Problem Solving with Equilibrium
Now that we have learned the basic rules of dealing with equilibrium mathematics, let’s take a look at the typical types of problems that you will encounter throughout the unit.
At Equilibrium
In many problems, the information will tell you that the system is already at equilibrium. If this is the case,
nothing further needs to be done other than plugging-and-chugging into your Kc or Kp equation.
Example 1:
At equilibrium, the following system is found to have [N2] = 2.4x10-5, [O2] = 3.1x10-3, and [NO]= 5.2x103.
Calculate Kc for the equation:
N2 (g) + O2 (g) <——> 2 NO (g)
Based upon the above equation:
Kc =
[NO]2
[N2][O2]
Since the system is already at equilibrium, plug in the appropriate numbers in their spots in the equation:
Kc =
[NO]2
Kc =
[N2][O2]
[5.2x103]2
= 3.6x1014
[2.4x10-5][3.1x10-3]
Example 2:
At equilibrium, the following system is found to have PCl2 = 3.4x10-3, PH2O = 5.1x10-2, and PO2 = 7.2x10-4. If
Kp = 0.45, calculate the PHCl.
2 Cl2 (g) + 2 H2O (g) <——> 4 HCl (g) + O2 (g)
Kp =
P4HCl PO2
P2Cl2 P2H2O
0.45 =
P4HCl (7.x10-4)
PHCl= 6.61x10-2
(3.4x10-3)2(5.1x10-2)2
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Dissociation
Instead of telling you what the equilibrium conditions are, you may instead be given starting information and
be told by what percent the system dissociates. Dissociation is simply how much of the starting material reacts
over to the other side. It is usually best to use what is called an ICE table.
ICE stands for
Initial Change Equilibrium
Or to put it another way, how much you start with (initial), how much it changes by (change) and what is left
over once equilibrium is reached (equilibrium). Let’s see how this works.
Example 1:
For the equilibrium system:
PCl5 (g) <——> PCl3 (g) + Cl2 (g)
If you start with 2 atm of PCl5 and the system dissociates 20% to reach equilibrium, calculate Kp.
Set up an ICE table
PCl5
Initial
2
Change
Equilibrium
<——>
PCl3 +
0
Cl2
0
The table is set up so that the correct amounts
are underneath the correct species. Notice that
we are starting only with 2 atm of PCl5 and
none of the other species as the problem didn’t
say we had any.
Now let’s make the system undergo the required dissociation. Since we are starting with 2 atm and the problem says we dissociate by 20%, the system must lose:
20% (2 atm) = 0.4 atm
Initial
Change
Equilibrium
PCl5
2
-0.4
<——>
PCl3 +
0
Cl2
0
We have now dissociated by 20% of the starting amount
But keep in mind that stoichiometry is still in effect here. If we lose 0.4 from the left side we must gain an appropriate amount on the right side:
PCl5 <——>
PCl3 +
Cl2
Stoichiometric principles are now satisfied
Initial
2
0
0
Change
-0.4
+0.4
+0.4
Equilibrium
So now calculate what is left at equilibrium:
PCl5 <——>
PCl3 +
Initial
2
0
Change
-0.4
+0.4
Equilibrium 1.6
0.4
Now that we have a value for each equilibrium spot,
plug the numbers into your Kp equation and solve:
Cl2
0
+0.4
0.4
Stoichiometric principles are now satisfied
Kp = PPCl3PCl2
PPCl5
=
(0.4)(0.4)
= 0.1
1.6
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Example 2:
For the equilibrium system:
2 H2S (g) <——> 2 H2 (g) + S2 (g)
If you start with 0.25 atm of H2S and the system dissociates 4% to reach equilibrium, calculate Kp.
Set up an ICE table
2 H2S <——>
Initial
0.25
Change
Equilibrium
Calculate the dissociation:
2 H2 +
0
S2
0
4% (0.25) = 0.01
But note that the stoichiometry in the above equation is not a 1:1:1 ratio. Note that you only get 1 S2 mole for
every 2 H2S moles. You must take this into account just like we did in the stoichiometry section of the class:
Initial
Change
Equilibrium
2 H2S <——>
0.25
-0.01
0.24
2 H2 +
0
+0.01
0.01
Kp = P2H2PS2
P2H2S
=
S2
0
+0.005
0.005
Note that you only get 1/2 as much S2 as
H2 or 1/2 as much as the H2S dissociated.
Stoichiometric principles must be maintained.
(0.01)2(0.005)
= 8.68x10-6
(0.24)2
Example 3:
For the equilibrium system: 2 NH3(g) <——> N2 (g) + 3 H2 (g)
If you start with 0.015 M of NH3 and by the time equilibrium is established, there is 0.0009 M of H2 present,
calculate the % dissociation of the NH3.
Set up an ICE table
2 NH3 <——>
Initial
0.015
Change
Equilibrium
Initial
Change
Equilibrium
2 NH3 <——>
0.015
N2 +
0
3 H2
0
N2 +
0
3 H2
0
0.0009
Initial Conditions
But note that at equilibrium there must be
0.0009 M H2 present
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Since you started with 0 and ended with 0.009 M H2 it must have increased by 0.009 M
Initial
Change
Equilibrium
2 NH3 <——>
0.015
N2 +
0
3 H2
0
+0.0009
0.0009
According to the problem this must have
happened
Now apply stoichiometric principles in the correct direction. You only get 1 N2 for every 3 H2 moles and you
only use 2 NH3 moles for every 3 moles of H2
Initial
Change
Equilibrium
2 NH3 <——>
0.015
-0.0006
N2 +
0
+0.0003
3 H2
0
+0.0009
0.0009
2 NH3 <——>
0.015
-0.0006
0.0144
N2 +
0
+0.0003
0.0003
3 H2
0
+0.0009
0.0009
Now apply stoichiometry in the correct
direction
That results in:
Initial
Change
Equilibrium
Here are the equilibrium results
The problem is asking you for the % dissociation. Remember that a % is simply the part over the total so a %
dissociation is the dissociation (change) over the total (initial):
% dissociation = Change
Initial
=
0.0006
*100% = 4%
0.015
Example 4:
A system starts with 3 atm of N2O4. At equilibrium, there is 0.50 atm of NO2 present. What is the % dissociation of N2O4?
Initial
Change
Equilibrium
N2O4 <——>
3
-0.25
2.75
2 NO2
0
+0.5
0.5
% dissociation = Change
Initial
=
0.25
*100% = 8.3%
3
Again note here that you only lose 1/2 as
much N2O4 as you gained of the NO2 because of the stoichiometry
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Q vs. K
All of the problems we have seen so far have not been at equilibrium to begin with and it has been obvious
why not. If you examine every previous problem, at least one of the species present has had an initial value of
zero. A system cannot be at equilibrium if a species is at zero so the reaction must go towards that side to
make at least a little bit of that species. But what if you are given initial conditions where the species are all
non-zero values? Is the system at equilibrium? How we will know? If it is not at equilibrium, what direction
will the reaction proceed, towards the reactants or the products? Let’s examine a typical problem of this sort:
Example 1: For the reaction seen below, the initial conditions are made so that
PCO = 2.5 atm PBr2 = 1.4 atm PCOBr2 = 5.2 atm
CO (g) + Br2 (g) <——> COBr2 (g)
Kp = 14.5
Is the system at equilibrium? If not, which way will the reaction proceed; left or right?
First, how do we know if a system is at equilibrium? The ratio of products to reactants will equal K, right?
Let’s do a test case called Q.
Q = reaction quotient
It is a test case for K
In all respects, Q is the same thing as K; it just might not be at equilibrium. Let’s take a look:
CO (g) +
Br2 (g) <——>
COBr2 (g)
Initial
2.5
1.4
5.2
Calculate Q (which is the same format as Kp) so
Q = PCOBr
PCOPBr2
=
5.2
= 1.49
(2.5)(1.4)
Is the system at equilibrium? If it was at equilibrium, the value would have to be 14.5 as stated above in the
problem. The system is clearly not at equilibrium because 1.49 does not equal 14.5 or:
Q≠K
For the reaction to be at equilibrium, Q must equal K. So what, then, has to happen to the equation? Does Q
have to get bigger or smaller? Q must get bigger because 1.49 < 14.5
Since Q and K are both:
Products
the only way for Q to get bigger is to go to the products side
Reactants
By going to the products side, the numerator gets bigger, the denominator gets smaller and the ratio of products to reactants increases until Q = K. If Q was instead too BIG, then the reaction would proceed from the
products towards the reactants, thus making the numerator smaller and the denominator bigger so that Q got
smaller until again, Q = K. To summarize:
Q = reaction quotient; equilibrium test case
Q=K
Reaction is at equilibrium
Q<K
Reaction proceeds to the right
Q>K
Reaction proceeds to the left
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Example 2:
For the reaction:
H2 (g) + I2 (g) <——> 2 HI (g)
Is the reaction at equilibrium if
[H2] = 0.42 [I2] = 0.67
If not at equilibrium which way will the reaction proceed?
First step is to calculate Q
Q=
[HI]2
[H2][I2]
Kc = 0.754
[HI] = 0.96
0.962
=
= 3.28
(0.42)(0.67)
We see that Q > K
because 3.28 > 0.754 from the given information
Thus the system is NOT at equilibrium and because Q is too big, the system will proceed to the left side
Example 3:
For the reaction:
HN3 (aq) + H2O (l) <——> H3O+1 (aq) + N3-1 (aq)
Kc = 4x10-6
Is the reaction at equilibrium if
[HN3] = 2x10-6
[H3O+1] = 1x10-9
[N3-1] = 2x10-4
If not at equilibrium which way will the reaction proceed?
First step is to calculate Q
Q = [H3O+1][N3-1]
[HN3]
=
(1x10-9)(2x10-4) = 1x10-7
2x10-6
We see that Q < K
because 1x10-7 < 4x10-6 from the given information
Thus the system is NOT at equilibrium and because Q is too small, the system will proceed to the right
side
Example 4:
For the reaction:
H3PO4 (aq) + 3 H2O (l) <——> 3 H3O+1 (aq) + PO4-3 (aq)
[H3O+1] = 3x10-3 [PO4-3] = 5x10-24
Is the reaction at equilibrium if:
Kc = 5x10-22
[H3PO4] = 0.01
If not at equilibrium, which way will the reaction proceed?
First step is to calculate Q
Q=
[H3O+]3[PO4-3]
[H3PO4]
We see that Q < K
=
(3x10-3)3(5x10-24)
= 1.35x10-29
0.01
because 1.35x10-29 < 5x10-22 from the given information
Thus the system is NOT at equilibrium and because Q is too small, the system will proceed to the right
side
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Calculating Equilibrium Conditions
We now know how to determine if a system is at equilibrium or not. We can even determine which way the
system goes if we have to. But what about the last part; the hardest part? If a system is NOT at equilibrium,
what conditions can it shift to so that it IS at equilibrium? Let’s examine a simple example:
Example 1:
The molecule butane, C4H10 (g) can isomerize into the molecule methyl propane C3H7CH3 (g) by the following
equation:
C4H10 (g) <——> C3H7CH3 (g)
Kc = 3
If you start with 2 M C4H10 (g), what are the concentrations of all species at equilibrium?
To do this, we need to use all the skills we learned before so let’s set up an ICE table:
Initial
Change
Equilibrium
C4H10 (g) <——>
2
C3H7CH3 (g)
0
The system is clearly NOT at equilibrium
as this value is 0
We can clearly see that this equation is not at equilibrium because the products are starting at a value of 0. So
we know the reaction will proceed to the right; what we don’t know is how FAR the reaction will proceed to
the right. Because of this, we can set the change value of the problem as “x” because we don’t know! So:
Initial
Change
Equilibrium
C4H10 (g) <——>
2
-x
2-x
C3H7CH3 (g)
0
+x
x
We don’t know how far the reaction will
go so we insert the value of x
We now have values for each of the equilibrium substances. We can also write the Kc equation for this and
insert its value from the given information above so:
KC =
[C3H7CH3]
x
[C4H10]
2-x
=3
Re-arranging the equation we get:
x = 6 – 3x
or
4x = 6
or x = 1.5
Plugging our value of x = 1.5 into the ICE table above we get our equilibrium answers:
C4H10 (g) <——>
C3H7CH3 (g)
We can always check to see if we’re right. Plug
Initial
2
0
the values back into the K equation and it should
Change
-1.5
+1.5
equal the given K value:
Equilibrium
0.5
1.5
K = [C3H7CH3] = 1.5 = 3
[C4H10]
0.5
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Example 2:
The system below has a Kp = 20. Calculate the equilibrium pressures of all species if
PN2 = 0.25 and PO2 = 0.25
N2 (g) + O2 (g) <——> 2 NO (g)
Initial
Change
Equilibrium
Kp =
N2
+
0.25
-x
0.25 –x
P2NO
O2
<——>
0.25
-x
0.25 –x
20 =
PN2PO2
2 NO
0
+2x
2x
(2x)2
This looks really awful to solve but you
can get around it by taking the square root
of everything
(0.25 –x)2
(2x)
4.47 =
Now solve for x like normal
(0.25 –x)
1.12—4.47 x = 2x
Note that you have to remember that if the
left goes down by x, the right has to go up
by 2x in this case because of the stoichiometry
1.12 = 6.47x
x = 0.173
Plugging x into our equation above we get:
Initial
Change
Equilibrium
N2
+
0.25
-0.173
0.077
O2
<——>
0.25
-0.173
0.077
2 NO
0
+2(0.173)
0.346
Here are our answers
Double-check to make sure you have the right answers by plugging back into K:
Kp =
P2NO
PN2PO2
Kp =
(0.346)2
(0.077)2
= 20.2
This is really darn close to the expected
value of 20. It is within rounding errors.
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Example 3:
For the equation:
SO2Cl2 (g) <——> SO2 (g) + Cl2 (g)
Kp = 3.4
If enough SO2Cl2 is put into a container so its pressure is 5 atm, what is the total pressure in the container?
Initial
Change
Equilibrium
SO2Cl2 <——>
5
-x
5-x
PSO2PCl2
Kp =
PSO2Cl2
17-3.4x = x2
3.4 =
SO2
0
+x
x
+
Cl2
0
+x
x
x2
No easy way out of this one. We have to
do a quadratic equation. Need to get it
into correct format
5-x
0 = x2 + 3.4x +17
Now solve for x using quadratic equation
x = 2.76
Initial
Change
Equilibrium
SO2Cl2 <——>
5
-2.76
2.24
SO2 +
0
+2.76
2.76
Cl2
0
+2.76
2.76
Here are our answers
But the problem asks for the total pressure in the system at equilibrium so:
Ptot = 2.24 + 2.76 + 2.76 = 7.76 atm
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Example 4:
For the equation:
NH4HS (s) <——> NH3 (g) + H2S (g)
Kp = 4.8
If 100 grams of NH4HS are placed into a 15 L container at 500 K, how many grams of the solid remain at equilibrium?
We’re going to have to use our knowledge of gas laws for this problem. First it is important to recognize that
the solid NH4HS will dissociate by some amount x so that the products are each made. What is strange is that
the x for the NH4HS will not actually be used in our calculations for K, though, because it is a solid. Let’s set
up the ICE table:
Initial
Change
Equilibrium
NH4HS (s) <——> NH3 (g) +
some
0
-x
+x
x
H2S (g)
0
+x
x
Kp = PNH3PH2S =
x2 = 4.8
x = 2.19
It’s key to note here that the NH4HS must decrease
by x but we ignore it mathematically because it is a
solid
Since x = 2.19 atm, we need to be able to turn this into grams of NH4HS.
Using our gas laws knowledge and the equation PV = nRT we can do this:
PV = nRT
(2.19 atm) (15 L) = (n) (0.0821 Latm/molK)(500 K)
n = 0.80 moles NH4HS
Knowing the molar mass of NH4HS is 51.1 grams/mole we get:
(0.80 moles)(51.1 g/mole) = 40.9 grams NH4HS reacted away
Since there was 100 grams of NH4HS to begin with an 40.9 g of it reacted there is 100-40.9 g = 59.1 g left
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