The Haber process proceeds as follows 2NH 3 g 92KJ N 2 g 3H 2 g If the equilibrium concentrations are NH 3 31x10 2 M N 2 85x10 1 M H 2 ID: 778732
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Slide1
Equilibrium Follow-up
Slide2Equilibrium Problems
The Haber process proceeds as follows:
2NH
3 (g)
+ 92KJ ↔ N
2 (g)
+ 3H
2 (g)
If the equilibrium concentrations are:
[NH
3
] = 3.1x10
-2
M
[N
2
] = 8.5x10
-1
M
[H
2
] = 3.1x10
-3
M
Then what is the value for K?
Is the forward reaction favored?
How can we increase the amount of product?
Slide3Answer
K =
[H
2
]
3
[N
2
]
[NH
3
]
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 NH
3
, or removing H
2
or N
2
.
Slide4Kp
And
K
c
In the reaction at 25
o
C :
2NO
(g)
+ Cl
2(g)
↔ 2NOCl
(g)
The equilibrium pressures were found to be:
P
NOCl
= 1.2atm
P
NO
= 5.0x10
-2
atm
P
Cl2
= 3.0x10
-1
atm
What is the
K
p
and
K
c
for this reaction?
Slide5Answer
K
p
=
P
NOCl
2
(P
NO
)
2
(P
Cl2
)
K
p
=
1.2
2
(5.0x10
-2
)
2
(3.0x10
-1
)
K
p
= 1.9x10
3
K
p
=
K
c
(RT)
∆n
∆n = 2-3 = -1
So K
P
=
K
c
/ RT or
K
c
=
K
p
(RT)
K
c
= 1.9x10
3
(0.0821)(298)
= 4.6x10
4
Slide6The 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.
Slide7You try!
Haber again!
N
2 (g)
+ 3H
2 (g)
↔ 2NH
3 (g)
Give the concentrations:
[NH
3
] = 1.0x10
-3
M
[N
2
] = 1.0x10
-5
M
[H
2
] = 2.0x10
-3
M
Calculate the value for Q and predict which direction the equilibrium will shift if the K is 6.0x10
-2
.
Slide8Answer
Q =
(1.0x10
-3
)
2
(1.0x10
-5
)(2.0x10
-3
)
3
Q = 1.3x10
7
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.
N
2 (g)
+ 3H
2 (g)
← 2NH
3 (g)
Slide9Calculating Equilibrium Pressures
Lets start easy:
In the reaction N
2
O
4
↔ 2NO
2
has a
K
p
value of 0.133. If the pressure of N
2
O
4
at equilibrium is 2.71atm, then what is the Equilibrium pressure of NO
2
?
Tougher:
A 1.00L flask initially contained 0.298mol of PCl
3
and 8.70x10
-3
mol
of PCl
5
. After the system reached equilibrium, 2.00x10
-3
mol
of Cl
2
was found. Calculate K
(c)
and the Equilibrium concentrations.
PCl
5
↔ PCl
3
+ Cl
2
Slide10Answers
K
p
= P
NO2
2
/P
N2O4
0.133 = P
NO2
2
/ 2.71
P
NO2
2
= 0.360
P
NO2
= 0.600
Slide11Answer Part 2
K = [Cl
2
][PCl
3
]/[PCl
5
]
Concentrations must be found first.
[Cl
2
]
o
= 0
[PCl
3
]
o
= 0.298mol/1L = 0.298M
[PCl
5
]
o
= 8.70x10
-3
M
Next we find the change to reach equilibrium.
∆[Cl
2
] = +2.00x10
-3
M
Since they all have coefficients of 1, then
∆[PCl
3
] = +2.00x10
-3
M
∆[PCl
5
] = -2.00x10
-3
M
Slide12Part 2 finish
Final concentrations of gases:
[Cl
2
] = 2x10
-3
M
[PCl
3
] = 0.300M
[PCl
5
] = 6.70x10
-3
M
K = (2x10
-3
)(0.300)/(6.70x10
-3
)
K =
8.96x10
-2
Slide13ICE
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
Slide14ICE problem
For the reaction:
CO + H
2
O ↔ CO
2
+ H
2
O
The K value is 5.10. Calculate the equilibrium concentrations if the initial concentrations of all components is 1M.
Slide15Answer
[CO]
[H
2
O]
[CO
2
]
[H
2
]
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],[H
2
O] = 0.613M [CO
2
],[H
2
] = 1.387M
Slide16You try!
H
2
+ F
2
↔ 2HF
If the equilibrium Constant is 1.15x10
2
then if 3.00 moles of each substance were added to a 1.5L flask, what will the equilibrium concentrations be?
Slide17answer
Must find Q first to know which way reaction will shift.
Q = [HF]
2
/[H
2
][F
2
]
All are 2M gas solutions so
Q = 2
2
/(2)(2) = 1
Since Q is less then K reaction shifts right.
[H
2
]
[F
2
]
[HF]
I
2M
2M
2M
C
-x
-x
+2x
E2M-x
2M-x2M + 2x
Slide18Answer finish
X = 1.528 so:
[H
2
] = [F
2
] = 0.472M
[HF] = 5.056M
Slide19Small 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.
Slide20You Try
2NOCl
(g)
↔
2NO
(g)
+
Cl
2(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?
Slide21A
nswer
Initial concentration of
NOCl
is 1mol/2L or 0.5M.
2NOCl
↔ 2NO +
Cl
2
K = [NO]
2
[Cl
2
]/[
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.
[
NOCl
]
[NO]
[Cl2]
I0.50
0C-2x+2x
+xE0.5-2x
+2x+x
Slide22Answer Cont.
So the equations becomes:
1.6x10
-5
= 4x
3
/(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
-2
M and [Cl
2
] = 1.0x10
-2
M
Slide23One More time with Haber!
N
2 (g)
+ 3H
2 (g)
↔ 2NH
3 (g)
If the equilibrium concentrations of N
2
, H
2
, and NH
3
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?
Slide24A
nswer
First use the given equilibrium concentrations to find the value for K.
K = 5.96x10
-2
Clearly since N
2
was added, then the reaction will shift right.
5.96x10
-2
= (0.202-2x)
2
/((1.399-x)(1.197-3x)
3
)
[N
2
]
[H
2
]
[NH
3
]
I
1.399
1.197
0.202
C
-x-3x
+2xE1.399-x
1.197-3x0.202+2x