CHAPTER 15 EQUILIBRIUM: THE EXTENT
OF CHEMICAL REACTIONS
CHEMICAL CONNECTIONS BOXED READING PROBLEM
B15.1
Plan: To control the pathways, the first enzyme specific for a branch is inhibited by the end product of that
branch.
Solution:
a) The enzyme that is inhibited by F is the first enzyme in that branch, which is enzyme 3.
b) Enzyme 6 is inhibited by I.
c) If F inhibited enzyme 1, then neither branch of the reaction would take place once enough F was produced.
d) If F inhibited enzyme 6, then the second branch would not take place when enough F was made.
END–OF–CHAPTER PROBLEMS
15.1
If the rate of the forward reaction exceeds the rate of reverse reaction, products are formed faster than they are
consumed. The change in reaction conditions results in more products and less reactants. A change in reaction
conditions can result from a change in concentration or a change in temperature. If concentration changes,
product concentration increases while reactant concentration decreases, but the Kc remains unchanged because the
ratio of products and reactants remains the same. If the increase in the forward rate is due to a change in
temperature, the rate of the reverse reaction also increases. The equilibrium ratio of product concentration to
reactant concentration is no longer the same. Since the rate of the forward reaction increases more than the rate of
the reverse reaction, Kc increases (numerator, [products], is larger and denominator, [reactants], is smaller).
 products 
Kc =
 reactants 
15.2
The faster the rate and greater the yield, the more useful the reaction will be to the manufacturing process.
15.3
A system at equilibrium continues to be very dynamic at the molecular level. Reactant molecules continue to form
products, but at the same rate that the products decompose to re-form the reactants.
15.4
If K is very large, the reaction goes nearly to completion. A large value of K means that the numerator is much
larger than the denominator in the K expression. A large numerator, relative to the denominator, indicates that
 products 
most of the reactants have reacted to become products. K =
 reactants 
15.5
One cannot say with certainty whether the value of K for the phosphorus plus oxygen reaction is large or small
(although it likely is large). However, it is certain that the reaction proceeds very fast.
15.6
No, the value of Q is determined by the mass action expression with arbitrary concentrations for products and
reactants. Thus, its value is not constant.
15.7
The equilibrium constant expression is K = pO2 (remember the activity of solids is 1). If the temperature remains
constant, K remains constant. If the initial amount of Li2O2 present was sufficient to reach equilibrium, the
pressure of O2 obtained will be constant, regardless of how much Li2O2(s) is present.
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15.8
a) On the graph, the concentration of HI increases at twice the rate that H2 decreases because the stoichiometric
ratio in the balanced equation is 1H2: 2HI. Q for a reaction is the ratio of concentrations of products to
concentrations of reactants. As the reaction progresses the concentration of reactants H2 and I2 decrease and the
concentration of product HI increases, which means that Q increases as a function of time.
2
pHI
H2(g) + I2(g)  2HI(g) Q =
pH 2 pI 2
The value of Q increases as a function of time until it reaches the value of K.
b) No, Q would still increase with time because the [I2] would decrease in exactly the same way as [H2] decreases.
15.9
A homogeneous equilibrium reaction exists when all the components of the reaction are in the same phase
(i.e., gas, liquid, solid, aqueous).
2NO(g) + O2(g)  2NO2(g)
A heterogeneous equilibrium reaction exists when the components of the reaction are in different phases.
Ca(HCO3)2(aq)  CaCO3(s) + H2O(l) + CO2(g)
15.10
Plan: Write the reaction and then the expression for Q
1/2N2(g) + 1/2O2(g)  NO(g)
p NO
Q(form) =
1
1
p N2 pO2
2
2
NO(g)  1/2N2(g) + 1/2O2(g)
1
Q(decomp) =
1
pN2 pO2
2
2
pNO
Q(decomp) = 1/Q(form), so the constants do differ (they are the reciprocal of each other).
15.11
Plan: Write the reaction and then the expression for Q.
Solution:
The balanced equation for the first reaction is
3/2H2(g) + 1/2N2(g)  NH3(g)
(1)
The coefficient in front of NH3 is fixed at 1 mole according to the description. The reaction quotient for this
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reaction is Q1 =
pNH3
3
1
pH2 pN2
2
2
.
In the second reaction, the coefficient in front of N2 is fixed at 1 mole.
3H2(g) + N2(g)  2NH3(g)
(2)
The reaction quotient for this reaction is Q2 =
2
pNH
3
pH3 2 pN 2
Q2 is equal to Q12.
15.12
Plan: Remember that Q =
aCc aDd
where A and B are reactants, C and D are products, and a, b, c, and d are the
aAa aBb
stoichiometric coefficients in the balanced equation.
Solution:
a) 4NO(g) + O2(g)  2N2O3(g)
Q=
pN2 2O3
4
pNO
pO2
b) SF6(g) + 2SO3(g)  3SO2F2(g)
Q=
3
pSO
2 F2
2
pSF6 pSO
3
c) 2SC1F5(g) + H2(g)  S2F10(g) + 2HCl(g)
Q=
15.13
2
pS2 F10 pHCl
2
pSClF
p
5 H2
a) 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
Q=
4
pCO
p6
2 H 2O
pC22 H6 pO7 2
b) CH4(g) + 4F2(g)  CF4(g) + 4HF(g)
Q=
4
pCF4 pHF
pCH 4 pF42
c) 2SO3(g)  2SO2(g) + O2(g)
Q=
15.14
2
pSO
p
2 O2
2
pSO
3
Plan: Remember that Q will be determined using pressures in place of activities when the species in the chemical
reaction are all gases.
Solution:
a) 2NO2Cl(g)  2NO2(g) + Cl2(g)
Q=
2
pNO
p
2 Cl2
2
pNO
2 Cl
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b) 2POCl3(g)  2PCl3(g) + O2(g)
Q=
2
pPCl
p
3 O2
2
pPOCl
3
c) 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)
Q=
15.15
pN2 2 pH6 2O
4
pNH
p3
3 O2
a) 3O2(g)  2O3(g)
Q=
pO2 3
pO3 2
b) NO(g) + O3(g)  NO2(g) + O2(g)
Q=
pNO2 pO2
pNO pO3
c) N2O(g) + 4H2(g)  2NH3(g) + H2O(g)
Q=
15.16
2
pNH
p
3 H 2O
pN2O pH4 2
Plan: Compare each equation with the reference equation to see how the direction and coefficients have changed.
If a reaction has been reversed, the K value is the reciprocal of the K value for the reference reaction. If the
coefficients have been changed by a factor n, the K value is equal to the original K value raised to the nth power.
Solution:
2
a) The K for the original reaction is K =
pH 2 pS2
pH2 2S
The given reaction 1/2S2(g) + H2(g)  H2S(g) is the reverse reaction of the original reaction and the coefficients
of the original reaction have been multiplied by a factor of 1/2. The equilibrium constant for the reverse reaction
is the reciprocal (1/K) of the original constant. The K value of the original reaction is raised to the 1/2 power.
p H 2S
K (a) = (1/K)1/2 =
pH 2 pS12/ 2
K (a) = (1/1.6x10–2)1/2 = 7.90569 = 7.9
b) The given reaction 5H2S(g)  5H2(g) + 5/2S2(g) is the original reaction multiplied by 5/2. Take the original
K to the 5/2 power to find K of given reaction.
pH5 2 pS52/ 2
K (b) = (Kc)5/2 =
pH5 2S
K (b) = (1.6x10–2)5/2 = 3.23817x10–5 = 3.2x10–5
15.17
K=
pN 2 pH2 2 O
2
pNO
pH2 2
a) K (a) = [K]
1/2
=
p1N/22 pH 2O
pNO pH2
Thus, K (a) = [K]1/2 = (6.5x102)1/2 = 25.495 = 25
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b) K = [K]–2 =
15.18
4
pNO
pH4 2
pN2 2 pH4 2 O
K = [K]–2 = (6.5x102)–2 = 2.36686x10–6 = 2.4x10–6
Plan: The activity of pure solids and pure liquids is 1, so they do not appear
in the reaction quotient expression. Remember that stoichiometric coefficients are used as exponents in the
expression for the reaction quotient.
Solution:
a) 2Na2O2(s) + 2CO2(g)  2Na2CO3(s) + O2(g)
Q=
pO2
2
pCO
2
b) H2O(l)  H2O(g)
Q = pH2 O
c) NH4Cl(s)  NH3(g) + HCl(g)
Q = pNH3 pHCl
15.19
a) H2O(l) + SO3(g)  H2SO4(aq)
Q=
 H 2SO4 
pSO3
b) 2KNO3(s)  2KNO2(s) + O2(g)
Q = pO2
c) S8(s) + 24F2(g)  8SF6(g)
Q=
15.20
8
pSF
6
pF242
Plan: The activity of pure solids and pure liquids is 1, so they do not appear
in the reaction quotient expression. Remember that stoichiometric coefficients are used as exponents in the
expression for the reaction quotient.
Solution:
a) 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g)
Q = pCO2 pH2 O
b) SnO2(s) + 2H2(g)  Sn(s) + 2H2O(g)
Q=
pH2 2O
pH2 2
c) H2SO4(l) + SO3(g)  H2S2O7(l)
Q=
15.21
1
pSO3
a) 2Al(s) + 2NaOH(aq) + 6H2O(l)  2Na[Al(OH)4](aq) + 3H2(g)
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2
 Na  Al  OH    pH3
4 
2

Q= 
 NaOH 2
b) CO2(s)  CO2(g)
Q = pCO2
c) 2N2O5(s)  4NO2(g) + O2(g)
4
Q = p NO
p
2 O2
15.22
Write balanced chemical equations for each reaction, and then write the appropriate equilibrium expression.
a) 4HCl(g) + O2(g)  2Cl2(g) + 2H2O(g)
Q=
2
pCl
p2
2 H 2O
4
pHCl
pO 2
b) 2As2O3(s) + 10F2(g)  4AsF5(l) + 3O2(g)
Q=
pO3 2
p10
F2
c) SF4(g) + 2H2O(l)  SO2(g) + 4HF(g)
Q=
4
pSO2 pHF
pSF4
d) 2MoO3(s) + 6XeF2(g)  2MoF6(l) + 6Xe(g) + 3O2(g)
Q=
15.23
6
pXe
pO3 2
6
pXeF
2
Plan: Add the two equations, canceling substances that appear on both sides of the equation. Write the Qc
expression for each of the steps and for the overall equation. Since the individual steps are added, their Qc’s are
multiplied and common terms are canceled to obtain the overall Qc.
Solution:
a) The balanced equations and corresponding reaction quotients are given below. Note the second equation must
be multiplied by 2 to get the appropriate overall equation.
2
pClF
(1) Cl2(g) + F2(g)  2ClF(g)
Q1 =
pCl2 pF2
(2) 2ClF(g) + 2F2(g)  2ClF3(g)
Overall: Cl2(g) + 3F2(g)  2ClF3(g)
Q2 =
2
pClF
3
2
pClF
pF22
Qoverall =
2
pClF
3
pCl2 pF32
b) The reaction quotient for the overall reaction, Qoverall, determined from the reaction is:
2
pClF
3
Qoverall =
pCl2 pF32
 p2

ClF 
Qoverall = Q1Q22 = 
 pCl pF 
2
2 

2
 pClF

p2
3

 = ClF3
2
 pClF
pF22  pCl2 pF32

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15.24
According to the ideal gas equation, pV = nRT. Concentration and pressure of gas are directly proportional
as long as the temperature is constant: c= n/V = p/RT.
15.25
Kc and K are related by the equation K = Kc(RT)n, where n represents the change in amount (mol) of gas in the
reaction (amount (mol) gaseous products – amount (mol) gaseous reactants). When n is zero (no change in
amount (mol) of gas), the term (RT)n equals 1 and Kc = K. When n is not zero, meaning that there is a
change in the amount (mol) of gas in the reaction, then Kc  K.
15.26
a) K = Kc(RT)n. Since n = amount (mol) gaseous products – amount (mol) gaseous reactants, n is a positive
integer for this reaction. If n is a positive integer, then (RT)n is greater than 1. Thus, Kc is multiplied by a
number that is greater than 1 to give K. Kc is smaller than K.
b) Assuming that RT > 1 (which occurs when T > 12.0 K, because 0.08314 (R) x 12.0 = 1), K > Kc if the amount
(mol) of gaseous products exceeds the amount (mol) of gaseous reactants. K < Kc when the amount (mol) of
gaseous reactants exceeds the amount (mol) of gaseous product.
Plan: ngas = moles gaseous products – moles gaseous reactants.
Solution:
a) Amount (mol) of gaseous reactants = 0; amount (mol) of gaseous products = 3; ngas = 3 – 0 = 3
b) Amount (mol) of gaseous reactants = 1; amount (mol) of gaseous products = 0; ngas = 0 – 1 = –1
c) Amount (mol) of gaseous reactants = 0; amount (mol) of gaseous products = 3; ngas = 3 – 0 = 3
15.27
15.28
a) ngas = 1
b) ngas = –3
c) ngas = 1
15.29
Plan: First, determine n for the reaction and then calculate Kc using K = Kc(RT)n.
Solution:
a) n = moles gaseous products – moles gaseous reactants = 1 – 2 = –1
K = Kc(RT)n
3.9x102
K
Kc =
=
= 3.24246 = 3.2
[(0.08314)(1000.)]1
( RT ) n
b) n = moles gaseous products – moles gaseous reactants = 1 – 1 = 0
K
28.5
Kc =
=
= 28.5
( RT ) n
[(0.08314)(500.)]0
15.30
First, determine n for the reaction and then calculate Kc using K = Kc(RT)n.
a) n = moles gaseous products – moles gaseous reactants = 2 – 2 = 0
K
49
Kc =
=
= 49
n
( RT )
[(0.08314)(730.)]0
b) n = moles gaseous products – moles gaseous reactants = 2 – 3 = –1
2.5x1010
K
Kc =
=
= 1.03925x1012 = 1.0x1012
[(0.08314)(500.)]1
( RT ) n
15.31
Plan: First, determine n for the reaction and then calculate K using K = Kc(RT)n.
Solution:
a) n = moles gaseous products – moles gaseous reactants = 2 – 1 = 1
K = Kc(RT)n = (6.1x10–3)[(0.08314)(298)]1 = 0.15113 = 0.15
b) n = moles gaseous products – moles gaseous reactants = 2 – 4 = – 2
K = Kc(RT)n = (2.4x10–3)[(0.08314)(1000.)]–2 = 3.4721x10–7 = 3.5x10–7
15.32
First, determine n for the reaction and then calculate K using K = Kc(RT)n.
a) n = moles gaseous products – moles gaseous reactants = 2 – 2 = 0
K = Kc(RT)n = (0.77)[(0.08314)(1020.)]0 = 0.77
b) n = moles gaseous products – moles gaseous reactants = 2 – 3 = –1
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K = Kc(RT)n = (1.8x10–56)[(0.08314) (570.)]–1 = 3.7983x10–58 = 3.8x10–58
15.33
When Q < K, the reaction proceeds to the right to form more products. The reaction quotient and equilibrium
constant are determined by [products]/[reactants]. For Q to increase and reach the value of K, the concentration
of products (numerator) must increase in relation to the concentration of reactants (denominator).
15.34
a) The reaction is 2D ↔ E and Kc =
[E]
[D]2
.
 0.0100 mol   1 
Concentration of D = Concentration of E = 3 spheres 

 = 0.0300 mol/L
 1 sphere   1.00 L 
[0.0300]
[E]
K=
=
= 33.3333 = 33.3
2
[D]
[0.0300]2
b) In Scene B the concentrations of D and E are both 0.0300 mol/0.500 L = 0.0600 mol/L
[0.0600]
[E]
Q=
=
= 16.66666 = 16.7
[D]2
[0.0600]2
B is not at equilibrium. Since Q < K, the reaction will proceed to the right.
In Scene C, the concentration of D is still 0.0600 mol/L and the concentration of E is 0.0600 mol/0.500 L = 0.120
mol/L
[0.120]
[E]
Q=
=
= 33.3333 = 33.3
2
[D]
[0.0600]2
Since Q = K in Scene C, the reaction is at equilibrium.
15.35
Plan: To decide if the reaction is at equilibrium, calculate Q and compare it to K. If Q = K, then the reaction is at
equilibrium. If Q > K, then the reaction proceeds to the left to produce more reactants. If Q < K, then the reaction
proceeds to the right to produce more products.
Solution:
pH 2 pBr2
(0.010)(0.010)
Q=
=
= 2.5x10–3 > K = 4.18x10–9
2
pHBr
(0.20) 2
Q > K, thus, the reaction is not at equilibrium and will proceed to the left (towards the reactants). Thus,
the numerator will decrease in size as products are consumed and the denominator will increase in size as more
reactant is produced. Q will decrease until Q = K.
15.36
Q=
15.37
n = moles gaseous products – moles gaseous reactants = 2 – 2 = 0
Since n = 0, K = Kc = 2.7 (Note: If n had any other value, we could not finish the calculation without the
temperature.)
CO 2 H 2  = 0.62 /2.00.43/2.0 = 3.662 > K = 2.7
Q=
CO H 2 O 0.13/2.00.56/2.0
2
pNO
pBr2
=
(0.10) 2 (0.10)
= 0.10 < K = 60.6
(0.10) 2
Q < K Thus, the reaction is not at equilibrium and will proceed to the right (towards the products).
2
pNOBr
Q > K Thus, the reaction is not at equilibrium and will proceed to the left (towards the reactants).
15.38
At equilibrium, equal concentrations of CFCl3 and HCl exist, regardless of starting reactant concentrations. The
equilibrium concentrations of CFCl3 and HCl would still be equal if unequal concentrations of CCl4 and HF were
used. This occurs only when the two products have the same coefficients in the balanced equation. Otherwise,
more of the product with the larger coefficient will be produced.
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15.39
When x mol of CH4 reacts, 2x mol of H2O also reacts to form x mol of CO2 and 4x mol of H2. This is based
on the 1:2:1:4 mole ratio in the reaction. The final (equilibrium) concentration or pressure of each reactant is the
initial concentration or pressure minus the amount that reacts. The final (equilibrium) concentration or pressure of
each product is the initial concentration or pressure plus the amount that forms.
15.40
a) The approximation applies when the change in concentration from initial to equilibrium is so small that it is
insignificant. This occurs when K is small and initial concentration is large.
b) This approximation will not work when the change in concentration is greater than 5%. This can occur when
[reactant]initial is very small, or when [reactant]change is relatively large due to a large K.
15.41
Plan: Since all equilibrium concentrations are given in mol/L and the reaction is balanced, construct an
equilibrium expression and substitute the equilibrium concentrations to find Kc.
Solution:
2
1.87x103 
HI 2 =


Kc =
= 50.753 = 50.8
H 2 I 2  6.50x105  1.06x103 
N 2 H 2 3 = 0.1140.3423 = 9.0077875 = 9.01
2
0.02252
NH 3 
15.42
Kc =
15.43
Plan: Calculate the initial concentration of PCl5 from the given amount (mol) and the container volume; the
reaction is proceeding to the right, consuming PCl5 and producing products. There is a 1:1:1 mole ratio between
the reactants and products.
Solution:
Initial [PCl5] = 0.15 mol/2.0 L = 0.075 mol/L
Since there is a 1:1:1 mole ratio in this reaction:
x = [PCl5] reacting (–x), and the amount of PCl3 and of Cl2 forming (+x).
Concentration (M)
PCl5(g)

PCl3(g)
+
Cl2(g)
Initial
Change
Equilibrium
15.44
0
+x
x
0
+x
x
The reaction table requires that the initial [H2] and [F2] be calculated: [H2] = 0.10 mol/0.50 L = 0.20 mol/L;
[F2] = 0.050 mol/0.50 L = 0.10 mol/L.
x = [H2] = [F2] reacting (–x); 2x = [HF] forming (+2x)
Concentration (mol/L)
H2(g)
+
F2(g)

2HF(g)
Initial
Change
Equilibrium
15.45
0.075
–x
0.075 – x
0.20
–x
0.20 – x
0.10
–x
0.10 – x
0
+2x
2x
Plan: Two of the three equilibrium pressures are known, as is K. Construct an equilibrium expression and solve
for pNOCl.
Solution:
p2
K = 6.5x104 = 2 NOCl
pNO pCl2
6.5x104 =
PNOCl =
P 2 NOCl
(0.35)2 (0.10)
6.5x10   0.35 0.10  = 28.2179 bar = 28 bar
4
2
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A high pressure for NOCl is expected because the large value of K indicates that the reaction proceeds largely to
the right, i.e., to the formation of products.
15.46
C(s) + 2H2(g)  CH4(g)
K=
pCH4
pH2 2
= 0.262
pCH4 = K  pH2 2 = (0.262)(1.22)2 = 0.38996 bar = 0.390 bar
15.47
Plan: Use the balanced equation to write an equilibrium expression and to define x. Set up a reaction table,
substitute into the K expression, and solve for x.
Solution:
NH4HS(s)  H2S(g) + NH3(g)
x = [NH4HS] reacting (–x), and the amount of H2S and of NH3 forming (+x) since there is a 1:1:1 mole
ratio between the reactant and products.
(It is not necessary to include the NH4HS as it is a solid with activity=1).
Pressure (bar)
NH4HS(s)
H2S(g)
+
NH3(g)
Initial
Change
Equilibrium



0
+x
x
K = 0.11 = ( pH2S )( pNH3 )
0
+x
x
(The solid NH4HS is not included.)
0.11 = (x)(x)
x = pNH3 = 0.33166 = 0.33 bar
15.48
2H2S(g)  2H2(g) + S2(g)
H2S = 0.45 mol/3.0 L = 0.15 mol/L
Concentration (mol/L)
2H2S(g)
Initial
Change
Equilibrium
0.15
–2x
0.15 – 2x
Kc = 9.30x10–8 =

2H2(g)
0
+2x
2x
+
S2(g)
0
+x
x
 H 2 2 S2  = 2 x 2 x 
 H 2S2
0.15  2 x 2
Assuming 0.15 mol/LM – 2 x  0.15 mol/LM
9.30x10–8 =
2 x 2 x  = 4 x 3
0.152
0.152
x = 8.0575x10–4 mol/L
H 2  = 2x = 2 (8.0575x10–4 mol/L) = 1.6115x10–3 mol/L= 1.6x10–3 mol/L
(Since (1.6x10–3)/(0.15) < 0.05, the assumption is OK.)
15.49
Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of each reactant
from the given amounts and container volume, use the balanced equation to define x and set up a reaction table,
substitute into the equilibrium expression, and solve for x, from which the concentration of NO is calculated.
Solution:
The initial concentrations of N2 and O2 are (0.20 mol/1.0 L) = 0.20 mol/L and (0.15 mol/1.0 L) = 0.15 mol/L,
respectively.
N2(g) + O2(g)  2NO(g)
There is a 1:1:2 mole ratio between reactants and products.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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15-10
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
Concentration (mol/L)
Initial
Change
Equilibrium
N2(g)
0.20
–x
0.20 – x
Kc = 4.10x10–4 =
O2(g)

0.15
–x
0.15 – x
+
2NO(g)
0
+2x
2x
(1:1:2 mole ratio)
NO2 =
2 x 2
N 2 O2  0.20  x 0.15  x 
Assume 0.20 mol/L – x  0.20 mol/L
4 x2
4.10x10–4 =
0.200.15
0.15 mol/L – x  0.15 mol/L
and
x = 1.753568x10–3 mol/L
[NO] = 2x = 2(1.753568x10–3 mol/L) = 3.507136x10–3 mol/L= 3.5x10–3 mol/L
(Since (1.8x10–3)/(0.15) < 0.05, the assumption is OK.)
15.50
2NO2(g)  2NO(g) + O2(g)
Pressure (bar)
There is a 2:2:1 mole ratio between reactants and products.

2NO2(g)
2NO(g)
+
Initial
Change
Equilibrium
0.75
0
– 2x
+2x
0.75 – 2x
2x
2
2
p
p
2x  x 
NO O 2
K = 4.48x10–13 =
=
2
2
pNO2
0.75  2x 
O2(g)
0
+x
x
(2:2:1 mole ratio)
Assume 0.75 bar – 2x  0.75 bar
4x x  = 4x 
2
4.48x10–13 =
3
0.752
0.752
x = 3.979x10–5 bar = 4.0x10–5 bar O2
(assumption is justified)
pNO = 2x = 2(3.979x10–5 bar) = 7.958x10–5 bar = 8.0x10–5 bar NO
15.51
Plan: Find the initial concentration of each reactant and product from the given amounts and container volume,
use the balanced equation to define x, and set up a reaction table. The equilibrium concentration of H 2 is known,
so x can be calculated and used to find the other equilibrium concentrations.
Solution:
Initial concentrations:
[HI] = (0.0244 mol)/(1.50 L) = 0.0162667 mol/L
[H2] = (0.00623 mol)/(1.50 L) = 0.0041533 mol/L
[I2] = (0.00414 mol)/(1.50 L) = 0.00276 mol/L
Equilibrium concentration of H2 is greater than the initial, so the reaction moves in the forward direction.
2 HI(g) H2(g) + I2(g)
There is a 2:1:1 mole ratio between reactants and products.
Concentration (mol/L)
2 HI(g)

H2(g)
+
I2(g)
Initial
0.0162667
0.0041533
0.00276
Change
–2x
+x
+x
(2:1:1 mole ratio)
Equilibrium
0.0162667 – 2x
0.0041533 + x 0.00276 + x
[H2]eq = 0.00467 = 0.0041533 + x
x = 0.0005167 mol/L
[I2]eq = 0.00276 + x = 0.00276 + 0.0005167 = 0.0032767 mol/L= 0.00328 mol/L I2
[HI]eq = 0.0162667 – 2x = 0.0162667 – 2(0.0005167) = 0.0152333 mol/L= 0.0152 mol/L HI
15.52
Initial concentrations:
[A] = (1.75x10–3 mol)/(1.00 L) = 1.75x10–3 mol/L
[B] = (1.25x10–3 mol)/(1.00 L) = 1.25x10–3 mol/L
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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15-11
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
[C] = (6.50x10–4 mol)/(1.00 L) = 6.50x10–4 mol/L
Concentration (mol/L)
A(g)

2B(g)
–3
Initial
1.75x10
Change
–x
Equilibrium
1.75x10–3 – x
[A]eq = 2.15x10–3 = 1.75x10–3 – x
x = –0.00040
[B]eq = 1.25x10–3 + 2x = 4.5x10–4 mol/L
[C]eq = 6.50x10–4 + x = 2.5x10–4 mol/L
15.53
+
–3
1.25x10
+ 2x
1.25x10–3 + 2x
C(g)
6.50x10–4
+x
6.50x10–4 + x
Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of ICl from
the given amount and container volume, use the balanced equation to define x and set up a reaction table,
substitute into the equilibrium expression, and solve for x, from which the equilibrium concentrations can be
calculated.
Solution:
[ICl]init = (0.500 mol/5.00 L) = 0.100 mol/L
Concentration (mol/L)
2ICl(g)

I2(g)
+
Cl2(g)
Initial
0.100
0
0
Change
–2x
+x
+x
(2:1:1 mole ratio)
Equilibrium
0.100 – 2x
x
x
I2 Cl2  =  x  x 
Kc = 0.110 =
 0.100  2 x 2
ICl2
 x 2
0.110 =
Take the square root of each side:
 0.100  2 x 2
x
0.331662 =
 0.100  2 x 
x = 0.0331662 – 0.663324x
1.663324x = 0.0331662
x = 0.0199397
[I2]eq = [Cl2]eq = x = 0.0199397 mol/L= 0.0200 mol/L
[ICl]eq = 0.100 – 2x = 0.100 – 2(0.0199397) = 0.0601206 mol/L= 0.060 mol/L ICl
15.54
Concentration (mol/L)
Initial
Change
Equilibrium
SCl2(g)
+
0.675
–x
0.675 – x
2C2H4(g)
 S(CH2CH2Cl)2(g)
0.973
– 2x
0.973 – 2x
0
+x
x
[S(CH2CH2Cl)2]eq = x = 0.350 mol/L
[SCl2]eq = 0.675 – x = 0.675 – 0.350 = 0.325 mol/L
[C2H4]eq = 0.973 – 2x = 0.973 – 2(0.350) = 0.273 mol/L
 0.350
S(CH 2 CH 2Cl) 2  =
Kc =
= 14.4497
2
 0.325 0.2732
SCl2 C2 H 4 
K= Kc(RT)n
n = 1 mol – 3 mol = –2
K = 14.4497   0.08314  273.2  20.0 
15.55
2
= 0.0237617 = 0.0238
Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of each reactant
from the given amounts and container volume, use the balanced equation to define x, and set up a reaction table.
The equilibrium concentration of N2 is known, so x can be calculated and used to find the other equilibrium
concentrations. Substitute the equilibrium concentrations into the equilibrium expression to find Kc.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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15-12
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
Solution:
4NH3(g) + 3O2(g)  2N2(g) + 6H2O(g)
Initial [NH3] = Initial [O2] = (0.0150 mol)/(1.00 L) = 0.0150 mol/L
Concentration (mol/L)
4NH3(g)
+
3O2(g)

Initial
0.0150
0.0150
Change
–4x
–3x
Equilibrium
0.0150 – 4x
0.0150 – 3x
[N2] eq = 2x = 1.96x10–3 mol/L
x = (1.96x10–3 mol/L)/2 = 9.80x10–4 mol/L
[H2O]eq = 6x = 6(9.80x10–4) = 5.8800x10–3 mol/L
[NH3]eq = 0.0150 – 4x = 0.0150 – 4(9.80x10–4 ) = 1.1080x10–2 mol/L
[O2]eq = 0.0150 – 3x = 0.0150 – 3(9.80x10–4 ) = 1.2060x10–2 mol/L
2N2(g)
+
6H2O(g)
0
+2x
+2x
0
+6x
+6x
2
6
1.96x10  5.8800x10  = 6.005859x10–6 = 6.01x10–6
N 2  H 2 O 

Kc =
=
4
3
4
3
NH3  O2 
1.1080x102  1.2060x102 
3
15.56
Pressure (bar)
FeO(s)
+
2
CO(g)
3

6
Fe(s)
Initial
—
1.00
—
Change
–x
Equilibrium
1.00 – x
pCO2
x
K=
= 0.403 =
1.00  x
pCO
x = 0.28724 = 0.287 bar CO2
1.00 – x = 1.00 – 0.28724 = 0.71276 bar = 0.71 bar CO
+
CO2(g)
0
+x
x
15.57
A change in equilibrium conditions such as a change in concentration of a component, a change in pressure
(volume), or a change in temperature.
15.58
Equilibrium position refers to the specific concentrations or pressures of reactants and products that exist
at equilibrium, whereas equilibrium constant refers to the overall ratio of equilibrium concentrations and not to
specific concentrations. Changes in reactant concentration cause changes in the specific equilibrium
concentrations of reactants and products (equilibrium position), but not in the equilibrium constant.
15.59
A positive rH indicates that the reaction is endothermic, and that heat is consumed in the reaction:
NH4Cl(s) + heat  NH3(g) + HCl(g)
a) The addition of heat (high temperature) causes the reaction to proceed to the right to counterbalance the effect
of the added heat. Therefore, more products form at a higher temperature and container (B) with the largest
number of product molecules best represents the mixture.
b) When heat is removed (low temperature), the reaction shifts to the left to produce heat to offset that
disturbance. Therefore, NH3 and HCl molecules combine to form more reactant and container (A) with the
smallest number of product gas molecules best represents the mixture.
15.60
Equilibrium component concentration values may change but the mass action expression of these concentrations
is a constant as long as temperature remains constant. Changes in component amounts, pressures (volumes), or
addition of a catalyst will not change the value of the equilibrium constant.
15.61
a) Ratef = kf[reactants]x. An increase in reactant concentration shifts the equilibrium to the right by increasing
the initial forward rate. Since Keq = kf /kr and kf and kr are not changed by changes in concentration, Keq remains
constant.
b) A decrease in volume causes an increase in concentrations of gases. The reaction rate for the formation of
fewer moles of gases is increased to a greater extent. Again, the kf and kr values are unchanged.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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15-13
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
c) An increase in temperature increases kr to a greater extent for an exothermic reaction and thus lowers the Keq
value.
d) An endothermic reaction can be written as: reactants + heat  products. A rise in temperature (increase in
heat) favors the forward direction of the reaction, i.e., the formation of products and consumption of reactants.
Since K = [products]/[reactants], the addition of heat increases the numerator and decreases the denominator,
making K2
larger than K1.
15.62
XY(s) X(g) + Y(s) Since product Y is a solid substance, addition of solid Y has no effect on the
equilibrium position (as long as some Y is present). Scene A best represents the system at equilibrium after the
addition of two formula units of Y. More Y is present but the amounts of X and XY do not change.
15.63
Plan: If the concentration of a substance in the reaction increases, the equilibrium position will shift to consume
some of it. If the concentration of a substance in the reaction decreases, the equilibrium position will shift to
produce more of it.
Solution:
a) Equilibrium position shifts towards products. Adding a reactant (CO) causes production of more products as
the system will act to reduce the increase in reactant by proceeding toward the product side, thereby consuming
additional CO.
b) Equilibrium position shifts towards products. Removing a product (CO2) causes production of more products
as the system acts to replace the removed product.
c) Equilibrium position does not shift. The amount of a solid reactant or product does not impact the equilibrium
as long as there is some solid present.
d) Equilibrium position shifts towards reactants. When product is added, the system will act to reduce the
increase in product by proceeding toward the reactant side, thereby consuming additional CO 2; dry ice is solid
carbon dioxide that sublimes to carbon dioxide gas. At very low temperatures, CO2 solid will not sublime, but
since the reaction lists carbon dioxide as a gas, the assumption that sublimation takes place is reasonable.
15.64
a) no change
c) shifts towards the products
15.65
Plan: An increase in container volume results in a decrease in pressure (Boyle’s law). Le Châtelier’s principle
states that the equilibrium will shift in the direction that forms more moles of gas to offset the decrease in
pressure.
Solution:
a) More F forms (two moles of gas) and less F2 (one mole of gas) is present as the reaction shifts towards the
right.
b) More C2H2 and H2 form (four moles of gas) and less CH4 (two moles of gas) is present as the reaction shifts
towards the right.
15.66
a) less CH3OH(l); more CH3OH(g)
b) less CH4 and NH3; more HCN and H2
15.67
Plan: Decreasing container volume increases the pressure (Boyle’s law). Le Châtelier’s principle states that the
equilibrium will shift in the direction that forms a smaller amount (mol) of gas to offset the increase in pressure.
Solution:
a) There are two moles of reactant gas (H2 and Cl2) and two moles of product gas (HCl). Since there is the same
amount (mol) of reactant and product gas , there is no effect on the amounts of reactants or products.
b) There are three moles of reactant gases (H2 and O2) and zero moles of product gas. The reaction will shift to the
right to produce a smaller amount (mol) of gas to offset the increase in pressure. H2 and O2 will decrease from
their initial values before the volume was changed. More H2O will form because of the shift in equilibrium
position.
15.68
a) more CO2 and H2O; less C3H8 and O2
b) more NH3 and O2; less N2 and H2O
b) no change
d) shifts towards the reactants
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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15-14
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
15.69
Plan: The purpose of adjusting the volume is to cause a shift in equilibrium to the right for increased product
yield. Increasing the volume of the container results in a shift in the direction that forms a larger amount (mol) of
gas, while decreasing the container volume results in a shift in the direction that forms a smaller amount (mol) of
gas.
Solution:
a) Because the amount (mol) of reactant gas (4H2) equals the amount (mol) of product gas (4H2O), a change in
volume will have no effect on the yield.
b) The moles of gaseous product (2CO) exceed the moles of gaseous reactant (1O2). A decrease in pressure favors
the reaction direction that forms more moles of gas, so increase the reaction vessel volume.
15.70
a) increase volume
15.71
Plan: An increase in temperature (addition of heat) causes a shift in the equilibrium away from the side of
b) decrease volume
the reaction with heat. Recall that a negative value of  r H  indicates an exothermic reaction, while a positive
value of  r H  indicates an endothermic reaction.
Solution:
a) CO(g) + 2H2(g)  CH3OH(g) + heat
 r H  = –90.7 kJ/mol
The reaction is exothermic, so heat is written as a product. The equilibrium shifts to the left, away from heat,
towards the reactants, so amount of product decreases.
b) C(s) + H2O(g) + heat  CO(g) + H2(g)
 r H  = 131 kJ/mol
The reaction is endothermic, so heat is written as a reactant. The equilibrium shifts to the right, away from heat,
towards the products, so amounts of products increase.
c) 2NO2(g) + heat  2NO(g) + O2(g)
The reaction is endothermic, so heat is written as a reactant. The equilibrium shifts to the right, away from heat,
towards the product, so amounts of products increase.
d) 2C(s) + O2(g)  2CO(g) + heat
The reaction is exothermic, so heat is written as a product. The equilibrium shifts to the left, away from heat,
towards the reactants; amount of product decreases.
15.72
a) increase
b) decrease
c) decrease
d) increase
15.73
Plan: The van’t Hoff equation shows how the equilibrium constant is affected by a change in temperature.
Substitute the given variables into the equation and solve for K2.
Solution:
K298 = K1 = 1.80
T1 = 298 K
K500 = K2 = ?
T2 = 500. K
R = 8.314 J/mol•K
3
 10 J 
 0.32 kJ 
r H  = 
 2 DH  
 = 6.4x102 J/mol

1
kJ
 1 mol DH 


ln
K2
 H  1
1
=  r



K1
R  T2
T1 
ln
K2
6.4x10 2 J / mol  1
1 
= 



1.80
8.314 J/mol•K  500. K
298 K 
K2
= 0.104360
1.80
K2
= 1.110
1.80
K2 = (1.80)(1.110) = 1.998 = 2.0
ln
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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15-15
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
15.74
The van’t Hoff equation shows how the equilibrium constant is affected by a change in temperature. Substitute
the given variables into the equation and solve for K2.
K298 = K1 = 2.25x104
K0 = K2 = ?
T1 = 298 K
T2 = (273 + 0.) = 273 K
 r H  = (–128 kJ/mol)(103 J/1 kJ) = –1.28x105 J/mol
 r H  = –128 kJ/mol
R = 8.314 J/mol•K
ln
K2
 H  1
1
=  r



K1
R  T2
T1 
ln
1.28x105 J / mol  1
1 
K2
= 



4
8.314 J/ mol•K  273 K 298 K 
2.25x10
K2
= 4.731088
2.25x104
K2
= 1.134189x102
2.25x104
K2 = (2.25x104)(1.134189x102) = 2.551925x106 = 2.55x106
ln
15.75
4Fe3O4(s) + O2(g)  6Fe2O3(s)
a) K =
K = 2.5x1087 at 298 K
1
= 2.5x1087
pO2
pO2 = 4.0x10–88 bar
b) Q =
1
= 1/(0.21) = 4.7619
PO 2
K> Q thus, the reaction will proceed to the right.
c) K = Kc(RT)n
Kc = K/(RT)n
n = 0 – 1 = –1
Kc = (2.5x1087)/[(0.08314)(298)]–1 = 6.19393x1088 = 6.2x1088
15.76
Plan: An increase in temperature (addition of heat) causes a shift in the equilibrium away from the side of the
reaction with heat, while a decrease in temperature (removal of heat) causes a shift in the equilibrium towards the
side with heat. Increasing the volume of the container (pressure decreases) results in a shift in the direction that
forms a larger amount (mol) of gas, while decreasing the container volume (pressure increases) results in a shift in
the direction that forms a smaller amount (mol) of gas. Adding a reactant causes a shift in the direction of
products.
Solution:
a) SO2(g) + 1/2O2(g)  SO3(g) + heat
The forward reaction is exothermic (  r H  is negative), so it is favored by lower temperatures. Lower
temperatures will cause a shift to the right, the heat side of the reaction. The amount (mol) of gas as products
(1SO3) is smaller than as reactants (1SO2(g) + 1/2O2), so products are favored by higher pressure. High
pressure will cause a shift in equilibrium to the side with the smaller amount (mol) of gas.
pSO3
b) Addition of O2 would decrease Q since Q =
, and have no impact on K.
1/2
pSO2 pO2
c) To enhance yield of SO3, a low temperature is used. Reaction rates are slower at lower temperatures, so a
catalyst is used to speed up the reaction.
15.77
3H2(g) + N2(g)  2NH3(g)
pNH3 = (41.49%/100%)(110. bar) = 45.639 bar
100.00% – 41.49% = 58.51% N2 + H2
pH2 + pN2 = (58.51%/100%)(110. bar) = 64.361 bar
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
Page
15-16
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
pH2 = (3/4)(64.361 bar) = 48.27075 bar
pN2 = (1/4)(64.361 bar) = 16.09025 bar
p  =
 45.639 
K=
= 1.15095x10 = 1.15x10
48.27075
16.09025




p  p 
2
2
NH3
–3
3
H2
15.78
–3
3
N2
a) 3H2(g) + N2(g)  2NH3(g)
The mole ratio H2:N2 = 3:1; at equilibrium, if N2 = x, H2 = 3x;
pNH3 = 50. bar
 p  = 1.00x10
K=
 p  p 
2
NH3
–4
3
N2
H2
 50. = 1.00x10–4
 x  3x 3
2
K=
x = 31.02016 = 31 bar N2
3x = 3(31.02016) = 93.06049 = 93 bar H2
ptotal = pnitrogen + phydrogen + pammonia = (31.02016 bar) + (93.06049 bar) + (50. bar)
= 174.08065 bar= 174 bar total
b) The mole ratio H2:N2 = 6:1; at equilibrium, if N2 = x, H2 = 6x; pNH3 = 50. bar
K=
 50.2 = 1.00x10–4
 x  6x 3
x = 18.445 = 18 bar N2
6x = 6(18.445) = 110.67 = 111 bar H2
ptotal = pnitrogen + phydrogen + pammonia = (18.445 bar) + (110.67 bar) + (50. bar)
= 179.115 bar= 179 bar total
This is not a valid argument. The total pressure in b) is greater than in a) to produce the same amount of NH 3.
15.79
a) More CaCO3. Because the forward reaction is exothermic, decreasing the temperature will cause an increase
in the amount of CaCO3 formed as the reaction shifts to the right to produce more heat.
b) Less CaCO3. The only gas in the equation is a reactant. Increasing the volume (decreasing the pressure) will
cause the equilibrium to shift toward the reactant side and the amount of CaCO 3 formed decreases.
c) More CaCO3. Increasing the partial pressure of CO2 will cause more CaCO3 to be formed as the reaction
shifts to the right to consume the added CO2.
d) No change. Removing half of the initial CaCO3 will have no effect on the amount of CaCO3 formed, because
CaCO3 is a solid with an activity of 1.
15.80
a) Q =
2
pXY
; since n=0, we can use either concentration or pressure terms.
pX 2 pY2
b)
Scene A: Q =
 02 = 0
0.40.4
0.42 = 4
Scene B: Q =
0.20.2
0.62 = 36 = 4x101
Scenes C–E: Q =
 0.1 0.1
c) Time is progressing to the right. Frame A must be the earliest time.
d) K = 4x101
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
Page
15-17
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
e) Scene B. At higher temperatures, the reaction shifts to the left (forming more X2 and Y2).
f) None. Volume (pressure) has no effect on the position of the equilibrium since there are two moles of gas on
each side.
15.81
Plan: Use the balanced equation to write an equilibrium expression and to define x. Set up a reaction table,
substitute into the Kc expression, and solve for x. Once the total concentration of the gases at equilibrium is
known, the pressure can be found with pV = nRT.
Solution:
Concentration (mol/L)
NH2COONH4(s)
 2NH3(g)
+
CO2(g)
Initial
7.80 g
0
0
Change
—
+2x
+x
Equilibrium
—
2x
x
The solid has an activity of 1 and, as long as some is present, is not included in the Kc expression.
Kc = [NH3]2[CO2]
Kc = 1.58x10–8 = (2x)2(x)
x = 1.580759x10–3 mol/L
Total concentration of gases = 2x + x = 2(1.580759x10–3 mol/L) + 1.580759x10–3 mol/L = 4.742277x10–3 mol/L
To find total pressure use the ideal gas equation: PV = nRT
nRT
n
p=
=   RT = cRT
V
V 
P = (4.742277x10–3 mol/L)(0.08314 L•bar/mol•K)(273 + 250.)K = 0.206205 bar = 0.206 bar
15.82
a)
(1)
2Ni3S2(s) + 7O2(g)  6NiO(s) + 4SO2(g)
(2)
6NiO(s) + 6H2(g)  6Ni(s) + 6H2O(g)
(3)
6Ni(s) + 24CO(g)  6Ni(CO)4(g)
Overall: 2Ni3S2(s) + 7O2(g) + 6H2(g) + 24CO(g)  4SO2(g) + 6H2O(g) + 6Ni(CO)4(g)
b) As always, the solid is not included in the Q expression.
SO2 4  H 2 O6  Ni  CO 4 
Qc(overall) =
 O2 7  H 2 6  CO 24
6
4
6
 Ni  CO  
SO 2 
H 2 O


4

Q1 · Q2 · Q3 =


7
6
24
O 2 
H2 
 CO
15.83
6
SO2 4  H 2 O6  Ni  CO 4 
=
 O2 7  H 2 6  CO 24
6
a) Since the volume is 1.00 L, the concentration (mol/L) equals the amount (mol) present.
Concentration (mol/L)
2NH3(g)

N2(g)
+
3H2(g)
Initial
0
1.30
1.65
Change
+2x
–x
–3x
Equilibrium
2x = 0.100 mol/L
1.30 – x
1.65 – 3x
x = 0.0500 mol
[N2]eq = (1.30 – 0.0500) mol/L = 1.25 mol/L N2
[H2]eq = [1.65 – 3(0.0500)] mol/L = 1.50 mol/L H2
 N2  H2 3
Kc =
2
NH3 
1
3
=
(1.25)(1.50)
(0.100)2
1
3
= 421.875 = 422
3
 N 2  2  H 2  2 = (1.50) 2 (1.25) 2 = 20.523177 = 20.5
b) K =
c
8.34x102
NH 3 
c) Kc in a) is the square of Kc in b). The balanced equations are different; therefore, the values of Kc are different.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
Page
15-18
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
15.84
Plan: Write the equilibrium expression. You are given a value of Kc but the amounts of reactant and product are
given in units of pressure. Convert Kc to K and use the equilibrium pressures of C2H5OH and H2O to obtain the
equilibrium pressure of C2H4. An increase in temperature (addition of heat) causes a shift in the equilibrium away
from the side of the reaction with heat, while a decrease in temperature (removal of heat) causes a shift in the
equilibrium towards the side with heat. Increasing the volume of the container (pressure decreases) results in a
shift in the direction that forms more moles of gas, while decreasing the container volume (pressure increases)
results in a shift in the direction that forms fewer moles of gas. The van’t Hoff equation shows how the
equilibrium constant is affected by a change in temperature. Substitute the given variables into the equation and
solve for K at 450. K.
Solution:
a) K = Kc(RT)n
n = amount (mol) gaseous products – amount (mol) gaseous reactants = 1 – 2 = –1 (one mol of product,
C2H5OH, and two mol of reactants, C2H4 + H2O)
K = Kc(RT)–1 = (9x103)[(0.08314 L•bar/mol•K)(600. K)]–1 = 1.8042x102
Substitute the given values into the equilibrium expression and solve for pC2H4 .
K=
pC2 H5OH
pC2 H4 pH 2 O
=
200.
= 1.8042x102
PC2 H 4  400.
PC2 H4 = 2.7713x10–3 = 3x10–3 bar
b) Since  r H  is negative, the reaction is exothermic and heat is written as a product. To shift the reaction
towards the right to yield more ethanol, heat must be removed. A low temperature favors an exothermic
reaction. The forward direction, towards the production of ethanol, produces the smaller amount (mol) of gas and
is favored by high pressure.
 103 J 
c) K1 = 9x103
T1 = 600. K
 r H  =  47.8 kJ/mol  
 = –4.78x104 J/mol
 1 kJ 
K2 = ?
T2 = 450. K
R = 8.314 J/mol•K
K2
r H   1
1
ln
= 



K1
R  T2
T1 
ln
4.78x104 J / mol  1
1 
K2
= 



3
8.314 J/mol•K  450. K
600. K 
9x10
K2
= 3.1940769
9x103
K2
= 24.38765
9x103
K2 = (9x103)(24.38765) = 2.1949x105 = 2x105
d) No, condensing the C2H5OH would not increase the yield. Ethanol has a lower boiling point (78.5°C) than
water (100°C). Decreasing the temperature to condense the ethanol would also condense the water, so moles of
gas from each side of the reaction are removed. The direction of equilibrium (yield) is unaffected when there is no
net change in the amount (mol) of gas.
ln
15.85
n/V = c = p/RT =
Initial
Change
Equil:
 2.0 bar 
= 0.08067 mol/L each gas
L•bar 

 0.08314 mol•K    273.2  25.0  K 


H2(g)
+
CO2(g)
 H2O(g)
+
CO(g)
0.08067
–x
0.08067 – x
0.08067
–x
0.08067 – x
0
+x
x
0
+x
x
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
Page
15-19
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
Kc =
 H 2 O  CO = 0.534 =
(x)(x)
(x) 2
=
(0.08067  x)(0.08067  x)
 H 2 CO2 
(0.08067  x)2
(0.534)1/2 = 0.730753 =
15.86
(x)
(0.08067  x)
x = 0.03406 mol/L
c of H2 at equilibrium = 0.08067 – x = 0.08067– 0.03406 = 0.0466098 mol/L
 0.0466098 mol   2.016 g 
Mass (g) of H2 = 1.00 L  
  1 mol  = 0.093965 = 0.094 g H2
L



To get the two equations to sum to the desired equation, the first equation must be reversed and doubled. This will
result in squaring the reciprocal of its Kc value. The other equation does not need to be changed. Adding the two
equations means the new Kc value will be the product of the individual Kc values.
2NO(g)  N2(g) + O2(g)
K1 = (Kc)–2 = 4.340x1018
2NO2(g)  2NO(g) + O2(g)
Overall: 2NO2(g)  N2(g) + 2O2(g)
15.87
K2 = Kc = 1.1x10–5
Kc (overall) = K1K2 = 4.774x1013 = 4.8x1013
Plan: Write the equilibrium expression. You are given a value of Kc but the amounts of reactants and product
are given in units of pressure. Convert Kc to K and use the equilibrium pressures of SO3 and O2 to obtain the
equilibrium pressure of SO2. For part b), set up a reaction table and solve for x. The equilibrium concentrations
can then be used to find the Kc value at the higher temperature. The concentration of SO2 is converted to pressure
using the ideal gas law, pV = nRT.
Solution:
a) K = Kc(RT)n
n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 2 – 3 = –1 (two mol of product,
SO3, and three mol of reactants, 2 SO2 + O2)
K = Kc(RT)n = Kc(RT)–1 = (1.7x108)[(0.08314 L•bar/mol•K)(600. K)]–1 = 3.4079x106
2
2
pSO
300.

3
K= 2
= 2
= 3.4079x106
pSO2 pO2
pSO2 100.
pSO2 = 0.016251 = 0.016 bar
b) Create a reaction table that describes the reaction conditions. Since the volume is 1.0 L, the amount (mol)
equals the concentration (mol/L). Note the 2:1:2 mole ratio between SO2:O2:SO3.
Concentration (mol/L)
2SO2(g)
+
O2(g)

2SO3(g)
Initial
0.0040
0.0028
0
Change
–2x
–x
+2x
(2:1:2 mole ratio)
Equilibrium
0.0040 – 2x
0.0028 – x
2x = 0.0020 (given)
x = 0.0010, therefore:
[SO2] = 0.0040 – 2x = 0.0040 – 2(0.0010) = 0.0020 mol/L
[O2] = 0.0028 – x = 0.0028 – 0.0010 = 0.0018 mol/L
[SO3] = 2(0.0010) = 0.0020 mol/L
Substitute equilibrium concentrations into the equilibrium expression and solve for Kc.
2
Kc =
SO3 
SO2 2 O 2 
=
(0.0020) 2
(0.0020) 2 (0.0018)
= 555.5556 = 5.6x102
The pressure of SO2 is estimated using the concentration of SO2 and the ideal gas law (although the ideal gas law
is not well behaved at high pressures and temperatures).
pV = nRT
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
Page
15-20
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
pSO2 =
15.88
nRT
=
V
 0.0020 mol   0.08314

L•bar 
1000. K 
mol•K 
1.0 L 
= 0.1663 = 0.17 bar
The original concentrations are: (0.350 mol/0.500 L) = 0.700 mol/L for CO and Cl 2.
Concentration (mol/L)
CO(g) +
Cl2(g)

COCl2(g)
Initial
0.700
0.700
0
Change
–x
–x
+x
Equilibrium
0.700 – x
0.700 – x
x
COCl
x
x

 
 
2
Qc =
=
=
= 4.95
CO Cl2  0.700  x 0.700  x  0.490  1.400x  x 2 
4.95x2 – 7.93x + 2.4255 = 0
a = 4.95
b = – 7.93
c = 2.4255
x=
x=
b 
b 2  4ac
2a
( 7.93) 
 7.932  4  4.95  2.4255 
2  4.95
x = 1.19039 or 0.41162959
(The 1.19039 value is not possible because 0.700 – x would be negative.)
[CO] = [Cl2] =0.700 – x = 0.700 – 0.41162959 = 0.28837041 = 0.288 mol/L
[COCl2] = x = 0.41162959 = 0.412 mol/L
15.89
Plan: Set up a reaction table to find the equilibrium amount of CaCO3 after the first equilibrium is established and
then the equilibrium amount after the second equilibrium is established.
Solution:
The equilibrium pressure of CO2 = pCO2 = 0.220 bar.
CaCO3(s)

CaO(s)
+
CO2(g)
Initial
0.100 mol
0.100 mol
0
Change
–x
–x
+x
Equilibrium
0.100 – x
0.100 – x
x = 0.220 bar (given)
The amount of calcium carbonate solid in the container at the first equilibrium equals the original amount,
0.100 mol, minus the amount reacted to form 0.220 bar of carbon dioxide. The amount (mol) of CaCO 3 reacted is
equal to the amount (mol) of carbon dioxide produced. Use the pressure of CO 2 and the ideal gas equation to
calculate the amount (mol) of CO2 produced:
pV = nRT
pV
Amount (mol) of CO2 = n =
RT
0.220 bar 10.0 L 

n=
= 0.068731 mol CO2
L•bar 

 0.08314 mol•K   385 K 


Amount (mol) of CaCO3 reacted = amount (mol) of CO2 produced = 0.068731 mol
Amount (mol) of CaCO3 remaining = initial amount (mol) amount (mol) reacted = 0.100 mol CaCO3 – 0.068731
mol CaCO3
= 0.0313 mol CaCO3 at first equilibrium
As more carbon dioxide gas is added, the system returns to equilibrium by shifting to the left to convert the added
carbon dioxide to calcium carbonate to maintain the partial pressure of carbon dioxide at 0.220 bar (K). Convert
the added 0.300 bar of CO2 to moles using the ideal gas equation. The amount (mol) of CO2 reacted equals the
amount (mol) of CaCO3 formed.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
Page
15-21
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
Amount (mol) of CO2 = n =
 0.300 bar 10.0 L 
n=
pV
RT
= 0.09372 mol CO2
L•bar 

0.08314
385
K



mol•K 

Amount (mol) of CaCO3 produced = amount (mol) of CO2 reacted = 0.09372 mol CaCO3
Add the amount (mol) of CaCO3 formed in the second equilibrium to the amount (mol) of CaCO3 at the first
equilibrium position.
Amount (mol) of CaCO3 = amount (mol) at first equilibrium + amount (mol) formed in second equilibrium
= 0.0313 mol + 0.09372 = 0.12502mol CaCO3
 100.09 g CaCO3 
Mass (g) of CaCO3 =  0.12502 mol CaCO3  
 = 12.514 = 12.5 g CaCO3
 1 mol CaCO3 
15.90
a) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)
b) 4NO2(g) + 6H2O(g)  4NH3(g) + 7O2(g)
15.91
C2H2(g) + H2(g)  C2H4(g)
 r H  = m  products H  – n  reactants H 
= {1  f H  [C2H4(g)]} – {1  f H  [C2H2(g)] + 1  f H  [H2(g)]}
= [(52.47 kJ/mol)] – [(227 kJ/mol) + (0.0 kJ/mol)]
= –174.53 kJ/mol
K 300
 H  1
1
ln
=  r



K 2000
R  T2
T1 
ln
K 300
= 
8
2.9x10
ln
 174.53 kJ /mol  
  103 J 
1
1




J   300. K
2000. K   1 kJ 

 8.314 mol•K 


K 300
= 59.478189
2.9x108
K 300
= 6.77719x1025
2.9x108
K300 = (2.9x108)(6.77719x1025) = 1.9654x1034 = 2.0x1034
2
MN 

Kc =
 M2 [N2 ]
15.92
M2(g) + N2(g) 2MN(g)

Scene A: Concentrations: [M2] = [N2] = 0.20 mol/L; [MN] = 0.40 mol/L
Kc =

 0.402 = 4.0
0.20[0.20]
Scene B:
Concentration (mol/L)
Initial
Change
Equilibrium
Kc = 4.0 =
M2(g)
0.60
–x
0.60 – x
+
N2(g)
0.30
–x
0.30 – x

2MN(g)
0
+2x
2x
 2x 2
0.60  x [0.30  x]
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
Page
15-22
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
4x 2
4.0 =
0.18  0.90 x + x 2
4x = 0.72 – 3.6x + 4x2
3.6x = 0.72
x = 0.20 mol/L
[M2] = 0.60 – x = 0.60 – 0.20 = 0.40 mol/L
[N2] = 0.30 – x = 0.30 – 0.20 = 0.10 mol/L
[MN] = 2x = 2(0.20 mol/L) = 0.40 mol/L
2
15.93
Plan: Use the balanced reaction to write the equilibrium expression. The equilibrium concentration of S 2F10 is
used to write an expression for the equilibrium concentrations of SF4 and SF6.
Solution:
S2F10(g)  SF4(g) + SF6(g)
The reaction is described by the following equilibrium expression:
SF4 SF6 
Kc =
S2 F10 
At the first equilibrium, [S2F10] = 0.50 mol/L and [SF4] = [SF6] = x ([SF4]:[SF6] = 1:1).
SF4 SF6  = (x)(x)
Kc =
(0.50)
S2 F10 
x2 = 0.50Kc
[SF4] = [SF6] = x =
0.50K c
At the second equilibrium, [S2F10] = 2.5 mol/L and [SF4] = [SF6] = x.
SF4 SF6  = (x)(x)
Kc =
(2.5)
S2 F10 
x2 = 2.5Kc
[SF4] = [SF6] = x =
2.5K c
Thus, the concentrations of SF4 and SF6 increase by a factor of:
2.5Kc
2.5
=
= 2.236 = 2.2
0.50Kc
0.50
15.94
Calculate Kc.
Kc =
 CO 2  H 2  = (0.40)(0.10) = 4.0
CO H 2 O (0.10)(0.10)
Calculate new concentrations.
New H2 = 0.10 mol/L + (0.60 mol/2.0 L) = 0.40 mol/L
Concentration (mol/L)
CO(g) +
H2O(g)
 CO2(g) +
H2(g)
Initial
Change
Equilibrium
0.10
0.10
0.40
0.40
+x
+x
–x
–x
0.10 + x
0.10 + x
0.40 – x
0.40 – x
2
 CO 2  H 2  = (0.40  x)(0.40  x) = (0.40  x) = 4.0 (take the sq root of both sides)
Kc =
CO H 2 O (0.10  x)(0.10  x) (0.10  x)2
(0.40  x)
= 2.0
(0.10  x)
x = 0.066667
[CO] = [H2O] = 0.10 + x = 0.10 + 0.066667 = 0.166667 = 0.17 mol/L
[CO2] = [H2] = 0.40 – x = 0.40 – 0.066667 = 0.333333 = 0.33 mol/L
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
15.95
Plan: Use the volume fraction of O2 and CO2 to find the partial pressure of each gas and substitute these
pressures into the equilibrium expression to find the partial pressure of CO. Use pV = nRT to convert the partial
pressure of CO to moles per liter and then convert to pg/L.
Solution:
a) Calculate the partial pressures of oxygen and carbon dioxide because volumes are proportional to the amount
(mol) of gas, so volume fraction equals mole fraction. Assume that the amount of carbon monoxide gas is small
relative to the other gases, so the total volume of gases equals VCO2 + VO2 + VN2 = 10.0 + 1.00 + 50.0 = 61.0.
 10.0 mol CO 2 
pCO2 = 
  4.0 bar  = 0.6557377 bar
 61.0 mol gas 
 1.00 mol O2 
pO2 = 
  4.0 bar  = 0.06557377 bar
 61.0 mol gas 
Use the partial pressures and given K to find pCO.
2CO2(g)  2CO(g) + O2(g)
K=
2
pCO
pO2
=
2
pCO
2
2
pCO
 0.06557377 
 0.6557377 
2
pCO = 3.0299x10–14 = 3.0x10–14 bar
b) pV = nRT

= 1.4x10–28

3.0299x1014 bar
nCO
p
=
=
= 4.55542x10–16 mol/L
V
RT
L•bar 

 0.08314 mol•K   800 K 


 4.55572x1016 mol CO   28.01 g CO   1 pg 
Concentration (pg/L) of CO = 
 
  12  = 0.01276 pg/L = 0.013 pg
L

  1 mol CO   10 g 
CO/L
15.96
Although the yield is favored by low temperature, the rate of formation is not. In fact, ammonia forms so slowly at
low temperatures that the process becomes uneconomical. In practice, a compromise is achieved that optimizes
yield and rate (high pressure, continual removal of NH3, increasing the temperature).
15.97
Plan: Write a reaction table given that pCH4 (init) = pCO2 (init) = 10.0 bar, substitute equilibrium values into the
equilibrium expression, and solve for pH2 .
Solution:
a) Pressure (bar)
CH4(g)
Initial
Change
Equilibrium
10.0
–x
10.0 – x
K=
2
pCO
pH2 2
pCH4 pCO2
 2x 2
10.0  x 
=
+
CO2(g)

10.0
–x
10.0 – x
2CO(g)
0
+2x
2x
+
2H2(g)
0
+2x
2x
 2x 2  2x 2
 2x 4 = 3.548x106 (take square root of each side)
=
10.0  x 10.0  x  10.0  x 2
= 1.8836135x103
A quadratic is necessary:
4x2 + (1.8836135x103 x) – 1.8836135x104 = 0
a = 4 b = 1.8836135x103 c = – 1.8836135x104
x=
b 
b 2  4ac
2a
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
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1.8836135x103 
x=
1.8836135x10   4  4   1.8836135x10 
3 2
4
2 4
x = 9.796209
pH2 = 2x = 2(9.796209) = 19.592419 bar
If the reaction proceeded entirely to completion, the partial pressure of H2 would be 20.0 bar (pressure is
proportional to amount (mol), and twice as many moles of H2 form for each mole of CH4 or CO2 that
reacts).
19.592418 bar
The percent yield is
100%  = 97.96209 %= 98.0%.
20.0 bar
b) Repeat the calculations for part a) with the new K value. The reaction table is the same.
K=
2
pCO
pH2 2
pCH4 pCO2
 2x 2
10.0  x 
=
 2x 2  2x 2
 2x 4 = 2.626x107
=
10.0  x 10.0  x  10.0  x 2
= 5.124451x103
A quadratic is needed:
4x2 + (5.124451x103 x) – 5.124451x104 = 0
a=4
b = 5.124451x103
c = – 5.124451x104
x=
5.124451x103 
5.124451x10   4  4   5.124451x10 
3 2
4
2 4
x = 9.923144
pH2 = 2x = 2(9.923144) = 19.84629 bar
If the reaction proceeded entirely to completion, the partial pressure of H 2 would be 20.0 bar (pressure is
proportional to moles, and twice as many moles of H2 form for each mole of CH4 or CO2 that reacts).
19.84629 bar
The percent yield is
100%  = 99.23145 % = 99.0%.
20.0 bar
c) van’t Hoff equation:
K1 = 3.548x106
K2 = 2.626x107
ln
ln
K2
 H  1
=  r

K1
R  T2
T1 = 1200. K
T2 = 1300. K
1


T1 
r H  = ?
R = 8.314 J/mol•K
2.626x107


r H 
1
1
=




6
J   1200. K
1300. K 
3.548x10

 8.314 mol•K 


2.0016628 =  r H  (7.710195x10–6)mol/J
 r H  = 2.0016628/(7.710195x10–6 mol/J )= 2.5961247x105 J/mol= 2.60x105 J/mol
(The subtraction of the 1/T terms limits the answer to three significant figures.)
15.98
a)
C3H8(g) + 3H2O(g)  3CO(g) + 7H2(g)
K'1 = K1 = 8.175x1015
3CO(g) + 3H2O(g)  3CO2(g) + 3H2(g)
K'2 = (K2)3 =(0.6944)3 = 0.33483368
(Overall): C3H8(g) + 6H2O(g)  3CO2(g) + 10H2(g)
b) K (overall) = K'1 x K'2 = (8.175x1015)(0.33483368) = 2.737265x1015 = 2.737x1015
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
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p  p 
c) K =
 p  p 
3
10
CO2
H2
C3 H 8
H2O
6
The partial pressures of each reactant are proportional to the amount (mol), and the limiting reactant may
be determined from the partial pressures.
pC3H8 (initial) = (1.00/5.00) x 5.0 bar = 1.0 bar
pH2O(initial) = (4.00/5.00) x 5.0 bar = 4.0 bar
(limiting reactant)
pCO2(formed) = 4.0 bar H2O x (3 mol CO2/6 mol H2O) = 2.0 bar
pH2(formed) = 4.0 bar H2O x (10 mol H2/6 mol H2O) = 6.6667 bar
pC3H8 (remaining) = 1.0 bar C3H8 – [4.0 bar H2O x (1 mol C3H8/6 mol H2O)] = 0.3333 bar
pH2O(remaining) = 0.00 bar (limiting reactant)
pTotal = pCO2 + pH2 + pC3H8 + pH2O = 2.0 bar + 6.6667 bar + 0.3333 bar + 0.00 bar = 9.0 bar
d) Percent C3H8(unreacted) = [0.3333 bar/1.0 bar] x 100% = 33.33% = 33%
15.99
Plan: Add the two reactions to obtain the overall reaction. Multiply the second equation by 2 to cancel the
amount (mol) of CO produced in the first reaction. K for the second reaction is then (K )2. K for the overall
reaction is equal to the product of the K values for the two individual reactions. Calculate Kc using K = Kc(RT)n.
Solution:
a)
2CH4(g) + O2(g)  2CO(g) + 4H2(g)
K = 9.34x1028
2CO(g) + 2H2O(g)  2CO2(g) + 2H2(g)
K =(1.374)2 = 1.888
2CH4(g) + O2(g) + 2H2O(g)  2CO2(g) + 6H2(g)
28
b) K = (9.34x10 )(1.888) = 1.76339x1029 = 1.76x1029
c) n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 8 – 5 = 3
(8 moles of product gas – 5 moles of reactant gas)
K = Kc(RT)n
1.76339x1029
K
Kc =
=
= 3.01713x1023 = 3.02x1023
n
[(0.08314 bar•L/mol•K)(1000)]3
( RT )
d) The initial total pressure is given as 30. bar. To find the final pressure use the relationship between pressure and
amount (mol) of gas: ninitial/Pinitial = nfinal/Pfinal
Total amount (mol) of gas initial = 2.0 mol CH4 + 1.0 mol O2 + 2.0 mol H2O = 5.0 mol
Total amount (mol) of gas final = 2.0 mol CO2 + 6.0 mol H2 = 8.0 mol (from mole ratios)
 8 mol products 
pfinal =  30. bar reactants  
 = 48 bar
 5 mol reactants 
15.100 Plan: Write an equilibrium expression. Use the balanced equation to define x and set up a reaction table,
substitute into the equilibrium expression, and solve for x, from which the pressure of N or H is calculated.
Convert log K to K. Convert pressures to amount (mol) using the ideal gas law, pV = nRT. Convert amount (mol)
to atoms using Avogadro’s number.
Solution:
a) The initial pressure of N2 is 200. bar.
log K = –43.10;
K =10–43.10 = 7.94328x10–44
Pressure (bar)
N2(g)

2N(g)
Initial
Change
Equilibrium
200.
–x
200 – x
K=
 pN 2
 
pN 2
0
+2x
2x
= 7.94328x10–44
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
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 2x 2
Assume 200. – x  200.
= 7.94328x10–44
 200.  x 
 2x 2
= 7.94328x10–44
 200 
4x2 = 1.588656x10–41
x = 1.992897x10–21
pN = 2x = 2(1.992897x10–21) = 3.985795x10–21 = 4.0x10–21 bar
b) Log K = –17.30;
K = 10–17.30 = 5.01187x10–18
Pressure (bar)
H2(g)

2H(g)
Initial
Change
Equilibrium
600.
–x
600 – x
K=
 pH 2
0
+2x
2x
= 5.01187x10–18
p 
H2
 2x 2
Assume 600. – x  600.
= 5.01187x10–18
 600.  x 
 2x 2
= 5.01187x10–18
 600
4x2 = 3.007122x10–15
x = 2.741862x10–8
pH = 2x = 2(2.741862x10–8) = 5.48372x10–8 = 5.5x10–8 bar
c) pV = nRT
Moles of N atoms =


3.985795x1021 bar 1.00 L 
pV
=
= 4.794076x10–23 mol
RT
L•bar 

 0.08314 mol•K  1000.K 


 6.022x1023 N atoms 
23
Number of N atoms = 4.794076x10 mol N atoms 
 = 28.8699 atoms/L = 29 N
 1 mol N atoms 
atoms/L




5.48372x108 bar 1.00 L 
pV
Moles of H atoms =
=
= 6.595766x10–10 mol
RT
L•bar 

 0.08314 mol•K  1000.K 


 6.022x1023 H atoms 
10
mol H atoms 
Number of H atoms = 6.595766x10
 1 mol H atoms 


= 3.97197x1014 = 4.0x1014 H atoms/L
d) The more reasonable step is N2(g) + H(g)  NH(g) + N(g). With only twenty-nine N atoms in 1.0 L, the first
reaction would produce virtually no NH(g) molecules. There are orders of magnitude more N2 molecules than N
atoms, so the second reaction is the more reasonable step.


15.101 a) Scenes B and D represent equilibrium.
b) C, A, B = D
 0.025 mol   1 
c) [Y] =  4 spheres  

 = 0.25 mol/L
 1 sphere   0.40 L 
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Instructor’s Solution Manual
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 0.025 mol   1 
[Z] =  8 spheres  

 = 0.50 mol/L
 1 sphere   0.40 L 
Kc =
[Z]2
[0.50]2
=
= 1.0
[Y]
[0.25]
15.102 The K is very small, thus the reaction will shift to the right to reach equilibrium. To simplify the calculations,
assume the equilibrium shifts entirely to the left, and then a little material reacts to reach equilibrium. Shifting
entirely to the left gives [H2S] = 0.600 mol/L, and [H2] = [S2] = 0 mol/L.
Concentration
2H2S(g)
 2H2(g) +
S2(g)
Initial
Change
Equilibrium
0.600 mol/L
–2x
0.600 – 2x
Kc =
0 mol/L
+2x
2x
0 mol/L
+x
x
 H 2 2 S2  = 9.0x10–8
 H 2S2
(2x) 2 (x)
(0.600  2x) 2
(2x)2 (x)
2
= 9.0x10–8
Assume 2x is small compared to 0.600 mol/L.
= 9.0x10–8
(0.600)
x = 2.008x10–3
(assumption justified)
[H2S] = 0.600 – 2x = 0.600 – 2(2.008x10–3) = 0.595984 = 0.596 mol/L H2S
[H2] = 2x = 2(2.008x10–3) = 4.016x10–3 = 4.0x10–3 mol/L H2
[S2] = x = 2.008x10–3 = 2.0x10–3 mol/L S2
15.103 Plan: Write an equilibrium expression. Use the balanced equation to define x and set up a reaction table,
substitute into the equilibrium expression, and solve for x, from which the equilibrium pressures of the gases are
calculated. Add the equilibrium pressures of the three gases to obtain the total pressure. Use the relationship
K = Kc(RT)n to find Kc.
Solution:
a)
Pressure (bar)
N2(g) +
O2(g) 
2NO(g)
Initial
Change
Equilibrium
K=
0.780
–x
0.780 – x
 pNO 2
 p  p 
N2
0.210
–x
0.210 – x
0
+2x
2x
= 4.35x10–31
O2
 2 x 2
= 4.35x10–31 Assume x is small because K is small.
 0.780  x  0.210  x 
 2 x 2
= 4.35x10–31
 0.780 0.210
x = 1.33466x10–16
Based on the small amount of nitrogen monoxide formed, the assumption that the partial pressures of nitrogen and
oxygen change to an insignificant degree holds.
pnitrogen (equilibrium) = (0.780 – 1.33466x10–16) bar = 0.780 bar N2
poxygen (equilibrium) = (0.210 – 1.33466x10–16) bar = 0.210 bar O2 (assumption justified)
pNO (equilibrium) = 2(1.33466x10–16) bar = 2.66933x10–16 = 2.67x10–16 bar NO
b) The total pressure is the sum of the three partial pressures:
0.780 bar + 0.210 bar + 2.67x10–16 bar = 0.990 bar
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Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
c) K = Kc(RT)n
n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 2 – 2 = 0
(two moles of product NO and two moles of reactants N2 and O2)
K = Kc(RT)
Kc = K = 4.35x10–31 because there is no net increase or decrease in the amount (mol) of gas in the course of the
reaction.
15.104 a) K = Kc(RT)n n = 2 – (2 + 1) = –1
Kc = K/(RT)n
1.3x10 
4
Kc =
 0.08314  457 
1
= 4.939347x105 = 4.9x105
b)  r H  = [ f(products) H  ] – [  f(reactants) H  ]
 r H  = {2  f H  [NO2(g)]} – {2  f H  [NO(g)] +   f H  [O2(g)]}
= [(2)(33.2 kJ/mol)] – [(2 l)(90.29 kJ/mol) +(0.0 kJ/mol)]
= –114.18 kJ/mol = – 114.2 kJ/mol
K
 H  1
1
c) ln 2 =  r



K1
R  T2
T1 
ln
6.4x109
4.939347x10
= 
5
114.18 kJ / mol  1
1   103 J 




8.314 J/mol•K  T2
457 K   1 kJ 
 1
1 
9.4694052 = 13,733 K 


457 K 
 T2
T2 = 347.500 K= 3.5x102 K
15.105 Plan: Use the equation K = Kc(RT)n to find K. The value of Kc for the formation of HI is the reciprocal of the Kc
value for the decomposition of HI. Use the equation  r H  = [ f(products) H  ] – [  f(reactants) H  ] to find the
value of  r H  . Use the van’t Hoff equation as a second method of calculating  r H  .
Solution:
a)K = Kc(RT)n
n = amount (mol) of gaseous products – amount (mol) of gaseous reactants = 2 – 2 = 0
(2 mol product (1H2 + 1I2) – 2 mol reactant (HI) = 0)
K = Kc(RT)0 = 1.26x10–3(RT)0 = 1.26x10–3
b) The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the
forward reaction:
1
1
Kformation =
=
= 793.65 = 794
K decomposition
1.26x103
c)  r H  = [ f(products) H  ] – [  f(reactants) H  ]
 r H  = {1  f H  [H2(g)] +1  f H  [I2(g)]} – {2  f H  [HI(g)]}
 r H  = [(0 kJ/mol) + (62.442 kJ/mol)] – [(2)(25.9 kJ/mol)]
 r H  = 10.6 kJ/mol
K2
 H  1
1
=  r



K1
R  T2
T1 
K1 = 1.26x10–3; K2 = 2.0x10–2, T1 = 298 K; T2 = 729 K
d) ln
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
ln
 1
r H 
1 
2.0x10 2
=




8.314 J/mol•K  729 K
298 K 
1.26x103
2.764621 = (2.38629x10–4 mol/J)  r H 
 r H  = 1.1585x104 J/mol= 1.2x104 J/mol
15.106 C5H11OH + CH3COOH  CH3COOC5H11 + H2O
Removing water should help to increase the yield of banana oil. Both isopentyl alcohol and acetic acid are more
soluble in water than isopentyl acetate. Thus, removing water will increase the concentration of both reactants and
cause a shift in equilibrium towards the products. Also, according to le Châtelier’s principle, if a product is
removed, the reaction will shift to produce more of the product, as observed here.
[R]
15.107 Q(g) R(g)
K=
[Q]
For Scene A at equilibrium:
[R]
[2]
K=
=
= 0.33
[Q]
[6]
For Scene B:
Q(g)

R(g)
Initial
10
2
Change
–x
+x
Equilibrium
10 – x
2+x
[2 + x]
0.33 =
[10  x]
x = 0.977 = 1
Q = 10 – x = 10 – 1 = 9; R = 2 + x = 2 + 1 = 3

15.108 a) K = pH 2O
 = 4.08x10
10
–25
pH 2 O = 10 4.08x1025 = 3.6397x10–3 = 3.64x10–3 bar
b) (i) Adding more Na2SO4(s) will decrease the ratio of hydrated form/anhydrous form merely because you are
increasing the value of the denominator, not because the equilibrium shifts.
(ii) Reducing the container size will increase the pressure (concentration) of the water vapour, which will shift
the equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase.
(iii) Adding more water vapour will increase the concentration of the water vapour, which will shift the
equilibrium to the reactant side. The ratio of hydrated form/anhydrous form will increase.
(iv) Adding N2 gas will not change the partial pressure of the water vapour, so the ratio of hydrated
form/anhydrous form will not change.
15.109 Plan: Use the balanced equation to write an equilibrium expression. Find the initial concentration of each reactant
from the given amounts and container volume, use the balanced equation to define x, and set up a reaction table.
The equilibrium concentration of CO is known, so x can be calculated and used to find the other equilibrium
concentrations. Substitute the equilibrium concentrations into the equilibrium expression to find Kc. Add the
molarities of all of the gases at equilibrium, use (c)(V) to find the total amount (mol), and then use pV = nRT to
find the total pressure. To find [CO]eq after the pressure is doubled, set up another reaction table in which the
initial concentrations are equal to the final concentrations from part a) and add in the additional CO.
Solution:
The reaction is: CO(g) + H2O(g)  CO2(g) + H2(g)
a) Initial [CO] and initial [H2O] = 0.100 mol/20.00 L = 0.00500 mol/L.
concentrations
CO
H 2O

CO2
Initial
0.00500 mol/L
0.00500 mol/L
0
H2
0
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
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Change
–x
–x
+x
Equilibrium
0.00500 – x
0.00500 – x
x
[CO]equilibrium = 0.00500 – x = 2.24x10–3 mol/L = [H2O] (given in problem)
x = 0.00276 mol/L = [CO2] = [H2]
 CO 2  H 2  = (0.00276)(0.00276) = 1.518176 = 1.52
Kc =
CO H 2 O (0.00224)(0.00224)
+x
x
b) ctotal = [CO] + [H2O] + [CO2] + [H2] = (0.00224 mol/L) + (0.00224 mol/L) + (0.00276 mol/L) + (0.00276
mol/L)
= 0.01000 mol/L
ntotal = (ctotal)(V) = (0.01000 mol/L)(20.00 L) = 0.2000 mol total
pV = nRT
L•bar 
 0.2000 mol   0.08314
  273  900. K 
mol•K 

ptotal = ntotalRT/V =
= 0.9752322 bar = 0.975 bar
 20.00 L 
c) Initially, an equal amount (mol) must be added = 0.2000 mol CO
d) Set up a table with the initial concentrations equal to the final concentrations from part a), and then add
0.2000 mol CO/20.00 L = 0.01000 mol/L to compensate for the added CO.
CO
H 2O
CO2
H2
Initial
0.00224 mol/L 0.00224 mol/L 0.00276 mol/L 0.00276 mol/L
Added CO
0.01000 mol/L
Change
–x
–x
+x
+x
Equilibrium
0.01224 – x
0.00224 – x
0.00276 + x
0.00276 + x
 CO 2  H 2  = (0.00276  x)(0.00276  x) = 1.518176
Kc =
CO H 2 O (0.00224  x)(0.00224  x)
7.6176x106  5.52x103 x  x 2
= 1.518176
2.74176x105  1.448x102 x  x 2
7.6176x10–6 + 5.52x10–3x + x2 = (1.518176)(2.74176x10–5 – 1.448x10–2x + x2)
7.6176x10–6 + 5.52x10–3x + x2 = 4.162474x10–5 – 0.021983x + 1.518176x2
0.518176x2 – 0.027503x + 3.400714x10–5 = 0
a = 0.518176 b = – 0.027503 c = 3.400714x10–5
x=
x=
b  b 2  4ac
2a
( 0.027503) 
 0.0275032  4  0.518176  3.400714x105 
2  0.518176 
x = 1.31277x10–3
[CO] = 0.01224 – x = 0.01224 – (1.31277x10–3) = 0.01092723 = 0.01093 mol/L
15.110 a) At point A the sign of H is negative for the reaction graphite  diamond. An increase in temperature at
constant pressure will cause the formation of more graphite. Therefore, the equation must look like this:
graphite  diamond + heat, and adding heat shifts the equilibrium to the reactant side.
b) Diamond is denser than graphite. The slope of the diamond-graphite line is positive. An increase in pressure
favors the formation of diamond.
15.111 Plan:
Prepare an ICE table. Set up the expression for K and then substitute values. Where possible, make the
relevant assumptions and then calculate the equilibrium pressures. Justify and explain all assumptions.
Solution:
3 H2 (g)
+
N2 (g)
⇌
2 NH3 (g)
Initial
1.50 bar
1.00 bar
-Change
-3x
-x
+2x
Equilibrium
1.50-3x
1.00-x
2x
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
K
2
pNH
3
pH3 2 pN2
 2x 

3
1.5  3x  1.00  x 
2
3.5  10
7
Since K is fairly small, we will make an assumption that 3x << 1.5 and therefore x << 1.00. Then, we could say
1.5-3x≈1.5 and 1.00-x ≈ 1.00. The equation thus becomes:
 2x 

3
1.5 1.00 
2
3.5  10
7
 2 x   1.18 106
2
x = 5.4×10-4 bar
pH2  1.5 bar  3 x  1.5 bar  3  5.4  10 4 bar   1.5 bar (2 sf)
pN 2  1.00 bar  x  1.5 bar   5.4 104 bar   1.0 bar (2 sf)
pNH3  2 x  2  5.4  104 bar   1.1 103 bar
We assumed that the amount dissociated was smaller (significantly!) than the starting amounts of reactants (given
that K was so small). This in turn meant that we assumed that the equilibrium pressures of the starting materials
would not differ significantly from their initial pressures. The small K value indicates that the reaction stays
mainly on the left, meaning the reactants dissociate only slightly. The values calculated justified the
approximation to the number of significant digits in the question.
15.112 Plan:
We need to calculate Q and then compare it to K to determine the direction in which the reaction will
proceed. Then we can set up the ICE table and find the equilibrium pressures. We can substitute them into the
expression for K and then solve the resulting quadratic for the value of x. We will use x to calculate the values of
all the equilibrium pressures. Finally, we will use le Chatelier’s principle to determine the direction in which the
reaction will shift due to the stresses applied.
Solution:
We will need to calculate the pressures of all four substances.
bar  L 
  900. K 
nRT
mol  K 

pA 

 3.74 bar
V
5.00 L
bar  L 

0.500 mol   0.08314

  900. K 
nRT
mol  K 

pB 

 7.48 bar
V
5.00 L
bar  L 

0.750 mol   0.08314

  900. K 
nRT
mol  K 

pC 

 11.2 bar
V
5.00 L
bar  L 

0.450 mol   0.08314

  900. K 
nRT
mol  K 

pD 

 6.73 bar
V
5.00 L
 0.250 mol   0.08314
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
Q
pC pD 11.2 bar  6.73 bar 

 2.70
p A pB  3.74 bar  7.48 bar 
QK
Therefore, the reaction moves in the REVERSE direction, towards reactants.
b)
A (g)
Initial
Change
Equilibrium
K
+ B (g)
3.74
+x
3.74+ x
⇌
7.48
+x
7.48+ x
C (g)
11.2
-x
11.2- x
+ D (g)
6.73
-x
6.73- x
pC pD 11.2  x  6.73  x 

 1.83
p A pB  3.74  x  7.48  x 
Rearranging and solving for x,
11.2  x  6.73  x   1.83
 3.74  x  7.48  x 
11.2  x  6.73  x   1.83 3.74  x  7.48  x 
75.38  17.93 x  x 2  51.19  20.53 x  1.83 x 2
0.83 x 2  38.46 x  24.19  0
The quadratic can now be solved for x,
b  b 2  4ac
x
2a

38.46 
 38.46   4  0.83 24.19 
2  0.83
2
 0.62 bar
Calculating equilibrium pressures:
pA  3.74 bar  x  3.74 bar  0.62 bar  4.36 bar
pB  7.48 bar  x  7.48 bar  0.62 bar  8.10 bar
pC  11.2 bar  x  11.2 bar  0.62 bar  10.6 bar
pD  6.73 bar  x  6.73 bar  0.62 bar  6.11 bar
c)
i)
if the container volume is doubled, all the pressures are halved. However, n=0, so no change would
occur.
ii)
if more C is added, the reaction will shift to the LEFT or REVERSE so as to produce more reactants.
iii)
If D is removed from the system, then more reactants will react to form D, so the reaction will shift
RIGHT or move FORWARD.
iv)
If the temperature is raised, then heat is added. The reaction is endothermic meaning heat is a reactant.
Added heat will push the reaction in the direction that uses up additional heat. The reaction will shift RIGHT or
move FORWARD to produce more product.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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15-33
Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
15.113 Plan: Use the relationship between Kc and K to find the value of K. Use the given enthalpies and the values in
Appendix B to find the enthalpy of reaction. Use the van’t Hoff equation to find the value of K at the second T.
Calculate Q using the given pressures. Compare Q to K. Set up an ICE table and solve for equilibrium pressures.
Solution:
a)
K  K c  RT 
Δn
 1.3  104  0.08314  298.2 
0
 1.3  104
(alternately, since Δn = 0, we can say K = Kc)
b)

 
o
Δ r Ho  Δ f HoClNO  Δ f HoNO2  Δ f HClNO
 Δ f HoNO
2

kJ
kJ  
kJ
kJ 

  51.71
 33.2
  12.13
 90.29

mol
mol  
mol
mol 

kJ
 17.51
mol
c)
ln
ln
K2
 Ho  1
1 
 r



K1
R  T1
T2 
K2
1.3  104

J
1 
mol  1


J
1346K 
 298.2K
8.314
mol  K
1.751  10 4
K2
 4.1  10 3
1.3  104
K 2  53
Q
d)
p ClNO p NO2
p ClNO2 p NO
e)

 15.8bar  8.90bar 
 21.7 < K Therefore, the reaction moves FORWARD (right).
 2.70bar  2.40bar 
ClNO2 (g) + NO (g) ⇌ ClNO (g) + NO2 (g)
Initial
Change
Equilibrium
K
pClNO p NO2
pClNO2 p NO
2.70
-x
2.70-x

2.40
-x
2.40- x
15.8
+x
15.8+ x
8.90
+x
8.90+ x
15.8  x 8.90  x   53
 2.70  x  2.40  x 
Rearranging and solving for x,
15.8  x 8.90  x   53
 2.70  x  2.40  x 
15.8  x  8.90  x   53  2.70  x  2.40  x 
141  24.7 x  x 2  343  270 x  53 x 2
52 x 2  295 x  203  0
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
The quadratic can now be solved for x,
x

b  b 2  4ac
2a
 295   4  52  203
2  52 
2
295 
 0.80 bar
Calculating equilibrium pressures:
pClNO2  2.70 bar  x  2.70 bar  0.80 bar  1.90 bar
pNO  2.40 bar  x  2.40 bar  0.80 bar  1.60 bar
pClNO  15.8 bar  x  15.8 bar  0.80 bar  16.6 bar
pNO2  8.90 bar  x  8.90 bar  0.80 bar  9.70 bar
f) The total pressure is :
g)
i)
ii)
iii)
iv)
pTOTAL  1.90 bar  1.60 bar  16.6 bar  9.70 bar = 29.8 bar
no change as Δn = 0
reaction will go right to replace NO2
No change on addition of an inert gas
the reaction will move forward as it is exothermic
15.114 Plan: Use the relation between K and Kc to find K. Use the data from Appendix B to calculate the enthalpy of
reaction. Then use the enthalpy and the van’t Hoff equation to find the value of K at the other temperature.
Finally, use le Châtelier’s principle to determine which way the reaction will shift.
Solution:
a)
K  K c  RT 
Δn
 2.0  0.08314  373.2 
1
 62
b)

 
o
Δ r Ho  2Δ f HoNO  Δ f HBr
 2Δ f HoNOBr
2

 
  
kJ 
kJ  
  2  90.29
 0    2  231.0


mol 
mol  
 
  
kJ
 642.58
mol
K
 Ho  1
1 
ln 2  r



K1
R  T1
T2 
J
6.4258  105
K2
1 
 1
mol
ln



J
62
523.1K 
 373.1K
8.314
mol  K
K2
 6.3  1025
62
K 2  3.9  1027
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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Instructor’s Solution Manual
© Copyright 2021 McGraw-Hill Ryerson Ltd.
c)
i)
ii)
iii)
the reaction will shift left as there is a smaller amount (mol) of gas on the left
the reaction will shift left in order to remove added bromine
the reaction will shift left as it is endothermic.
Silberberg, Amateis, Venkateswaran, and Chen, Chemistry: The Molecular Nature of Matter and Change, 3rd Canadian Edition
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© Copyright 2021 McGraw-Hill Ryerson Ltd.