Equilibrium Follow-up
Equilibrium Problems
• The Haber process proceeds as follows:
2NH3 (g) + 92KJ ↔ N2 (g) + 3H2 (g)
If the equilibrium concentrations are:
[NH3] = 3.1x10-2M
[N2] = 8.5x10-1M
[H2] = 3.1x10-3M
Then what is the value for K?
Is the forward reaction favored?
How can we increase the amount of product?
Answer
• K = [H2]3[N2]
[NH3]2
K = (3.1x10-3)3 (8.5x10-1)
(3.1x10-2)2
K = 2.6x10-5
-No it is not favored, the K value demonstrates
that the reactant is more likely to be present at
equilibrium than the products.
-Product can be increased by increasing the
temperature, decreasing the pressure, increasing
the amount of NH3, or removing H2 or N2.
Kp And Kc
• In the reaction at 25oC :
2NO(g) + Cl2(g) ↔ 2NOCl (g)
The equilibrium pressures were found to be:
PNOCl = 1.2atm
PNO = 5.0x10-2atm
PCl2 = 3.0x10-1atm
• What is the Kp and Kc for this reaction?
Answer
• Kp = PNOCl2
(PNO)2(PCl2)
1.22
(5.0x10-2)2(3.0x10-1)
Kp = 1.9x103
Kp = Kc (RT)∆n
∆n = 2-3 = -1
So KP = Kc / RT or Kc = Kp (RT)
Kc = 1.9x103 (0.0821)(298)
= 4.6x104
Kp =
The Reaction Quotient
• It is possible to measure the concentrations of the
reactants and products in a reaction vessel at any
time.
• They may not necessarily be at equilibrium yet.
• When plugging those values into the Law of Mass
Action, you will still get a value, but it might not be
K.
• We call this value the reaction quotient, and
comparing this value to K, we can see which
direction the reaction is favored to move.
You try!
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Haber again!
N2 (g) + 3H2 (g) ↔ 2NH3 (g)
Give the concentrations:
[NH3] = 1.0x10-3M
[N2] = 1.0x10-5M
[H2] = 2.0x10-3M
Calculate the value for Q and predict which
direction the equilibrium will shift if the K is
6.0x10-2.
Answer
• Q = (1.0x10-3)2
(1.0x10-5)(2.0x10-3)3
• Q = 1.3x107
Since K = 6.0x10-2 and Q is significantly higher, then
the concentrations of the products needs to go
down, and reactants up meaning the reaction will
shift left.
N2 (g) + 3H2 (g) ← 2NH3 (g)
Calculating Equilibrium Pressures
• Lets start easy:
• In the reaction N2O4 ↔ 2NO2 has a Kp value of
0.133. If the pressure of N2O4 at equilibrium is
2.71atm, then what is the Equilibrium pressure of
NO2?
• Tougher:
• A 1.00L flask initially contained 0.298mol of PCl3 and
8.70x10-3 mol of PCl5. After the system reached
equilibrium, 2.00x10-3 mol of Cl2 was found.
Calculate K(c) and the Equilibrium concentrations.
• PCl5 ↔ PCl3 + Cl2
Answers
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Kp = PNO22/PN2O4
0.133 = PNO22/ 2.71
PNO22 = 0.360
PNO2 = 0.600
Answer Part 2
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K = [Cl2][PCl3]/[PCl5]
Concentrations must be found first.
[Cl2]o = 0
[PCl3]o = 0.298mol/1L = 0.298M
[PCl5]o = 8.70x10-3M
Next we find the change to reach equilibrium.
∆[Cl2] = +2.00x10-3M
Since they all have coefficients of 1, then
∆[PCl3] = +2.00x10-3M
∆[PCl5] = -2.00x10-3M
Part 2 finish
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Final concentrations of gases:
[Cl2] = 2x10-3M
[PCl3] = 0.300M
[PCl5] = 6.70x10-3M
K = (2x10-3)(0.300)/(6.70x10-3)
K = 8.96x10-2
ICE
• By knowing starting concentrations and K values,
it is possible to determine the concentrations
once equilibrium is reached.
• It is based on the stoichiometry of the problem.
• “I” stands for initial concentration
• “C” stands for change in concentration
• “E” stands for equilibrium concentration
ICE problem
• For the reaction:
• CO + H2O ↔ CO2 + H2O
• The K value is 5.10. Calculate the equilibrium
concentrations if the initial concentrations of all
components is 1M.
Answer
[CO]
[H2O]
[CO2]
[H2]
Initial
1M
1M
1M
1M
Change
-x
-x
+x
+x
Equilibrium
1M-x
1M-x
1M+x
1M+x
5.10 = (1+X)(1+X)/((1-X)(1-X))
5.10 = (1+X)2/(1-X)2
2.26 = (1+X)/(1-X)
X = 0.387M
[CO],[H2O] = 0.613M [CO2],[H2] = 1.387M
You try!
• H2 + F2 ↔ 2HF
• If the equilibrium Constant is 1.15x102 then if
3.00 moles of each substance were added to a
1.5L flask, what will the equilibrium
concentrations be?
answer
• Must find Q first to know which way reaction
will shift.
• Q = [HF]2/[H2][F2]
• All are 2M gas solutions so
• Q = 22/(2)(2) = 1
• Since Q is less then K reaction shifts right.
[H2]
[F2]
[HF]
I
2M
2M
2M
C
-x
-x
+2x
E
2M-x
2M-x
2M + 2x
Answer finish
• X = 1.528 so:
• [H2] = [F2] = 0.472M
• [HF] = 5.056M
Small K values
• If K is very small, and concentrations of
substances are large, then the change is
insignificant.
• X can be dropped at that point from the
change of the reactants which makes the
problem easier to solve.
You Try
• 2NOCl(g) ↔ 2NO(g) + Cl2(g)
• If the K is 1.6x10-5, what will the equilibrium
concentrations be if 1.0mol of NOCl is added
to an empty 2.0L flask?
Answer
• Initial concentration of NOCl is 1mol/2L or 0.5M.
2NOCl ↔ 2NO + Cl2
[NOCl]
[NO]
[Cl2]
I
0.5
0
0
C
-2x
+2x
+x
E
0.5-2x
+2x
+x
K = [NO]2[Cl2]/[NOCl]2
1.6x10-5= (2x)2(x)/(0.5-2x)2
Since K is to the -5 and the starting concentration is
4 powers of ten greater, then we can ignore the
change in the reactant concentration.
Answer Cont.
• So the equations becomes:
1.6x10-5 = 4x3/(0.5)2
X = 1x10-2
As long as the change in concentration is smaller
than 5%, it is valid to ignore the change.
Plugging the change into the law of mass action and
demonstrating that the K hasn’t changed is also a
valid way to check.
So the final concentrations are [NOCl] = 0.5M, [NO]
= 2.0x10-2M and [Cl2] = 1.0x10-2M
One More time with Haber!
• N2 (g) + 3H2 (g) ↔ 2NH3 (g)
• If the equilibrium concentrations of N2, H2, and
NH3 were .399M, 1.197M, and 0.202M
respectively, then what would the new equilibrium
expression be under the same conditions if
enough nitrogen gas was added to the system in
order to raise the concentration of nitrogen to
1.399M?
Answer
• First use the given equilibrium concentrations to
find the value for K.
• K = 5.96x10-2
• Clearly since N2 was added, then the reaction will
shift right.
[N2]
[H2]
[NH3]
I
1.399
1.197
0.202
C
-x
-3x
+2x
E
1.399-x
1.197-3x
0.202+2x
• 5.96x10-2 = (0.202-2x)2/((1.399-x)(1.197-3x)3)