Chem
1B
Dr.
Abel
1
Reaction Quotient and Manipulating K Worksheet KEY
1. For the reaction: PCl5 (g) ⇄ PCl3 (g) + Cl2 (g), Kc = 0.0454 at 261°C. If a vessel is
filled with these gases such that the initial concentrations are [PCl3] = 0.50 M, [Cl2] =
0.20 M and [PCl5] = 4.50 M, in which direction will a net change occur?
Qc =
[PCl 3 ][Cl 2 ]
=
(0.50M )(0.20M )
= 0.022
(4.50M )
[PCl 5 ]
K c > Qc at 261 C
To reach equilibrium, the reaction proceed in the forward direction
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2. For the following reaction:
CO (g) + H2O (g) ⇄ CO2 (g) + H2 (g)
Kc = 1.00 at 1100 K. The following amounts of substances are brought together in a 1.0 L
container and allowed to react at this temperature: 1.0 mole CO, 1.0 mole H2O, 2.0 mole CO2
and 2.0 mole H2. Compared to their initial amounts, which of the substances will be
present in greater amount and which in a lesser amount when equilibrium is established?
Qc =
[CO2 ][H 2 ]
=
(2.0M )(2.0M )
= 4.0
(1.0M )(1.0M )
[CO][H 2O]
K c < Qc at 1100K
To reach equilibrium, more reactants need to be formed.
So, at equilibrium, there will be less products and more reactants
3. For the reaction: N2 (g)
+ 3 H2 (g)
2 NH3 (g)
Kc = 2.4 x 10-3 at 1000 K. What are the values of Kc for the following balanced reactions?
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a) 1/3 N2 (g) + H2 (g)
2/3 NH3 (g)
This reaction is 1/3 x the original reaction, thus
K'c = (Kc )1/ 3 = (2.4x10−3 )1/ 3 = 1.3x10−1
b) NH3 (g)
1/2 N2 (g) + 3/2 H2 (g)
This reaction is 1/2 x the reverse of the original so
K'c =
1
1
1
=
1/ 2
−3 1/ 2 = 2.0x10
(Kc )
(2.4x10 )
Chem
1B
Dr.
Abel
2
4.
At a typical operating temperature of an automobile's engine (~ 100 °C), N2 and O2
form NO. NO then combines with more O2 to form NO2, a toxic pollutant:
1. N2 (g) + O2 (g)
2 NO (g)
K1 = 4.3 x 10-25
2. 2 NO (g)
+
O2 (g)
N2 (g) + O2 (g)
+ 2 NO (g) + O2 (g)
N2 (g)
+
2 O2 (g)
K2 = 6.4 x 109
2 NO2 (g)
2 NO (g)
2 NO2 (g)
K1 = 4.3 x 10-25
K2 = 6.4 x 109
2 NO2 (g)
Knet = K1 x K2 = 2.8 x 10-15
b) Prove to yourself that the expression for the overall K is the same as the product of the
K expressions for each individual step
[NO]2
[NO2 ]2
K1 =
and K 2 =
[N 2 ][O2 ]
[NO]2 [O2 ]
[NO]2
[NO2 ]2
[NO2 ]2
K1 × K 2 =
×
=
[N 2 ][O2 ] [NO]2 [O2 ] [N 2 ][O2 ]2
[NO2 ]2
K net =
= K1 × K2 as shown above
[N 2 ][O2 ]2
c) Based on the magnitude of this value for the overall K, would you say that this reaction
is product favored or reactant favored? Explain your answer. Since Knet is very small,
the reaction is reactant favored, as a small amount of product but large amount of
reactant would result in a very small K value.
5. The following equilibrium constants were obtained for the reactions listed below at 500 K:
H2
+
Br2
H2
2 HBr
K2 = 4.81 x 10-41
2H
Br2
K1 = 7.9 x 1011
K3 = 2.2 x 10-15
2 Br
What is the value of K for the reaction: H + Br
HBr?
(NOTE this is sort of like a Hess’s Law type problem where you first have to figure out
how to sum the three reactions together)
We need to take 1/2 of the reverse of the second and third reactions and add
those to 1/2 of the first reaction:
H
1/2 H2
K’ = (1/K2)1/2 = 1.44 x 1020
Br
+ 1/2 H2 + 1/2 Br2
H + Br
1/2 Br2
HBr
HBr
K’ = (1/K3)1/2 = 2.1 x 107
K’ = (K1)1/2 = 8.9 x 105
K = (1.44 x 1020) x (2.1 x 107) x (8.9 x 105) = 2.7 x 1033