Worked Solutions
Chapter 18
Question 9
In a reaction at a particular temperature between nitrogen and hydrogen forming ammonia
N2 + 3H2
2NH3
it was found that at equilibrium the concentrations of N2, H2 and NH3 were 0.06 mol l-1, 0.07
mol l-1 and 0.02 mol l-1 respectively. Calculate the value of the equilibrium constant (Kc) for
this reaction at this temperature.
Answer:
Kc =
[NH3]2
[N2][H2]3
= (0.02)2
= 19.44 mol-2 l2
(0.06)(0.07)3
Question 10
Four moles of COCl2 are placed in a 4 litre flask at 668 K. The following reaction occurs:
COCl2
CO + Cl2
and, at equilibrium, 1.6 moles of COCl2 remain. Calculate Kc for the reaction.
Answer:
The equation for the reaction is:
COCl2
CO + Cl2
COCl2
CO
Cl2
Initial amount
4 moles
0 moles
0 moles
Change
-(4– 1.6) moles
+ 4 - 1.6 moles
+ 4 – 1.6 moles
Equilibrium amount
1.6 moles
2.4 moles
2.4 moles
Equilibrium
concentration
1.6 / 4 = 0.4 mol l-1
2.4 / 4 = 0.6 mol l-1
2.4 / 4 = 0.6mol l-1
Kc = [CO] [Cl2] = 0.6 x 0.6
[COCl2]
0.4
= 0.9 mol l-1
1
Question 11
In the reaction
C4H9COOH + C2H5OH
C4H9COOC2H5 + H2O
at 473 K, it is found that if an initial mixture containing 2 moles of
C4H9COOH and 2 moles of C2H5OH is allowed to come to
equilibrium, 1 mole of C4H9COOH remains. Calculate the
equilibrium constant for the reaction at 473 K.
Answer:
The equation for the reaction is:
C4H9COOH + C2H5OH
C4H9COOC2H5 + H2O
CH3COOH
C2H5OH
CH3COOC2H5
H2O
Initial amount
2 moles
2 moles
0 moles
0 moles
Change
2 - 1 moles
2 - 1 moles
+ 1 mole
+1 mole
Equilibrium
amount
Equilibrium
concentration
1 mole
1 mole
1 mole
1 mole
1 / V mol l-1
1 / V mol l
1 / V mol l-1
1 / V mol l-1
Since the V values cancel,
Kc = [CH3COOC2H5] [H2O] = (1 / V)(1 / V) = 1
[CH3COOH] [C2H5OH] (1 / V)(1 / V)
Question 12
Calculate Kc for the reaction
CH3COOCH3 + H2O
CH3COOH + CH3OH
if an initial mixture of 7.4 g of CH3COOCH3 and 2.0 g of water
yields 4.0 g of CH3COOH at equilibrium.
Answer:
The equation for the reaction is:
CH3COOCH3 + H2O
CH3COOH + CH3OH
Amounts present at the start of the experiment:
CH3COOC2H5
H2O
7.4 / 74 moles
2.0 / 18 moles
= 0.1 moles
= 0.111 moles
CH3COOH
C2H5OH
0 moles
0 moles
2
At equilibrium, there is 4.0 / 60 = 0.0667 moles of ethanoic acid.
CH3COOC2H5
H2O
CH3COOH
C2H5OH
Initial amount
0.1 moles
0.111 moles
0 moles
0 moles
Change
0.1 – 0.0667 moles
+ 0.0667 moles
Equilibrium
amount
Equilibrium
concentration
0.0333 mole
0.111 – 0.0667
moles
0.0443 mole
0.0333 / V mol l-1
0.0443 / V mol l-1
0.0667 / V mol l-1
+ 0.0667
moles
0.0667
moles
0.0667 / V
mol l-1
0.0667 moles
Since the V values cancel,
Kc = [CH3COOH] [CH3OH] = (0.0667) (0.0667) = 3.02
[CH3COOCH3] [H2O] (0.0333)(0.0443)
Question 13
At 760 K, the value of Kc for the reaction
PCl5
PCl3 + Cl2
is 33. Calculate the equilibrium concentrations of all species if 10
moles of PCl5 are placed in a 1 litre flask at 760 K and allowed to
reach equilibrium.
Answer:
The balanced equation for the reaction is
PCl5
PCl3 + Cl2
Amounts present at the start of the experiment:
PCl5
PCl3
Cl2
10 moles
0 moles
0 moles
Assume that x moles respectively of PCl3 and Cl2 are formed at equilibrium. From the
balanced equation, this means that x moles of PCl5 are used up when equilibrium is reached.
Amounts present at equilibrium:
PCl5
10-x moles
PCl3
Cl2
x moles
x moles
The volume of the reaction vessel is 1 litres.
3
Concentrations at equilibrium:
[PCl5]
[PCl3]
[Cl2]
(10-x ) / 1 moles l-1
x / 1 moles l-1
x / 1 moles l-1
The following table summarises the steps taken so far in the calculation:
PCl5
PCl3
Cl2
Initial amount
10 mole
0 moles
0 moles
Change
- x moles
+ x moles + x moles
Equilibrium amount
10-x moles x moles
Equilibrium concentration (mol l-1) (10-x) / 1
x/1
x moles
x/1
Kc = 33 mol l-1 = [PCl3] x [Cl2]
[PCl5]
Substituting values for the equilibrium concentrations:
33 mol l-1 = [PCl3] x [Cl2] mol l-1 = (x) x (x) mol l-1
[PCl5]
10-x
x2 = 33(10-x) = 330 – 33x
x2 + 33x – 330 = 0
Using the formula x = (- b ±  [b2 - 4ac]) / 2a, we obtain:
x = (-33 ±  [1089 + 1320]) / 2
= (-33 ± [2409]) / 2
= (-33 ± 49.082) / 2
= 16.082 / 2 or – 82.082 / 2
= 8.04 or – 41.04
4
The negative value of x can be disregarded.
Concentrations at equilibrium:
[PCl5]
10-x / 1 moles l
[PCl3]
-1
= 1.96 moles l-1
[Cl2]
x / 1 moles l
-1
x / 1 moles l-1
= 8.04 moles l-1
= 8.04 moles l-1
Question 14
For the reaction
H2 + CO2
H2O + CO
at 1200 K, the value of Kc for the reaction is 1.40. Calculate the concentrations of all species
at equilibrium after 1 mole of H2O and 1 mole of CO react together in a 4 litre flask.
Answer:
The balanced equation for the reaction is
H2 + CO2
H2O + CO
Amounts present at the start of the experiment:
H2
CO2
H2O
0 moles
0 moles
1 mole
CO
1 mole
Assume that x moles respectively of H2 and CO2 are formed at equilibrium. From the balanced
equation, this means that x moles of H2O and x moles of CO are used up when equilibrium is
reached.
Amounts present at equilibrium:
H2
x moles
CO2
x moles
H2O
1 - x moles
CO
1 – x moles
The volume of the reaction vessel is 4 litres.
Concentrations at equilibrium:
[H2]
(x ) / 4 moles l-1
[CO2]
x / 4 moles l-1
[H2O]
x / 4 moles l-1
[CO]
x / 4 moles l-1
5
H2
CO2
H2O
CO
Initial amount
0 moles
0 moles
1 moles
1 moles
Change
+ x moles
+ x moles
1 - x moles
1 - x moles
Equilibrium amount
x moles
x moles
1 - x moles
1 - x moles
Equilibrium
x/4
x/4
1-x/4
1-x/4
concentration (mol l-1)
Kc = [H2O] [CO] = (1-x / 4)(1-x / 4)
=4
[H2] [CO2] (x/ 4)(x/ 4)
Since the 4 values cancel,
Kc = 1.40 = (1-x)(1-x)
(x)(x)
From this is got the following quadratic equation:
0.4x2 +2x - 1 = 0
Using the formula x = (- b ±  [b2 - 4ac]) / 2a, we obtain:
x = (-2 ±  [4 + 1.6]) / 0.8
= (-2 ± [5.6]) / 0.8
= (-2 ± 2.366) / 0.8
= 0.366 / 0.8 or – 4.366 / 0.8
= 0.458 or – 5.458
The negative value of x can be disregarded.
6
Amounts at equilibrium
Equilibrium concentrations (mol l-1)
H2
CO2
H2O
CO
0.458
0.458
0.542
0.542
moles
moles
moles
moles
0.11
0.11
0.14
0.14
Question 15
Ethanoic acid reacts with propan-1-ol at 373 K according to the equation
CH3COOH + C3H7OH
CH3COOC3H7 + H2O
If the equilibrium constant for the reaction at this temperature is 6.25, calculate the
equilibrium amounts in moles of all species when 210 g of CH3COOH and 210 g of C3H7OH
are placed in a flask at 373 K.
Answer:
210 g CH3COOH = 210 / 60 = 3.5 moles
210 g C3H7OH = 210 / 60 = 3.5 moles
Let V litres be the total volume of the mixture at equilibrium. Assume that x moles of propyl
ethanoate and water respectively are formed at equilibrium.
CH3COOH
C3H7OH
CH3COOC3H7
H2O
Initial amount
3.5 moles
3.5 moles
0 moles
0 moles
Change
- x moles
- x moles
+x
+ x moles
Equilibrium amount
3.5-x moles
3.5 -x moles
+x
+ x moles
Equilibrium
(3.5-x) / V
(3.5-x) / V
x/V
x/V
concentration (moles l-1)
Kc = [CH3COOC3H7] [H2O] = (x / V)(x / V)
= 6.25
[CH3COOH] [C3H7OH] ((3.5-x)/ V)((3.5-x)/ V)
Since the V values cancel,
Kc = 6.25 = (x)(x)
(3.5-x)(3.5-x)
7
From this is got the following quadratic equation:
5.25x2 - 43.75x – 76.5625 = 0
Using the formula x = (- b ±  [b2 - 4ac]) / 2a, we obtain:
x = (+43.75 ±  [1914.0625 - 1607.8125]) / 10.5
= (+43.75 ± [306.25]) / 10.5
= (+43.75 ± 17.5) / 10.5
= 26.25/ 10.5 or + 61.25/ 10.5
= 2.5 or 5.83333
Since x is the number of moles of propyl ethanoate or water formed, x = 5.8333 is impossible,
since the initial amounts (3.5 moles each) of ethanoic acid and propanol are not capable of
producing 5.83333 moles of propyl ethanoate and water respectively.
Amounts at equilibrium:
CH3COOH
1 mole
C3H7OH
1 mole
CH3COOC3H7
2.5 moles
H2O
2.5 moles
Question 16
Benzoic acid reacts with ethanol at 473 K according to the equation
C6H5COOH + C2H5OH
C6H5COOC2H5 + H2O
If the equilibrium constant for the reaction at this temperature is 2.73,
calculate the equilibrium amounts in moles of all species when 1 mole
of C6H5COOH and 3 moles of C2H5OH are placed in a flask at 473 K.
Answer:
Amounts present at the start of the experiment:
C6H5COOH + C2H5OH
1 mole
3 moles
C6H5COOC2H5 + H2O
0 moles
0 moles
8
Assume that x moles respectively of C6H5COOC2H5 and H2O are formed at equilibrium.
From the balanced equation, this means that x moles of C6H5COOH and x moles of C2H5OH
are used up when equilibrium is reached.
Amounts present at equilibrium:
C6H5COOH C2H5OH
1-x moles
C6H5COOC2H5 H2O
3 – x moles
x moles
x moles
Concentrations at equilibrium:
C6H5COOH
C2H5OH
C6H5COOC2H5
(1-x ) / V moles l-1 3 - x / V moles l-1
H2O
x / V moles l-1 x / V moles l-1
C6H5COOH
C2H5OH
C6H5COOC2H5
H2O
Initial amount
1 moles
3 moles
0 moles
0 moles
Change
1 - x moles
3 - x moles
+ x moles
+ x moles
Equilibrium amount
1-x moles
3-x moles
x moles
x moles
(1-x) / V
(3-x) / V
x/V
x/V
Equilibrium
-1
concentration (moles l )
Kc = [C6H5COOC2H5] [H2O] = (x / V)(x / V)
= 2.73
[C6H5COOH] [C2H5OH] ((1-x)/ V)((3-x)/ V)
Since the V values cancel,
Kc
=
2.73 =
(x)(x)
(1-x)(3-x)
From this is got the following quadratic equation:
2.73x2 – 10.92x + 8.19 = 0
Using the formula x = (- b ±  [b2 - 4ac]) / 2a, we obtain:
x = (10.92 ±  [119.2464 – 56.6748]) / 3.46
9
= (10.92 ± [62.5716]) / 3.46
= (10.92 ± 7.9102219) / 3.46
= 3.0097788 / 3.46 or 18.83022 / 3.46
= 0.87or 5.44
Since x is the number of moles of ethyl benzoate or water formed, x = 5.44 is impossible,
since the initial amounts of benzoic acid (1 mole) and ethanol (3 moles) are not capable of
producing 5.44 moles of propyl ethanoate and water respectively.
Concentrations at equilibrium:
C6H5COOH
0.13 moles l-1
C2H5OH
C6H5COOC2H5
2.13 moles l-1
0.87 moles l-1
H2O
0.87 moles l-1
Question 24
In an experiment, 280 cm3 of hydrogen (measured at s.t.p.) was placed in a 1 litre flask
containing 3.175 g of iodine. The flask was stoppered and heated for 30 minutes at 700 K. It
was then cooled rapidly and the mass of hydrogen iodide present was found to be 2.56g.
(a) Calculate the equilibrium constant (Kc) for this reaction at 700 K.
Answer:
The equation for the reaction is:
H2 + I2
2HI
Amounts present at the start of the experiment:
H2
I2
280 / 22400 moles
3.175 / 254 moles
= 0.0125 moles
= 0.0125 moles
HI
0 moles
Moles of hydrogen iodide present at equilibrium = 2.56 / 128 moles = 0.02 moles
From the equation, two moles of HI are formed when one mole of H2 reacts with one mole
of I2.
Therefore, 0.02 moles of HI are formed when 0.01 moles of H2 reacts with 0.01 moles of I2.
10
Amounts present at equilibrium:
H2
I2
HI
0.0125 – 0.01 moles
0.0125 – 0.01 moles
= 0.0025 moles
= 0.0025 moles
0.02 moles
Concentrations in mol l-1 present at equilibrium:
[H2]
[ I2]
[HI]
0.0025
0.0025
0.02
The following table summarises the steps taken so far in the calculation:
H2
I2
HI
Initial amount
0.0125 moles
0.0125 moles
0 moles
Change
- 0.01 moles
- 0.01 moles
+ 0.02 moles
Equilibrium amount
0.0025 moles
0.0025 moles
0.02 moles
Equilibrium
0.0025 mol l-1
0.0025 mol l-1
0.02 mol l-1
concentration
Kc = [HI]2
[H2] [I2]
= (0.02)2
= 64
0.0025 x 0.0025
Question 25
Ethanol reacts with ethanoic acid according to the equation:
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
(c) At 293 K, the value of Kc for the reaction is 4. If 44 g of CH3COOC2H5 and 9 g
H2O are mixed and heated at this temperature, calculate how many moles of (i)
CH3COOH and (ii) CH3COOC2H5 are present when equilibrium is reached.
Answer:
44 g CH3COOC2H5 = 44 / 88 = 0.5 moles
9 g H2O = 9 / 18 = 0.5 moles
Let V litres be the total volume of the mixture at equilibrium. Assume that x moles of ethyl
ethanoate and water respectively are formed at equilibrium.
CH3COOH
C2H5OH
CH3COOC2H5
H2O
11
Initial amount
0 moles
0 moles
0.5 moles
0.5 moles
Change
+ x moles
+ x moles
- x moles
- x moles
Equilibrium amount
x moles
x moles
0.5 - x moles
0.5 - x
moles
Equilibrium
x/V
x/V
0.5 - x / V
-1
concentration (mol l )
0.5 - x /
V
Kc = [CH3COOC2H5] [H2O] = (0.5 - x / V)(0.5 - x / V)
=4
[CH3COOH] [C2H5OH] (x/ V)x/ V)
Since the V values cancel,
Kc = 4 = (0.5-x)(0.5-x)
(x)(x)
From this is got the following quadratic equation:
3x2 + x - 0.25 = 0
Using the formula x = (- b ±  [b2 - 4ac]) / 2a, we obtain:
x = (- 1 ±  [1 + 3]) / 6
= (-1 ± [4]) / 6
= (-1 ± 2) / 6
= 1/ 6 or - 3/ 6
= 0.167 or – 0.5
The negative value of x, –0.5, can be disregarded.
x = 0.17.
12
Amounts at equilibrium:
CH3COOH
C2H5OH
0.17 moles
0.17 moles
CH3COOC2H5
0.33 moles
H2O
0.33 moles
Question 26
(a)Write the equilibrium constant expression for the reaction
NO2 + SO2
NO + SO3
(b) When 1.15 g of nitrogen dioxide and 1.92 g of sulfur dioxide were heated together in a
closed vessel at 800 K, it was found that at equilibrium 1.2 g of sulfur trioxide was present.
Calculate the equilibrium constant for the reaction at 800 K.
Answer:
Kc = [NO] [SO3]
[NO2] [SO2]
Amounts present at the start of the experiment:
NO2
1.15 / 46 moles
SO2
1.92 / 64 moles
= 0.025 moles
NO
SO3
0 moles
0 moles
= 0.03 moles
Moles of sulfur trioxide present at equilibrium = 1.2 /80 moles = 0.015 moles
NO2
SO2
NO
SO3
Initial amount
0.025 moles
0.03 moles
0 moles
0 moles
Change
0.025 – 0.015 moles
0.03 – 0.015
+ 0.015 moles
+ 0.015
moles
Equilibrium
0.01 mole
0.015 mole
moles
0. 0.015 moles
0. 0.015
amount
Equilibrium
moles
0.01 / V mol l
-1
0.015 / V mol
l-1
concentration
0.015 / V mol l
-1
0.015 / V
mol l-1
Since the V values cancel,
Kc = [NO] [SO3] = (0.015) (0. 0.015)
[NO2] [SO2]
= 1.5
(0.01)(0.015)
13
(d) 1.28 g of nitrogen dioxide was then added to the reaction mixture and equilibrium was
reached again, it was found that an additional 0.6 g of sulfur trioxide was formed. Calculate
(i) how many moles of each gas are present in the final equilibrium mixture (ii) the value of
the equilibrium constant.
Answer:
1.28 g nitrogen dioxide = 1.28 / 46 = 0.0278 moles
Amounts present now:
NO2
SO2
0.01 + 0.0278
0.015
NO
0.015 moles
SO3
0.015 moles
= 0.0378 moles
0.6 g sulfur trioxide = 0.6 / 80 = 0.0075 moles
NO2
SO2
NO
SO3
Initial amount
0.0378 moles
0.015 moles
0.015 moles
Change
0.0378 – 0.0075 moles
+ 0.0075 moles
Equilibrium
amount
Equilibrium
concentration
0.0303 moles
0.015 –
0.0075 moles
0.0075 moles
0.0075 / V
mol l
0.0225 / V mol l-1
0.015
moles
+ 0.0075
moles
0.0225
moles
0.0225 / V
mol l-1
0.0303 / V mol l-1
0.0225 moles
Since the V values cancel,
Kc = [NO] [SO3] = (0.0225)(0.0225)
[NO2] [SO2]
= 2.23
(0.0303) (0.0075)
Question 27
In the equilibrium
Br2 + Cl2
2BrCI
the forward reaction is exothermic.
(b)In an experiment, a 3 litre flask, containing 9 moles of BrCl only, at a temperature of
700 K was allowed to come to equilibrium. At equilibrium it was found that 3 moles of
chlorine were formed. Calculate the equilibrium constant for the reaction at this
temperature.
14
Answer:
The equation for the reaction is:
Br2 + Cl2
2BrCI
Br2
Cl2
BrCl
Initial amount
0 moles
0 moles
9 moles
Change
+ 3 moles
+ 3 moles
- 6 moles
Equilibrium amount
3 moles
3 moles
3 moles
Equilibrium concentration
3 / 3 = 1 mol l-1
3 /3 = 1 mol l-1
3 / 3 = 2 mol l-1
Kc = [BrCl]2
[Br2] [Cl2]
(c)
= (1)2
= 1
1x1
In a further experiment, 6 moles of BrCl and 6 moles of bromine, at the
same temperature, were allowed to come to equilibrium. Calculate the
amount in moles of each of the constituents of the mixture at equilibrium.
Answer:
The balanced equation for the reaction is
Br2 + Cl2
2BrCI
Amounts present at the start of the experiment:
Br2
6 moles
Cl2
BrCl
0 moles
6 moles
Assume that x moles respectively of Br2 and Cl2 are formed at equilibrium. From the balanced
equation, this means that 2x moles of BrCl are used up when equilibrium is reached.
Amounts present at equilibrium:
Br2
6 + x moles
Cl2
x moles
BrCl
6 - 2x moles
The volume of the reaction vessel is V litres.
Concentrations at equilibrium:
[Br2]
(6 + x ) / V moles l-1
[Cl2]
x / V moles l-1
[BrCl]
6 - 2 x / V moles l-1
15
The following table summarises the steps taken so far in the calculation:
Br2
Cl2
BrCl
Initial amount
6 mole
0 moles
6 moles
Change
6 + x moles
+ x moles
- 2x moles
Equilibrium amount
6 + x moles
x moles
6 - 2x moles
Equilibrium concentration (6 +x) / V moles l-1 x / V moles l-1 6 - 2x / V moles l-1
Kc = [BrCl]2
= (6 – 2x)2
= 1
[Br2] [Cl2] (6 + x)(x)
3x2 - 30x + 36 = 0
Using the formula x = (- b ±  [b2 - 4ac]) / 2a, we obtain:
x = (+30 ±  [900 - 432]) / 6
= (30 ± [468]) / 6
= (30 ± 21.633308) / 6
= 8.3666923 / 6 or 51.633308 / 6
= 1.39 or 8.61
8.61 moles is too big
Concentrations at equilibrium:
[Br2]
[Cl2]
6 + x moles l-1
x moles l-1
7.39 moles l-1
1.39 moles l-1
[BrCl]
6 - 2 x moles l-1
3.21 moles l-1
(e) Using the data in (b) above what is the value of the equilibrium constant at 700 K for
the reaction as represented by the equation
2BrCl
Br2 + Cl2?
16
Br2
Cl2
BrCl
Initial amount
0 moles
0 moles
9 moles
Change
+ 3 moles
+ 3 moles
- 6 moles
Equilibrium amount
3 moles
3 moles
3 moles
Equilibrium
3 / 3 = 1 mol l-1
3 /3 = 1 mol l-1
3 / 3 = 1 mol l-1
concentration
Kc = [Br2][Cl2]
[BrCl] 2
Kc = [1] [1]
=1
[1] 2
17