Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, September 2015
Exercises- General Chemistry II (Chem 2020)
Exercises-General chemistry II (Chem 2020)
Chapter-3-Chemical equilibrium
Exercise 1
Write the equilibrium expression for the following reaction:
4NH3 (g) + 7O2 (g) ⇌ 4NO2 (g) + 7H2 𝑂(g)
Solution
The equilibrium constant of the reaction is:
[NO2 ]4 [H2 O]7
K=
[NH3 ]4 [O2 ]7
Exercise 2
Write the equilibrium expression Kc for the following reactions:
(a) 2O3 (g) ⇌ 3O2 (g)
(b) 2NO (g) + Cl2 (g) ⇌ 2NOCl(g)
(c) Ag + (aq) + 2NH3 (g) ⇌ Ag(NH3 )+
2 (aq)
Solution
[O2 ]3
(𝑎) K c =
[O3 ]2
(𝑏) K c =
[NOCl]2
[NO]2 [Cl2 ]
(𝑐) K c =
[Ag(NH3 )+
2]
+
[Ag ][NH3 ]2
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Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, September 2015
Exercises- General Chemistry II (Chem 2020)
Exercise 3
Consider the following reaction:
3 H2 (g) + N2 (g) ⇌ 2NH3 (g)
𝐾𝑐,500 °𝐶 = 6.0 × 10−2
If 0.25 mol/L of H2 and 0.05 mol/L of NH3 are present at equilibrium, what is the concentration
of N2 at equilibrium?
Solution
By definition, the equilibrium constant of the reaction is:
[NH3 ]2
Kc =
[H2 ]3 [N2 ]
Thus,
[N2 ] =
[N2 ] =
[NH3 ]2
[H2 ]3 K c
(0.05 mol/L)2
(0.25 mol/L)3 × 6.0 × 10−2
[N2 ] = 2.67 mol/L
Exercise 4
An amount of PCl5 in vessel, 12 L, was heated. At the equilibrium, it was found that this vessel
contain of 0.21 mol of PCl5, 0.32 mol of PCl3 and 0.32 mol of Cl2. Calculate the equilibrium
constant for the decomposition of PCl5 at 25 °C.
PCl5 ⇌ PCl3 + Cl2
Solution
First, we calculate the molarities of PCl5, PCl3 and Cl2, the we calculate the equilibrium
constant.
[PCl5 ] =
n(PCl5 ) 0.21 mol
=
= 0.0175 mol/L
V
12 L
[PCl3 ] =
n(PCl3 ) 0.32 mol
=
= 0.0266 mol/L
V
12 L
[Cl2 ] =
n(Cl2 ) 0.32 mol
=
= 0.0266 mol/L
V
12 L
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Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, September 2015
Exercises- General Chemistry II (Chem 2020)
By definition, the equilibrium constant of the above reaction is:
Kc =
Kc =
[PCl3 ][Cl2 ]
[PCl5 ]
(0.0266 mol/L)(0.0266 mol/L)
0.0175 mol/L
𝐊 𝐜 = 𝟎. 𝟎𝟒
Exercise 5
The reaction for the formation of nitrosyl chloride was studied at 25 °C:
2NO(g) + Cl2 (g) ⇌ 2NOCl(g)
The partial pressures at equilibrium were found to be:
PNOCl = 1.2 atm; PNO = 0.05 atm; PCl2 = 0.3 atm
Calculate the equilibrium constant Kp for the formation of nitrosyl chloride at 25 °C.
Solution
By definition, the equilibrium constant of the above reaction is:
Kp =
Kp =
2
PNOCl
2
PNO
PCl2
(1.2)2
(0.05)2 × (0.3)
𝐊 𝐩 = 𝟏. 𝟗𝟐 × 𝟏𝟎𝟑
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Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, September 2015
Exercises- General Chemistry II (Chem 2020)
Exercise 6
Write the expression of K and Kp for the following processes:
(a) PCl5 (s) ⇌ PCl3 (l) + Cl2 (g)
(b) CuSO4 . 5H2 O(s) ⇌ CuSO4 (s) + 5H2 O(g)
Solution
The equilibrium expressions are:
(a) K = [Cl2 ] and K p = PCl2
(b) K = [H2 O]5 and K p = (PH2 O )
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Note: The concentration of solids are not considered [Solid] = 1.
Exercise 7
Calculate kp of the following reaction:
3 H2 (g) + N2 (g) ⇌ 2NH3 (g)
Solution
We have:
𝐾𝑐,300 𝐾 = 9.6
K p = K c (RT)∆n
Where,
∆n = ∑ 𝑛𝑔𝑎𝑠𝑒𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 − ∑ 𝑛𝑔𝑎𝑠𝑒𝑠 𝑟𝑒𝑎𝑐𝑡𝑒𝑑
For the above reaction
Thus,
∆n = 2 − (3 + 1) = −2
K p = 9.6 × (0.0821 × 300)−2
𝐊 𝐩 = 𝟏. 𝟔 × 𝟏𝟎−𝟐
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Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, September 2015
Exercises- General Chemistry II (Chem 2020)
Exercise 8
Calculate kp of the following reaction:
2NOCl(g) ⇌ 2NO(g) + Cl2 (g)
Solution
We have:
𝐾𝑝,500 𝐾 = 1.7 × 10−2
K c = K p (RT)−∆n
Where,
∆n = ∑ 𝑛𝑔𝑎𝑠𝑒𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 − ∑ 𝑛𝑔𝑎𝑠𝑒𝑠 𝑟𝑒𝑎𝑐𝑡𝑒𝑑
For the above reaction
Thus,
∆n = (2 + 1) − 2 = 1
K c = 1.7 × 10−2 × (0.0821 × 500)−1
𝐊 𝐜 = 𝟒. 𝟏𝟒 × 𝟏𝟎−𝟒
Exercise 9
Consider the reaction:
2H2 S(g) + CH4 (g) ⇌ 4H2 (g) + CS2 (g)
The equilibrium constant of the above reaction is: Kp = 3.31 × 10-4. Calculate Kp of the
following reaction is:
(a) 4H2 S(g) + 2CH4 (g) ⇌ 8H2 (g) + 2CS2 (g)
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1
(b) H2 S(g) + 2 CH4 (g) ⇌ 2H2 (g) + 2 CS2 (g)
(c) 4H2 (g) + CS2 (g) ⇌ 2H2 S(g) + CH4 (g)
Solution
(a) Kp = (kp)2
= (3.31 × 10-4)2 = 1.1 × 10-7
(a) Kp = (kp)1/2
= (3.31 × 10-4)1/2 = 1.8 × 10-2
1
(a) K p = k
=
p
1
3.31 × 10−4
= 3.02 × 103
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Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, September 2015
Exercises- General Chemistry II (Chem 2020)
Exercise 10
0.003 moles of H2 (g) and 0.006 moles of I2 (g) are introduce into 3 L vessel and allowed to
reach equilibrium at 448 °C.
H2 (g) + I2 (g) ⇌ 2HI(g)
Calculate kC if it is found that number of moles of HI (g) at equilibrium is 0.00561 moles.
Solution
First, we calculate the initial concentrations of reagents and products.
[H2 ]0 =
[I2 ]0 =
n (H2 ) 0.003
=
= 0.001 M
V
3
n (I2 ) 0.006
=
= 0.002 M
V
3
[HI2 ]0 = 0
To solve this problem, we need to construct a table as follow:
Equation
Eq. mol
Initial concentrations (M)
Change in concentrations (M)
Equilibrium concentrations (M)
𝐇𝟐 (𝐠) +
1
0.001
-x
0.001-x
𝐈𝟐 (𝐠)
1
0.002
-x
0.002-x
⇌
𝟐𝐇𝐈 (𝐠)
2
0
+2x
0 + 2x = 2x
At the equilibrium
[HI2 ]𝑒𝑞 =
0.00561
= 0.00187 M = 2x
3
Thus,
x=
Hence,
0.001871
= 9.35 × 10−4
2
[H2 ]𝑒𝑞 = 0.001 − x = 0.001 − 9.35 × 10−4 = 6.5 × 10−5 M
[I2 ]𝑒𝑞 = 0.002 − x = 0.001 − 9.35 × 10−4 = 1.065 × 10−3 M
The equilibrium constant of the formation of HI is:
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Chemistry Department
College of Science and Humanities Studies
Prince Sattam Bin Abdulaziz University
Al-kharj, September 2015
Exercises- General Chemistry II (Chem 2020)
[HI]2𝑒𝑞
Kc =
[H2 ]𝑒𝑞 [I2 ]𝑒𝑞
(1.87 × 10−3 )2
Kc =
(6.5 × 10−5 )(1.065 × 10−3 )
𝐊 𝐜 = 𝟓𝟎. 𝟓
Exercise 11
Consider the equilibrium synthesis of ammonia at 500 °C.
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
The equilibrium constant is 6.0 ×10-2.
Predict the direction in which the system will shift to reach equilibrium in each of the cases:
(a) [NH3]0 = 10-3 M; [N2]0 = 10-5 M; [H2]0 = 2 × 10-3 M
(b) [NH3]0 = 2 × 10-3 M; [N2]0 = 1.5 ×10-5 M; [H2]0 = 3.54 × 10-1 M
(c) [NH3]0 = 10-4 M; [N2]0 = 5.0 M; [H2]0 = 10-2 M
Solution
To predict the direction in each case, first we calculate Q and then we compared to the
equilibrium constant K.
Q=
[NH3 ]2
[NH3 ]20
=
[H2 ]3 [N2 ] [H2 ]30 [N2 ]0
(10−3 )2
(a) Q = (2×10−3 )3 (10−5 ) = 1.3 × 107 , Q > K => The system will shift to the left.
(2×10−4 )2
(b) Q = (3.54 ×10−1 )3 (1.5 × 10−5 ) = 6.0 × 10−2 , Q = K => The system is at equilibrium (no shift
will occur).
(10−4 )2
(c) Q = (10−2 )3 (5) = 2.0 × 10−3 , Q < K => The system will shift to the right.
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