O2(g)
A 1.00 mole sample Of CO2(g) is placed in a 1.00 L container and allowed to come to
equilibrium at 2500 K. When equilibrium is reached at 2500 K, 17.6% of the original
CO2(g) has decomposed to CO(g) and O2(g). Calculate the value of the equilibrium
constant, Kc, for the dissociation reaction at 2500 K.
2. Given the reaction:
NH4HS(s)  NH3(g) + H2S(g) H = +197.8 kJ
Suppose the substances in the reaction are at equilibrium at 500 K. State whether the
partial pressure of NH3(g) will have increased, decreased, or remained the same when
equilibrium is re-established after each of the following disturbances of the original
system. Justify each answer with a brief explanation.
(A) A small quantity of H2S(g) is added.
(B) The temperature of the system is increased.
(C) The volume of the system is increased.
(D) A quantity of N2(g) is added.
(E) A quantity of NH3(g) is added.
3. At 350 K, Kc is 0.14 for the reaction:
2 BrCl(g)  Br2(g) + Cl2(g)
An equilibrium mixture at this temperature contains equal concentrations of bromine and
chlorine, 0.0250 M. What is the equilibrium concentration of BrCl(g)?
4. Kc for the reaction: 2 ICl(g)  I2(g) + Cl2(g)
is 0.11 at a certain temperature. Suppose the initial concentrations (mol/L) of ICl(g), I2(g),
and Cl2(g), are 0.20 M, 0.00 M, and 0.00 M, respectively. Some of the ICl(g) decomposes
and the system reaches equilibrium. What is the equilibrium concentration of each
species?
5. The reaction:
CO(g) + H2O(g)  CO2(g) + H2(g)
has a Kc value of 0.40 at 500oC. Calculate the concentration of all species at equilibrium
starting with:
(A) [CO(g)] = [H2O(g)] = 0.100 M and [CO2(g)] = [H2(g)] = 0.00M
(B) [CO(g)] = [H2O(g)] = [CO2(g)] = [H2(g)] = 0.040 M
6. For the reaction:
CO(g) + H2O(g)  CO2(g) + H2(g)
equilibrium is established at a certain temperature when the concentration of CO(g),
H2O(g), CO2(g), and H2(g) are 0.010 M, 0.020 M, 0.012 M, and 0.012 M respectively.
(A) Calculate Kc.
(B) If enough CO(g) is added to raise its concentration temporarily to 0.020 M, what will
be the concentrations of all species after equilibrium has been re-established at the same
temperature?
7. 0.0100 moles of NH4Cl and 0.0100 moles of NH3 are place in a closed 2.00 L
container and heated to 603 K. At this temperature, all the NH4Cl vaporizes. When the
reaction NH4Cl(g)  NH3(g) + HCl(g) has reached equilibrium, 5.8 X 10-3 moles of HCl
are present. Calculate Kc and Kp for this reaction at 603 K.
8. For the reaction H2(g) + I2(g)  2 HI(g) at 450oC, the equilibrium constant is 65.0
Calculate the equilibrium pressure of all the species in a vessel that initially contained
0.500 atm HI.
9. For the reaction
SO2Cl2(g)  SO2(g) + Cl2(g) Kp = 2.4 at 375oC.
(A) Assume you place 6.70 g of SO2Cl2(g) in a 1.00 L closed flask and heat to 375oC.
What is the pressure of each compound in the system and the total pressure in the flask
when equilibrium is established? What fraction of SO2Cl2(g) has dissociated?
(B) What are the pressures of SO2Cl2(g), SO2(g), and Cl2(g) at equilibrium in the flask if
you begin with 6.70 g SO2Cl(g) and 1.00 atm Cl2(g)? What fraction of SO2Cl(g) has
dissociated?
(C) Do your answers to (A) and (B) agree with the expected results based on Le
Chatelier’s Principle? Explain.
10. The equilibrium constant Kp for the reaction
N2O4(g)  2 NO2(g) is 0.15 at
25oC. Calculate the equilibrium constant for the reaction
NO2(g)  ½ N2O4(g)
at the same temperature.
11. At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown
by the following equation: SbCl5(g)  SbCl3(g) + Cl2(g)
(a) An 89.7 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated
15.0 liter container at 182oC. 1. What is the concentration in moles per liter of
SbCl5 in the container before any decomposition occurs? 2. What is the pressure
in atmospheres of SbCl5 in the container before any decomposition occurs?
(b) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at
182oC, calculate the value for either equilibrium constant Kp or Kc, for this
decomposition reaction. Indicated whether you are calculating Kp or Kc.
(c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an
empty 2.00 liter container maintained at a temperature different from 182oC. At
this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this
container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?
Answers: Equilibrium Exercises
1. .004 M
2. a. decrease, reactions shifts left to consume H2S and also consumes the NH3
b. increase, reaction is endothermic and will shift toward products to consume heat
making more NH3
c. increase, increase volume decreases the pressure the reaction shift right to make more
moles of gas
d. remains the same, total pressure is changed but the partial pressures of the reactants
and products is not changed
e. increase, the reaction will shift left to consume NH3 but will not consume more than
was added so there is a net increase
3. [BrCl] = .0668M
4. [I2] and [Cl2] = .039M, [ICl] = .122M[CO2] & [H2] = ..031M, [CO] & [H2O] = .049M
5. a. [CO2] & [H2] = .061M, [CO] & [H2O] = .039M
b. [CO2] & [H2] = ..031M, [CO] & [H2O] = .049M
6. a. 72 b. . [CO2] & [H2] = .0147M, [CO] & [H2O] = .0173M
7. Kc = .0109, Kp = .54
8. [H2] & [I2] = .277M, [HI] = .223
9. a. [SO2Cl2] = 1.065 atm, [SO2] & [Cl2] = 1.595 atm, 4.255 atm., .600
b. [SO2Cl2] = 1.30atm, [SO2] = 1.34 atm [Cl2] = 2.34 atm, .51
c. reaction according to Le Chatelier shifts left resulting in more SO2Cl2
10. 2.58
11. 1988 A Answer:
1 mol
89.7g SbCl 5 x
0.300 mol SbCl 5
299.0g
(a) (1)
[ Sb Cl 5 ] in it . 
0.30 0 mol
15 .0 L
 0 . 20 0 M
(2) T = 182oC + 273 = 455K
P 
nRT
V

( 0 . 30 0 mol )0 . 08 21
L a t m
m ol K
( 455 K )
15 . 00 L
= 0.747 atm
(b) [SbCl3] = [Cl2] = (0.0200 mol/L)(0.292)
= 5.84 x 10-3M
[SbCl5] = (0.0200 mol/L)(0.708) = 1.42 x 10-2M
[SbCl 3][ Cl 2] ( 5.84 x103)2
K c 

 2.41 x103
[ SbCl 5]
1.42 x102
OR
PSbCl3 = PCl2 = (0.747 atm)(0.292) = 0.218 atm
PSbCl5= (0.747 atm)(0.708) = 0.529 atm
(PSbCl )( PCl ) ( 0.218)2
3
2
KP 

 8.98 x102
PSbCl
0.529
5
K 
(c)
[ Sb Cl 3 ][ Cl 2 ]
[ Sb Cl 5 ]
[ Sb Cl 5 ] eq u il . 
[ Sb Cl 3 ] eq u il . 
 0 . 11 7
(1 . 00  0 . 70 ) mol
2 . 00 L
0 . 70 0 mol
2 . 00 L
 0 . 15 M
 0 . 35 0 M
[Cl2]equil. = X M
K C 
( 0 . 35 0 )( X )
0 . 15
 0 . 11 7 ; X  0 . 05 0 M  [ Cl 2 ]
Moles Cl2(equil.) = (0.0050 mol/L)(2.00L)
= 0.10 mol Cl2
Moles Cl2 needed to make 0.300 mol SbCl3 into SbCl5 = 0.30 mol
Moles Cl2 that must be added = 0.40 mol