Keq Hand-In WORKSHEET
Remember to figure out what type of Keq problem it is before starting
1.
A reaction vessel with a capacity of 1.0 litre, in which the following reaction :
SO2 (g) + NO2 (g) <===> SO3 (g) + NO (g)
had reached a state of equilibrium, was found to contain 0.40 moles of SO3 , 0.30 moles of NO,
0.15 moles of NO2 , and 0.20 moles of SO2. Calculate the equilibrium constant for this
reaction.
[SO3] = 0.40mol/1.0L = 0.40M
[NO] = 0.30mol/1.0L = 0.30M
[NO2] = 0.15mol/1.0L = 0.15M
[SO2] = 0.20mol/1.0L = 0.20M
Keq =
[𝑆𝑂3][𝑁𝑂]
=
[𝑁𝑂2][𝑆𝑂2]
(0.40)(0.30)
(0.15)(0.20)
=
0.12
0.03
= 4.0
2.
Given the equilibrium 2 A(g) + B(g) <===> C(g) + 3 D(g)
and the concentrations [A] = 2.0 M, [B] = 0.50 M, [C] = 0.25 M, [D] = 1.5 M, calculate the value
for the equilibrium constant.
Keq =
3
3
[𝐶][𝐷]
=
2
[𝐴] [𝐵]
(0.25)(1.5)
2
(2.0) (0.50)
=
0.84375
2
= 0. 421875 = 0.42
3.
Consider the following equation H2 (g) + I2 (g) <===> 2 HI(g)
Calculate Keq if at 300oK the concentrations are [H2] = 0.40 M, [I2] = 0.45 M ; [HI] = 0.30 M
Keq =
2
2
[𝐻𝐼]
[𝐻2][𝐼2]
=
(0.30)
(0.40)(0.45)
=
4.
0.09
0.18
= 0.50
The equilibrium constant for the formation of ammonia by the reaction:
N2 (g) + 3 H2 (g) <===> 2 NH3 (g) is 2.0 at a certain temperature.
If the equilibrium concentration of N2 in a mixture is 0.50 M and H2 is 2.0 M, determine the
concentration of ammonia.
2
Keq =
[𝑁𝐻3]
3
[𝑁2][𝐻2]
3
𝐾𝑒𝑞[𝑁2][𝐻2] = [NH3]
3
(2. 0)(0. 50)(2. 0) = [NH3]
[NH3] = 2.828427125
[NH3] = 2.8 M
At 2000oK, a mixture of H2, S2, and H2S rapidly reaches a state of equilibrium
represented by the equation :
2 H2S (g) <===> 2 H 2(g) + S2 (g)
An analysis of the equilibrium mixture in a 2.00 litre flask reveals the presence of 2.00 moles of
H2S and 1.00 mole of H2.
If the Keq for this reaction at this temperature is 1.62 x 10-2,
how many moles of S2 can we expect to find in this flask ?
5.
[H2S] = 2.00mol/2.00L = 1.00M
[H2] = 1.00mol/2.00L = 0.500M
2
[𝐻2] [𝑆2]
Keq =
2
[𝐻2𝑆]
2
[S2] =
[S2] =
𝐾𝑒𝑞[𝐻2𝑆]
2
[𝐻2]
−2
2
(1.62 · 10 )(1.00)
2
(0.500)
[S2] = 0.0648 M
(
0.0648𝑚𝑜𝑙 𝑆2
1𝐿
)(2. 00𝐿) = 0.130 mol S2
6. The equilibrium constant for the formation of CO2 by the reaction 2 CO (g) + O2 (g)
<===> 2 CO2 (g) is 1.0 x 104. If, at equilibrium, the concentration of CO2 is 3.0 M and
the concentration of O2 is 9.0 M, calculate the concentration of CO.
2
Keq =
[𝐶𝑂2]
2
[𝐶𝑂] [𝑂2]
2
[𝐶𝑂] =
[CO] =
[𝐶𝑂2]
𝐾𝑒𝑞[𝑂2]
2
(3.0)
4
(1.0 · 10 )(9.0)
[CO] = 0.010 M
7. A sample containing 0.800 moles of POCl3 is enclosed in a 0.500 litre vessel at a certain
temperature. When the equilibrium for the dissociation reaction: POCl3 (g) <===> POCl
(g) + Cl2 (g) is attained, it is found that the vessel contains 0.259 moles of Cl2. Calculate
the equilibrium constant.
[POCl3] = 0.800mol/0.500L = 1.60M
[Cl2] = 0.259mol/0.500L = 0.518M
POCl3 (g)
⇌
POCl (g)
+
Cl2 (g)
INITIAL
1.60
0
0
CHANGE
-0.518
+0.518
+0.518
EQUIL.
1.082
0.518
0.518
Keq =
Keq =
[𝑃𝑂𝐶𝑙][𝐶𝑙2]
[𝑃𝑂𝐶𝑙3]
2
(0.518)
(1.082)
Keq = 0.2479889094
Keq = 0.248
8. The brown gas NO2 on cooling is converted into the colourless gas N2O4 as described by
the equation: 2 NO2(g) <===> N2O4(g) If the original concentration of NO2 is 0.90 M,
and at equilibrium its concentration is only 0.26 M, what is the equilibrium constant for
the reaction ?
2 NO2 (g)
⇌
N2O4 (g)
INITIAL
0.90
0
CHANGE
-0.64
+0.32
EQUIL.
0.26
0.32
Keq =
Keq =
[𝑁2𝑂4]
2
[𝑁𝑂2]
(0.32)
2
(0.26)
Keq = 4.733727811
Keq = 4.7
9.
SO3 (g) + NO (g) <====> NO2 (g) + SO2 (g)
If 0.300 mol of SO3 and 0.300 mol NO were placed in a 1.00 L flask and allowed to react, what
would be the equilibrium concentration of each gas ? (Keq = 0.500)
[SO3] = 0.300mol/1.00L = 0.300M
[NO] = 0.300mol/1.00L = 0.300M
SO3 (g)
+
NO (g)
⇌
NO2 (g)
+
SO2 (g)
INITIAL
0.300
0.300
0
0
CHANGE
-x
-x
+x
+x
0.300 - x
0.300 - x
x
x
EQUIL.
[𝑁𝑂2][𝑆𝑂2]
Keq = 0.500 =
0.500 =
[𝑆𝑂3][𝑁𝑂]
2
=
𝑥
2
(0.300 − 𝑥)
2
𝑥
2
(0.300 − 𝑥)
0. 500 =
𝑥
(0.300 − 𝑥)
0. 500(0.300 - x) = x
0. 500(0.300) - 0. 500x = x
0. 500(0.300) = (1 + 0. 500)x
0.500(0.300)
(1 + 0.500)
= 𝑥
x = 0.1242640687
EQUILIBRIUM CONCENTRATIONS:
[SO3] = [NO] = 0.300 - x = 0.300 - 0.1242640687 = 0.176 M
[NO2] = [SO2] = x = 0.124 M
10.
2 HCl (g) <====> H2 (g) + Cl2 (g)
If 2.00 mol HCl are placed in a 5.00 L flask and allowed to come to the equilibrium shown
above, what will the equilibrium [H2] be ? (Keq = 3.2 x 10-4)
[HCl] = 2.00mol/5.00L = 0.400M
2 HCl (g)
⇌
H2 (g)
+
Cl2 (g)
INITIAL
0.400
0
0
CHANGE
-2x
+x
+x
0.400 - 2x
x
x
EQUIL.
[𝐻2][𝐶𝑙2]
Keq = 3.2 x 10-4 =
−4
3. 2 · 10
=
2
[𝐻𝐶𝑙]
2
=
𝑥
2
(0.400 − 2𝑥)
𝑥
(0.400 − 𝑥)
−4
3. 2 · 10 (0. 400 − 2𝑥) = 𝑥
−4
−4
3. 2 · 10 (0. 400) − 2 3. 2 · 10 𝑥 = 𝑥
−4
−4
3. 2 · 10 (0. 400) = (1 + 2 3. 2 · 10 )𝑥
−4
𝑥=
3.2·10 (0.400)
−4
(1+2 3.2·10 )
x = 0.0069082601
EQUILIBRIUM CONCENTRATION:
[H2] = x = 0.00691 M
11.
H2 (g) + CO2 (g) <====> CO (g) + H2O (g)
If 1.50 mol of each chemical species are placed in a 3.00 L flask and allowed to achieve the
equilibrium above, what mass of CO will be present at equilibrium ? (Keq = 3.59)
[H2] = 1.50mol/3.00L = 0.500M
[CO2] = 1.50mol/3.00L = 0.500M
[CO2] = 1.50mol/3.00L = 0.500M
[H2O] = 1.50mol/3.00L = 0.500M
Q=
[𝐶𝑂][𝐻2𝑂]
2
=
[𝐻2][𝐶𝑂2]
(0.500)
2
(0.500)
= 1.00
Q = 1.00 < Keq = 3.59
Therefore, reaction will shift to the product side to increase Q value to that of Keq.
H2 (g)
+
CO2 (g)
⇌
CO (g)
H2O (g)
INITIAL
0.500
0.500
0.500
0.500
CHANGE
-x
-x
+x
+x
0.500 - x
0.500 - x
0.500 + x
0.500 + x
EQUIL.
Keq = 3.59 =
3. 59 =
2
(0.500 + 𝑥)
2
(0.500 − 𝑥)
(0.500 + 𝑥)
(0.500 − 𝑥)
3. 59(0.500 - x) = 0.500 + x
3. 59(0.500) - 3. 59x = 0.500 + x
3. 59(0.500) - 0.500 = (1 + 3. 59)x
( 3.59(0.500) − 0.500)
(1+ 3.59)
=x
x = 0.1545445822
[CO] = 0.500 + x = 0.500 + 0.1545445822 = 0.6545445822 M
(
+
0.6545445822𝑚𝑜𝑙 𝐶𝑂
1𝐿
)(3. 00𝐿)(
28.01𝑔 𝐶𝑂
1𝑚𝑜𝑙 𝐶𝑂
) = 55.0g CO