CHEMISTRY 12
10. Problem
Methane, ethyne, and hydrogen form the following equilibrium mixture.
2CH4(g) ⇀
↽ C2H2(g) + 3H2(g)
While studying this reaction mixture, a chemist analysed a 4.0 L sealed flask at
1700˚C. The chemist found 0.46 mol of CH4(g) , 0.64 mol of C2H2(g), and 0.92 mol
of H2(g) . What is the value of Kc for the reaction at 1700˚C?
What Is Required?
You need to calculate the value of Kc .
What Is Given?
You have the chemical equation and the amount of each substance at equilibrium
in a 4.0 L flask.
Plan Your Strategy
Step 1 Calculate the molar concentration of each compound at equilibrium.
Step 2 Write the equilibrium expression. Then substitute the equilibrium molar
concentrations into the expression.
Act on Your Strategy
Step 1 The reaction takes place in a 4.0 L flask. Calculate the molar concentrations
at equilibrium.
[CH4] = 0.46 mol = 0.115 mol/L
4.0 L
0.64
mol = 0.16 mol/L
[C2H2] =
4.0 L
[H2] = 0.92 mol = 0.23 mol/L
4.0 L
Step 2 Write the equilibrium expression. Substitute the equilibrium molar
concentrations into the expression.
3
Kc = [C2H2][H2 2]
[CH4]
3
= (0.16)(0.23)
2
(0.115)
= 0.15
Check Your Solution
The equilibrium expression has the product terms in the numerator and the reactant
terms in the denominator. The exponents in the equilibrium expression match the
corresponding coefficients in the chemical equation. The molar concentrations at
equilibrium were substituted into the expression. The answer has the correct number
of significant digits.
Solutions for Practice Problems
Student Textbook pages 347–348
11. Problem
At 25˚C the value of Kc for the following reaction is 82.
I2(g) + Cl2(g) ⇀
↽ 2ICl(g)
0.83 mol of I2(g) and 0.83 mol of Cl2(g) are placed in a 10 L container at 25˚C.
What are the concentrations of the three gases at equilibrium?
What Is Required?
You need to find [I2 ], [Cl2] and [ICl] at equilibrium.
What Is Given?
You are given the chemical equation. You know the initial amount of each gas,
the volume of the container, and the equilibrium constant.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Plan Your Strategy
Step 1 Calculate the initial concentrations.
Step 2 Set up an ICE table. Insert the initial concentrations you calculated in
Step 1 in your ICE table. Let the change in molar concentrations of the
reactants be x. Use the stoichiometry of the chemical equation to write
and record expressions for the equilibrium concentrations.
Step 3 Write the equilibrium expression. Substitute the expressions for the
equilibrium concentrations into the expression. Solve the equilibrium
expression for x.
Step 4 Calculate the equilibrium concentration of each gas.
Act on Your Strategy
Step 1 The initial amount of I2 is equal to the initial amount of Cl2.
[I2] = [Cl2] = 0.83 mol = 0.083 mol/L
10 L
Step 2 Set up an ICE table.
Concentration (mol/L)
Initial
Change
Equilibrium
I2(g)
Cl2(g)
2ICl(g)
0.083
0.083
0
−x
−x
+2x
0.083 − x
0.083 − x
2x
+
2
Kc = [ICI]
[I2][Cl2]
Step 3
(2x)2
(0.083 − x)(0.083 − x)
(2x)2
=
(0.083 − x)2
The right side of the equation is a perfect square.
(2x)
±9.06 =
(0.083 − x)
Solving for both values,
x = 0.068 and x = 0.11
Step 4 The value x = 0.11 is physically impossible because it is a larger concentration than was initially placed in the container. To achieve equilibrium, the
initial [I2 ] and [Cl2] must decrease.
[I2] = [Cl2] = 0.083 − 0.068 = 0.015 mol/L
[ICl] = 2 × 0.068 = 0.14 mol/L
82 =
Check Your Solution
The equilibrium expression has product concentrations in the numerator and reactant
concentrations in the denominator. Check Kc :
2
Kc = (0.14) 2 = 87
(0.015)
This is close to the given value, Kc = 82. The difference is due to mathematical
rounding.
12. Problem
At a certain temperature, Kc = 4.0 for the following reaction.
2HF(g) ⇀
↽ H2(g) + F2(g)
A 1.0 L reaction vessel contained 0.045 mol of F2(g) at equilibrium. What was the
initial amount of HF in the reaction vessel?
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
What Is Required?
You need to find the initial amount of HF(g) placed in the reaction vessel.
What Is Given?
You are given the chemical equation. You know the amount of F2(g) at equilibrium,
the volume of the container, and the equilibrium constant.
Plan Your Strategy
Step 1 Calculate the concentration of F2(g) at equilibrium and use the chemical
equation to find the equilibrium concentration of H2(g) .
Step 2 Set up an ICE table. Insert the equilibrium concentrations you calculated
in Step 1 in your ICE table. Let the change in molar concentrations of the
reactants be x. Use the stoichiometry of the chemical equation to write and
record expressions for the equilibrium concentrations.
Step 3 Write the equilibrium expression. Substitute the expressions for the
equilibrium concentrations into the expression. Solve the equilibrium
expression for the initial [HF].
Step 4 Calculate the initial amount of HF(g) .
Act on Your Strategy
Step 1 The volume of the reaction vessel is 1.0 L.
Therefore,
[F2] = 0.045 mol = 0.045 mol/L
1.0 L
[H2] = [F2] = 0.045 mol/L
Step 2 Set up an ICE table.
Concentration (mol/L)
Initial
Change
Equilibrium
2HF(g)
H2(g)
[HF]
0
0
−2x
+x
+x
([HF] − 2x)
0.045
0.045
+
F2(g)
Therefore, x = 0.045
Kc = [H2][F22]
Step 3
[HF]
(0.045)(0.045)
4.0 =
([HF] − 0.090)2
(0.045)2
=
([HF] − 0.090)2
The right side of the equation is a perfect square.
(0.045)
±2.0 =
([HF] − 0.090)
Solving for both values gives
[HF] = 0.112 and [HF] = 0.0675
Step 4 The value [HF] = 0.0675 is physically impossible because subtracting the
change in concentration, 0.090, results in a negative value for the initial
concentration of HF(g) placed in the container.
Therefore, [HF] = 0.11 mol/L.
Because the volume of the reaction vessel is 1.0 L, the initial amount of
HF(g) placed in the reaction vessel was 0.11 mol.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Check Your Solution
The equilibrium expression has product concentrations in the numerator and reactant
concentrations in the denominator. Check Kc :
(0.045)2
Kc =
= 5.1
(0.11 − 0.090)2
This is close to the given value, Kc = 4.0. The difference is due to mathematical
rounding.
13. Problem
A chemist was studying the following reaction.
↽ NO(g) + SO3(g)
SO2(g) + NO2(g) ⇀
In a 1.0 L container, the chemist added 1.7 × 10−1 mol of SO2(g) to 1.1 × 10−1 mol
of NO2(g). The value of Kc for the reaction at a certain temperature is 4.8. What is
the equilibrium concentration of SO3(g) at this temperature?
What Is Required?
You need to find the equilibrium [SO3].
What Is Given?
You are given the chemical equation. You know the equilibrium constant for the
reaction, Kc = 4.8. You also know the initial amounts of SO2(g) and NO2(g), and the
volume of the container.
Plan Your Strategy
Step 1 Calculate the initial concentrations of SO2(g) and NO2(g).
Step 2 Set up an ICE table. Insert the equilibrium concentrations you calculated
in Step 1 in your ICE table. Let the change in molar concentrations of the
reactants be x. Use the stoichiometry of the chemical equation to write and
record expressions for the equilibrium concentrations.
Step 3 Write the equilibrium expression. Substitute the expressions for the
equilibrium concentrations into the expression. Rearrange the equilibrium
expression into the form of a quadratic equation for x. Solve the quadratic
equation for x.
Step 4 Substitute x into the equilibrium line of the ICE table to find the equilibrium concentration of [SO3].
Act on Your Strategy
Step 1 The volume of the reaction vessel is 1.0 L.
Therefore,
−1
[SO2] = 1.7 × 10 mol = 1.7 × 10−1 mol/L
1.0 L
1.1
×
10−1 mol = 1.1 × 10−1 mol/L
[NO2] =
1.0 L
Step 2 Set up an ICE table.
Concentration (mol/L)
Initial
SO2(g)
Step 3
NO2(g)
NO(g) + SO3(g)
1.7 × 10−1
1.1 × 10−1
0
0
−x
−x
+x
+x
(1.7 × 10−1 − x) (1.1 × 10−1 − x)
x
x
Change
Equilibrium
+
[NO][SO3]
[SO2][NO2]
(x)(x)
4.8 =
(0.17 − x)(0.11 − x)
Kc =
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
This equation must be rearranged into a quadratic equation.
3.8x 2 − 1.344x + 0.08976 = 0
Recall that a quadratic equation of the form ax 2 + bx + c = 0 has the
following solution.
√
2
x = −b ± b − 4ac
2a
Therefore,
√
x = −(−1.344) ± 1.806 − 1.364
7.6
1.344
±
0.6648
=
7.6
x = 0.089 and x = 0.264
The value x = 0.264 is not physically possible. It would result in negative
concentration of SO2 and NO2 at equilibrium. The concentration of each
substance at equilibrium is found by substituting x = 0.089 into the last line
of the ICE table.
Concentration (mol/L)
Equilibrium
SO2(g)
+
NO2(g)
(0.17 − 0.089) (0.11 − 0.089)
NO(g) + SO3(g)
0.089
0.089
Applying the rules for subtraction involving measured values:
[SO2] = 0.08 mol/L
[NO2] = 0.02 mol/L
[NO] = 0.089 mol/L
[SO3] = 0.089 mol/L
At equilibrium the concentration of SO3(g) is 0.089 mol/L.
Check Your Solution
The equilibrium expression has product concentrations in the numerator and reactant
concentrations in the denominator. Check Kc :
2
Kc = (0.089)
=5
(0.08)(0.02)
Solving the quadratic equation gave a value of x known only to two decimal places.
Subtracting the value of x from [SO2] and [NO2] gave values only good to one significant figure. The calculated value of Kc is equal to the given value, 4.8, within the
error introduced by rounding.
14. Problem
Phosgene, COCl2(g) , is an extremely toxic gas. It was used during World War I.
Today, it is used to manufacture pesticides, pharmaceuticals, dyes, and polymers. It is
prepared by mixing carbon monoxide and chlorine gas.
CO(g) + Cl2(g) ⇀
↽ COCl2(g)
0.055 mol of CO(g) and 0.072 mol of Cl2(g) are placed in a 5.0 L container.
At 870 K, the equilibrium constant is 0.20. What are the equilibrium concentrations
of the mixture at 870 K?
What Is Required?
You need to find the equilibrium concentrations of CO, Cl2 and COCl.
What Is Given?
You are given the chemical equation. You know the equilibrium constant for the
reaction, Kc = 0.20. You also know the initial amounts of CO(g) and Cl2(g) , and the
volume of the container.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Plan Your Strategy
Step 1 Calculate the initial concentrations of CO(g) and Cl2(g) .
Step 2 Set up an ICE table. Insert the equilibrium concentrations you calculated
in Step 1 in your ICE table. Let the change in molar concentrations of the
reactants be x. Use the stoichiometry of the chemical equation to write and
record expressions for the equilibrium concentrations.
Step 3 Write the equilibrium expression. Substitute the expressions for the
equilibrium concentrations into the expression. Rearrange the equilibrium
expression into the form of a quadratic equation for x. Solve the quadratic
equation for x.
Step 4 Substitute x into the equilibrium line of the ICE table to find the
equilibrium concentrations of CO, Cl2, and COCl.
Act on Your Strategy
Step 1 The volume of the reaction vessel is 5.0 L.
Therefore,
[CO] = 0.055 mol = 0.011 mol/L
5.0 L
[Cl2] = 0.072 mol = 0.0144 mol/L
5.0 L
Step 2 Set up an ICE table.
Concentration (mol/L)
Initial
Change
Equilibrium
Step 3
Cl2(g)
COCl2(g)
0.011
0.0144
0
−x
−x
+x
(0.011 − x)
(0.0144 − x)
x
CO(g)
+
Kc = [COCl2]
[CO][Cl2]
x
0.20 =
(0.011 − x)(0.0144 − x)
This equation must be rearranged into a quadratic equation
x 2 − 5.025x + 1.58 × 10−4 = 0
Recall that a quadratic equation of the form ax 2 + bx + c = 0 has the
following solution.
√
−b
±
b 2 − 4ac
x =
2a
Therefore,
√
−4
x = −(−5.025) ± 25.25 − 6.32 × 10
2.0
5.025
±
5.025
=
2.0
x = 5.02 or x = 0.0
The value x = 5.02 is not physically possible. It would result in negative
concentration of CO and Cl2 at equilibrium. The concentration of reactants
remains essentially unchanged with very little formation of COCl2(g) .
Check Your Solution
Calculate the equilibrium [COCl2] to check that it is a very small value.
[COCl2]
Kc = 0.20 =
0.011 × 0.014
[COCl2] = 3.1 × 10−5 mol/L
The equilibrium concentration of [COCl2] is a very small value.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
15. Problem
Hydrogen bromide decomposes at 700 K.
Kc = 4.2 × 10−9
2HBr(g) ⇀
↽ H2(g) + Br2(g)
0.090 mol of HBr is placed in a 2.0 L reaction vessel and heated to 700 K. What is
the equilibrium concentration of each gas?
What Is Required?
You need to find [HBr], [H2], and [Br2] at equilibrium.
What Is Given?
You are given the chemical equation. You know the initial amount of HBr, the
volume of the container, and the equilibrium constant.
Plan Your Strategy
Step 1 Calculate the initial concentration of HBr.
Step 2 Set up an ICE table. Insert the initial concentrations you calculated in Step 1
in your ICE table. Let the change in molar concentrations of HBr be x. Use
the stoichiometry of the chemical equation to write and record expressions
for the equilibrium concentrations.
Step 3 Write the equilibrium expression. Substitute the expressions for the
equilibrium concentrations into the expression. Solve the equilibrium
expression for x.
Step 4 Calculate the equilibrium concentration of each gas.
Act on Your Strategy
Step 1
[HBr] = 0.090 mol = 0.045 mol/L
2.0 L
Step 2 Set up an ICE table.
Concentration (mol/L)
Initial
Change
Equilibrium
Step 3
2HBr(g)
H2(g)
0.045
0
0
−2x
+x
+x
0.045 − 2x
x
x
+
Br2(g)
Kc = [H2][Br22]
[HBr]
x2
4.2 × 10−9 =
(0.045 − 2x)2
The right side of the equation is a perfect square.
x
±6.48 × 10−5 =
0.045 − 2x
Solving for both values,
x = +2.92 × 10−6 and x = −2.92 × 10−6
Step 4 The value x = −2.92 × 10−6 is physically impossible because it would result
in negative concentration of both H2 and Br2 .
Therefore, [H2] = [Br2] = 2.9 × 10−6 mol/L .
[HBr] = 0.045 − (2 × 2.9 × 10−6) = 0.045 mol/L
Check Your Solution
The equilibrium expression has product concentrations in the numerator and reactant
concentrations in the denominator. The equilibrium concentrations are in mol/L.
Check Kc :
2
Kc = (2.9) 2 = 4.2 × 10−9
(0.045)
This is equal to the given value of Kc .
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Solutions for Practice Problems
Student Textbook pages 349–350
16. Problem
↽ 2HI(g) , the value of Kc is 25.0 at 1100 K and
For the reaction H2(g) + I2(g) ⇀
8.0 × 102 at room temperature, 300 K. Which temperature favours the dissociation
of HI(g) into its component gases?
What Is Required?
Choose the temperature that favours the greater dissociation of HI.
What Is Given?
Kc = 25.0 at 1100 K and 8.0 × 102 at 300 K
Plan Your Strategy
HI is on the right in the chemical equation. A greater dissociation of hydrogen iodide
corresponds to a smaller value of Kc .
Act on Your Strategy
The value of Kc is smaller at 1100 K. The position of equilibrium lies more to the left
and favours the dissociation of hydrogen iodide at the higher temperature.
Check Your Solution
The question asked to choose the conditions for a smaller concentration of a product.
Kc is a fraction with product terms in the numerator, so the smallest value of Kc must
correspond to the smallest concentration of product.
17. Problem
Three reactions, and their equilibrium constants, are given below.
↽ 2NO(g)
Kc = 4.7 × 10−31
I. N2(g) + O2(g) ⇀
↽ 2NO2(g)
Kc = 1.8 × 10−6
II. 2NO(g) + O2(g) ⇀
↽ 2NO2(g)
Kc = 0.025
III. N2O4(g) ⇀
Arrange these equations in the order of their tendency to form products.
What Is Required?
Place the reactions in the order of their tendency to form products.
What Is Given?
Kc is given for each reaction.
Plan Your Strategy
The equilibrium constant, Kc , is a ratio of product concentrations divided by reactant
concentrations. The larger the value of Kc , the greater the tendency to form products.
Act on Your Strategy
The largest value of Kc is 0.025, then 1.8 × 10−6 , and finally, 4.7 × 10−31. From
greatest to least tendency to form products, the order is III, II, I.
Check Your Solution
The question asked about the tendency to form products. Kc is a ratio of
product concentrations divided by reactant concentrations. The largest value of Kc
corresponds to the greatest tendency to form products.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
18. Problem
Identify each reaction as essentially going to completion or not taking place.
Kc = 3.0 × 1011
Kc = 9.5 × 10−13
Kc = 2.2 × 1059
↽ 2NCl3(g)
(a) N2(g) + 3Cl2(g) ⇀
↽ C2H6(g) + H2(g)
(b) 2CH4(g) ⇀
↽ N2(g) + 2CO2(g)
(c) 2NO(g) + 2CO(g) ⇀
What Is Required?
For each reaction, determine whether the equilibrium mixture consists mostly of
products or reactants.
What Is Given?
Kc is given for each reaction.
Plan Your Strategy
The equilibrium constant, Kc , is a ratio of product concentrations divided by reactant
concentrations. A large value of Kc corresponds to mostly products, and a small value
of Kc indicates mostly reactants.
Act on Your Strategy
Reaction (a): Kc = 3.0 × 1011 . At equilibrium, mostly products are present and the
reaction essentially goes to completion.
Reaction (b): Kc = 9.5 × 10−13 . At equilibrium, mostly reactants are present and
there is essentially no reaction.
Reaction (c): Kc = 2.2 × 1059 . At equilibrium, mostly products are present and the
reaction goes to completion.
Check Your Solution
Kc is a ratio of product concentrations divided by reactant concentrations. Reactions
that go to completion have very large values for Kc . Reactions that do not take place
have very small values for Kc .
19. Problem
Most metal ions combine with other ions in solution. For example, in aqueous
ammonia, silver(I) ions are in equilibrium with different complex ions.
[Ag(H2O)2]+(aq) + 2NH3(aq) ⇀
↽ [Ag(NH3)2]+(aq) + 2H2O(ℓ)
At room temperature, Kc for this reaction is 1 × 107. Which of the two silver
complex ions is the more stable? Explain your reasoning.
What Is Required?
Decide which silver ion is more stable.
What Is Given?
Kc is given for the reaction.
Plan Your Strategy
The value of Kc indicates the relative concentrations of products and reactants at
equilibrium.
Act on Your Strategy
Kc for this reaction is 1 × 107. Therefore, the concentration of products is much
greater than the concentration of reactants. Consequently, [Ag(NH3)2]+(aq) must be
the more stable silver ion.
Check Your Solution
Kc is a ratio of product concentrations divided by reactant concentrations. Kc for the
reaction is a large value, therefore the concentration of silver ion on the product side
of the equation must be greater than the concentration of silver ion on the reactant
side of the equation.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
20. Problem
Consider the following reaction.
Kc = 2.4 × 1033 at 25˚C
H2(g) + Cl2(g) ⇀
↽ 2HCl(g)
HCl(g) is placed into a reaction vessel. To what extent do you expect the equilibrium
mixture to dissociate into H2(g) and Cl2(g) ?
What Is Required?
Describe the extent of dissociation of HCl at 25˚C.
What Is Given?
Kc = 2.4 × 1033 at 25˚C
Plan Your Strategy
HCl is on the right in the chemical equation. A small dissociation of HCl
corresponds to a large [HCl] and a large value of Kc .
Act on Your Strategy
The value of Kc is large. The position of equilibrium lies to the right and very little
HCl dissociates at 25˚C.
Check Your Solution
Kc is a fraction with product terms in the numerator. A large value of Kc corresponds
to large concentration of product terms. Because HCl is on the product side of the
equation, essentially no dissociation takes place.
Solutions for Practice Problems
Student Textbook page 352
21. Problem
The following equation represents the equilibrium reaction for the dissociation
of phosgene gas.
COCl2(g) ⇀
↽ CO(g) + Cl2(g)
At 100˚C, the value of Kc for this reaction is 2.2 × 10−8 . The initial concentration
of COCl2(g) in a closed container at 100˚C is 1.5 mol/L. What are the equilibrium
concentrations of CO(g) and Cl2(g) ?
What Is Required?
You need to find [CO] and [Cl2] at equilibrium.
What Is Given?
You are given the chemical equation. You know the value of Kc , and the initial concentration of COCl2(g) .
Plan Your Strategy
Step 1 Divide the initial concentration of COCl2(g) by Kc to determine whether you
can ignore the change in concentration.
Step 2 Set up an ICE table. Let x represent the change in [COCl2].
Step 3 Write the equilibrium expression. Substitute the equilibrium concentrations
into the equilibrium expression. Solve the equilibrium expression for x.
Step 4 Calculate [CO] and [Cl2] at equilibrium.
Act on Your Strategy
initial concentration =
1.5
Step 1
Kc
2.2 × 10−8
= 6.8 × 107
Because this is well above 500, the change in [COCl2] can be ignored.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Step 2
Set up an ICE table.
COCl2(g)
CO(g)
Initial
1.5
0
0
Change
−x
+x
+x
1.5 − x ≈ 1.5
x
x
Concentration (mol/L)
Equilibrium
Step 3
+
Cl2(g)
Kc = [CO][Cl2]
[COCl2]
(x)(x)
2.2 × 10−8 =
1.5
p
x = 3.3 × 10−8
= 1.8 × 10−4
Step 4 [CO] = [Cl2] = 1.8 × 10−4 mol/L
Check Your Solution
The assumption that x is negligible compared with the initial concentration
is valid because, using the rules for subtracting measured quantities,
1.5 − (1.8 × 10−4) = 1.5 . Next, check the equilibrium values:
(1.8 × 10−4)2 = 2.2 × 10−8
1.5
This is equal to the equilibrium constant.
22. Problem
Hydrogen sulfide is a poisonous gas with a characteristic, offensive odour.
It dissociates at 1400˚C, with Kc equal to 2.4 × 10−4 .
H2S(g) ⇀
↽ 2H2(g) + S2(g)
4.0 mol of H2S is placed in a 3.0 L container. What is the equilibrium concentration
of H2(g) at 1400˚C?
What Is Required?
You need to find the concentration of H2(g) at equilibrium.
What Is Given?
You are given the chemical equation. You know the value of Kc , and the initial
amount of H2(g) . The volume of the container is given.
Plan Your Strategy
Step 1 Calculate the initial molar concentration of H2S(g).
Step 2 Divide the initial concentration of H2S(g) by Kc to determine whether you
can ignore the change in concentration.
Step 3 Set up an ICE table. Let x represent the change in [H2S(g)].
Step 4 Write the equilibrium expression. Substitute the equilibrium concentrations
into the equilibrium expression. Solve the equilibrium expression for x.
Step 5 Calculate [H2] at equilibrium.
Act on Your Strategy
4.0 mol = 1.33 mol/L
Step 1 [H2S] =
3.0 L
initial concentration =
1.33
Step 2
Kc
2.4 × 10−4
= 5.5 × 103
Because this is well above 500, the change in [H2S] can be ignored.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Set up an ICE table.
Concentration (mol/L)
2H2(g)
+
S2(g)
0
0
−x
+2x
+x
1.33 − x ≈ 1.33
2x
x
Change
Step 4
⇀
1.33
Initial
Equilibrium
H2S(g)
⇀
Step 3
2
Kc = [H2] [S2]
[H2S]
2
2.4 × 10−4 = (2x) (x)
1.33
p
3
x = 7.98 × 10−5
= 4.3 × 10−2
Step 4 [H2] = 2 × (4.3 × 10−2) = 8.6 × 10−2 mol/L
Check Your Solution
The assumption that x is negligible compared with the initial concentration is valid
because, using the rules for subtracting measured quantities,
1.3 − (4.3 × 10−2) = 1.3 . Next, check the equilibrium values:
(8.6 × 10−2)2(4.3 × 10−2) = 2.4 × 10−4
1.3
This is equal to the equilibrium constant.
23. Problem
At a certain temperature, the value of Kc for the following reaction is 3.3 × 10−12.
2NCl3(g) ⇀
↽ N2(g) + 3Cl2(g)
A certain amount of nitrogen trichloride, NCl3(g) is put in a 1.0 L reaction vessel at
this temperature. At equilibrium, 4.6 × 10−4 mol of N2(g) is present. What amount
of NCl3(g) was placed in the reaction vessel?
What Is Required?
You need to find the initial amount of NCl3(g) placed in the reaction vessel.
What Is Given?
You are given the chemical equation and the value of Kc . You know the volume of the
reaction vessel. You are told the amount of N2(g) at equilibrium.
Plan Your Strategy
Step 1 Calculate the equilibrium molar concentration of N2(g) .
Step 2 Kc is very small. Assume you can ignore the change in concentration of
NCl3(g) and set up an ICE table. Let x represent the change in [N2].
Step 3 Write the equilibrium expression. Substitute the equilibrium concentrations
into the equilibrium expression. Solve the equilibrium expression for [NCl3].
Step 4 Determine the amount of NCl3(g) put in the reaction vessel.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Act on Your Strategy
Step 1 The volume of the reaction vessel is 1.0 L. Therefore, the equilibrium
[N2] = 4.6 × 10−4 mol/L.
Step 2 Set up an ICE table.
⇀
N2(g)
+
3Cl2(g)
[NCl3]
0
0
−2x
+x
+3x
[NCl3] − 2x ≈ [NCl3]
4.6 × 10−4
3x = 3(4.6 × 10−4)
Initial
Change
Equilibrium
2NCl3(g)
⇀
Concentration (mol/L)
= 1.38 × 10−3
3
Kc = [N2][Cl22]
[NCl3]
Step 3
−4
−3 3
3.3 × 10−12 = (4.6 × 10 )(1.382 × 10 )
[NCl3]
p
[NCl3] = 3.66 × 10−1
= 6.1 × 10−1 mol/L
Step 4 The reaction vessel has a volume of 1.0 L. The amount of NCl3(g) put in the
vessel must have been 6.1 × 10−1 mol.
Check Your Solution
First, check your assumption that x is negligible compared with the initial
concentration of NCl3(g) . Using the rules for subtracting measured quantities,
6.1 × 10−1 − 2(4.6 × 10−4) = 6.1 × 10−1 . Next, check the equilibrium values:
(4.6 × 10−4)(1.4 × 10−3)3 = 3.4 × 10−12
(6.1 × 10−1)2
This is equal to the equilibrium constant within mathematical rounding error.
24. Problem
At a certain temperature, the value of Kc for the following reaction is 4.2 × 10−8 .
N2(g) + O2(g) ⇀
↽ 2NO(g)
0.45 mol of N2(g) and 0.26 mol of O2(g) are put in a 6.0 L reaction vessel. What is
the equilibrium concentration of NO(g) at this temperature?
What Is Required?
You need to find the concentration of NO(g) at equilibrium.
What Is Given?
You are given the chemical equation. You know the value of Kc , and the initial
amounts of N2(g) and O2(g) . The volume of the container is given.
Plan Your Strategy
Step 1 Calculate the initial molar concentrations of N2(g) and O2(g) .
Step 2 Divide the smallest initial concentration by Kc to determine whether you can
ignore the change in concentration.
Step 3 Set up an ICE table. Let x represent the change in [N2(g)].
Step 4 Write the equilibrium expression. Substitute the equilibrium concentrations
into the equilibrium expression. Solve the equilibrium expression for x.
Step 5 Calculate [NO] at equilibrium.
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CHEMISTRY 12
Act on Your Strategy
0.45 mol = 0.075 mol/L
Step 1 [N2] =
6.0 L
0.26
mol = 0.043 mol/L
[O2] =
6.0 L
Step 2 The smallest initial concentration is [O2].
initial concentration = 0.043
Kc
4.2 × 10−8
= 1.0 × 106
Because this is well above 500, the changes in [O2] and [N2] can be ignored.
Step 3 Set up an ICE table.
Concentration (mol/L)
Initial
Change
Equilibrium
Step 4
0.075 −x
O2(g)
2NO(g)
0.075
0.043
0
−x
−x
+2x
0.043
2x
N2(g)
0.075
+
0.043 −x
2
Kc = [NO]
[N2][O2]
(2x)2
4.2 × 10−8 =
(0.075)(0.043)
4x 2
=
3.22 × 10−3
p
x = 3.39 × 10−11
= 5.8 × 10−6
Step 5 [NO] = 2 × (5.8 × 10−6) = 1.2 × 10−5 mol/L
Check Your Solution
The assumption that x is negligible compared with the initial concentration is valid
because, using the rules for subtracting measured quantities,
0.043 − (5.8 × 10−6) = 0.043. Next, check the equilibrium values:
(1.2 × 10−5)2 = 4.5 × 10−8
(0.075)(0.043)
This is equal to the equilibrium constant, within the error introduced by
mathematical rounding.
25. Problem
At a particular temperature, Kc for the decomposition of carbon dioxide gas
is 2.0 × 10−6 .
2CO2(g) ⇀
↽ 2CO(g) + O2(g)
3.0 mol of CO2 is put in a 5.0 L container. Calculate the equilibrium concentration
of each gas.
What Is Required?
You need to find the concentration of CO2(g), CO(g) , and O2(g) at equilibrium.
What Is Given?
You are given the chemical equation. You know the value of Kc , and the initial
amount of CO2(g). The volume of the container is given.
Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR
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CHEMISTRY 12
Plan Your Strategy
Step 1 Calculate the initial molar concentration of CO2(g).
Step 2 Divide the initial concentration of CO2(g) by Kc to determine whether you
can ignore the change in concentration.
Step 3 Set up an ICE table. Let x represent the change in [O2(g)].
Step 4 Write the equilibrium expression. Substitute the equilibrium concentrations
into the equilibrium expression. Solve the equilibrium expression for x.
Step 5 Calculate the concentration of each gas at equilibrium.
Act on Your Strategy
3.0 mol = 0.60 mol/L
Step 1 [CO2] =
5.0 L
Step 2
0.60
initial concentration =
Kc
2.0 × 10−6
= 3.0 × 105
Because this is well above 500, the change in [CO2] can be ignored.
Step 3 Set up an ICE table.
2CO2(g)
2CO(g)
Initial
0.60
0
0
Change
−2x
+2x
+x
0.60
2x
x
Concentration (mol/L)
Equilibrium
Step 4
0.60 − 2x
+
O2(g)
2
Kc = [CO] [O2 2]
[CO2]
2
(2x)
(x)
2.0 × 10−6 =
0.60
√
3
x = 3.0 × 10−7
= 6.7 × 10−3
Step 4 At equilibrium:
[CO2] = 0.60 mol/L
[CO] = 2(6.7 × 10−3) = 1.3 × 10−2 mol/L
[O2] = 6.7 × 10−3 mol/L
Check Your Solution
The assumption that x is negligible compared with the initial concentration is valid
because, using the rules for subtracting measured quantities,
0.60 − 2(6.7 × 10−3) = 0.59 ≈ 0.60 . Next, check the equilibrium values:
(1.3 × 10−2)2(6.7 × 10−3) = 1.9 × 10−6
0.60
This is equal to the equilibrium constant, within the error introduced by
mathematical rounding.
Solutions for Practice Problems
Student Textbook page 356
26. Problem
The following reaction takes place inside a cylinder with a moveable piston.
2NO2(g) ⇀
↽ N2O4(g)
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