Answers Example Equilibrium Problems #1
1) The first step in the industrial synthesis of hydrogen is the reaction of steam and methane to give synthesis gas, a
mixture of carbon monoxide and hydrogen:
H2O(g) + CH4(g)
CO(g)
+ 3 H2(g) ;
Kc = 4.7 at 1400 K
A mixture of reactants and products at 1400 K contains 0.035 M H2O, 0.050 M CH4, 0.15 M CO, and 0.20 M H2. In which
direction does the reaction proceed to reach equilibrium?
Answer: Left to Right; I.e., forward reaction (as written) will occur to reach equilibrium.
Strategy: When faced with the question of "In which direction will reaction occur?", you need to
compare Q to K to see where the system “is” with respect to equilibrium. If Q is larger than
K, then there are too many products compared to reactants to be at equilibrium, and so
reverse reaction will occur. If Q is less than K, then there aren't enough products, and
forward reaction will occur.
Execution of Strategy:
For the reaction mixture in this problem, Q = 0.69:
(0.15)(0.20)
CO H 

 0.6857...  0.69 <
H O CH  (0.035)(0.050)
3
Q
3
2
2
K (4.7)
4
which is less than K, which is 4.7. Thus there are “not enough products (in the numerator) to be at
equilibrium), and thus forward reaction will occur (to get to equilibrium).
NOTE: As forward reaction occurs, Q increases, because you increase the concentration of products
and decrease the concentration of reactants. So yes, the system will reach equilibrium if forward
reaction occurs (as Q will increase until it equals K).
2) Assume you place 0.010 mol of N2O4(g) in a 2.0-L flask at 50. C. After the system reaches equilibrium, [N2O4] = 0.00090
M. What is the value of Kc for this equation (not reaction!) at 50. C:
N2O4(g)
2 NO2(g)
ANSWER: Kc = 0.075
Explanation. Before you really start the problem in earnest, note that you can always write the
equilibrium constant expression for a chemical equation (Law of Mass Action). For the one above, it is
simply:
Kc 
2
[ NO 2 ( g )]eq
[ N 2 O 4 ( g )]eq
Okay, let’s stop for a second and analyze. Do you know what you are trying to get to at this point? The
question asks for the value of the equilibrium constant. Well, looking at the expression above, in order
to get that, you will need just two things: the final (equilibrium) concentrations of N2O4 and NO2! You
are given the final (equilibrium) concentration of N2O4, so essentially this problem comes down to
finding just one thing: the concentration of NO2 after equilibrium is established! Do you see
this? Do the analysis before you start putting concentration values everywhere, or defining x’s, or trying
to follow some pat “procedure” or filling in some “table”! THINK FIRST.
So, in order to get the final concentration of NO2 all you really need to realize is that the amount of it
formed must be related to the amount of N2O4 used up because that is where it comes from! That is,
you use the balanced equation and do some stoichiometry!
Stoichiometry “warning”: Remember to choose either moles or concentrations to work with when doing the
“stoichiometry” part of this type of problem. Remember that strictly speaking, one should use moles when
doing stoichiometry calculations, but as long as the volume of the vessel is not changed during the reaction,
concentration values will be proportional to the mole values and so you can do the stoichiometric calculations
with concentrations.
Okay, let’s use concentration values here (that’s what I typically prefer since then you can put those
values directly into the equilibrium constant expression).
Initial concentrations: [N2O4]o = 0.010 mol/2.0 L = 0.0050 M (be careful with precision!)
[NO2]o = 0 (the problem didn’t mention that any was placed into the container, so you
can assume you do not have any!)
Final concentration of [N2O4]eq is 0.00090 M. This is smaller than the initial amount, which means that
some of it (net) must have reacted as equilibrium was being established. Well, how much of it
reacted? It is just the difference between what you started with and what you have left!
0.0050 M - 0.00090 M = 0.00410 M of N2O4 must have reacted.
From the balanced equation, the amount of NO2 formed must have been two times as much as the
amount of N2O4 that reacted (in moles or molar) (because from the equation, you get two moles of NO2
for every mole of N2O4 that reacts!). Thus [NO2] formed = 2(0.0041 M) = 0.0082 M. Since there was no
NO2 to start with, the equilibrium concentration of NO2 is simply 0.0082 M.
Now we can calculate Kc, the goal of the problem:
Kc 
2
[ NO 2 ( g )] eq
[ N 2 O 4 ( g )] eq

(0.0082) 2
 0.0747  0.075
0.00090
NOTE: We often use an "ICE" table to help sort out the various quantities in a problem like this. So that
is fine, and I’ll show that approach below. I just wanted you to make sure you are really understanding
what you are doing with it (and that it technically isn’t “necessary”)! In the problem above, the initial
situation might have looked like:
(I) Initial Concentrations
(C) Change ([ ]'s used up or formed)
(E)Equilibrium Concentrations
N2O4
0.0050 M
NO2
0
0.00090 M
NOTE: Just because you set up an ICE table does not mean you must define an “x”! You generally
(only) must do that for a problem where you do not know any of the equilibrium values. Here, since you
know the initial and equilibrium values for [N2O4], you can simply use stoichiometric logic to figure out
the amount of change of N2O4 and then the change for NO2 (as I did above "verbally").
You could finish off the table as follows:
(I) Initial Concentrations
(C) Change ([ ]'s used up or formed)
(E)Equilibrium Concentrations
N2O4
0.0050 M
NO2
0
0.00090 – 0.0050 =
C = E – I (because D = final – initial)
-0.0041
0.00090 M
Then:
(I) Initial Concentrations
(C) Change ([ ]'s used up or formed)
(E)Equilibrium Concentrations
N2O4
0.0050 M
NO2
0
0.00090 – 0.0050 =
-2(-0.0041)
-0.0041
0.00090 M
= 0.0082
[NO2] = -2 x [N2O4]
Then:
(I) Initial Concentrations
(C) Change ([ ]'s used up or formed)
(E)Equilibrium Concentrations
N2O4
0.0050 M
NO2
0
0.00090 – 0.0050 =
-2(-0.0041)
-0.0041
= 0.0082
0.00090 M
[NO2] = -2 x [N2O4]
0 + 0.0082
= 0.0082
3) (from Ebbing textbook): The equilibrium constant Kc for the equation:
N2(g) + 3 H2(g)
2 NH3(g)
at 450 C is 0.159. Calculate the equilibrium composition after 1.00 mol N2 is mixed with 3.00 mol H2 in a 2.00 L vessel.
Answers: [N2]eq 0.363 M; [H2]eq 1.09 M; [NH3]eq  0.273 M
Analysis/Reasoning:
The first thing you should do with problems like this is to get a picture of what is actually happening!! Don’t
just start writing expressions based on past problems. Think about the fact that what is happening is that
some reactants are put into a vessel. Since there are no products yet, the reverse rate is zero, so the
forward rate > reverse rate and the system cannot yet be at equilibrium. Reaction will occur in the forward
direction until the rates become equal, at which point equilibrium will be established and the concentrations
will not change anymore. What will these concentrations be? Well, that depends on how much reaction
occurs. In these problems, you don’t know that at first, so you pick one species and call the amount of that
species that reacts “x”. Because the stoichiometry of the reaction relates all concentration changes of
species to one another, all the concentrations at equilibrium can be expressed in terms of “x”. The
equilibrium constant expression can then be used to set up an equation so that x can be determined.
In the problem at hand, first figure out the initial concentrations of things by dividing the initial moles of
substances by the volume of the container (2.0 L). Thus the initial [N2] = 0.500 M and initial [H2] = 1.50 M.
Now you have to examine the balanced equation to see how the various concentrations change as reaction
occurs to establish equilibrium.
N2(g) + 3 H2(g)
2 NH3(g)
It turns out to be easiest (you avoid fractions) to pick the substance with a coefficient of “one” to be the one
whose concentration change you call “x”. So here I will say:
Let x = the concentration (in M) of N2 that reacts (goes away via the reaction) before the system reaches
equilibrium.
I seek the concentrations at equilibrium. The concentration of each species at equilibrium will equal
“what it was initially” plus “what changes occur as the reaction occurs to establish equilibrium”
(i.e., I + C = E).
What happens as equilibrium is established? Some of the reactants react and some products are formed!
How much of each?
Well, because of how I defined “x”, x M of N2 will react, thus leaving 0.500 M – x present at equilibrium.
Since 3 moles of H2 react for each mole of N2 used, then 3x M of H2 reacts, leaving 1.50 – 3x M present at
equilibrium. Since 2 moles of NH3 are produced for each mole of N2, then 2x M of NH3 will be produced and
since there was none to start with, 2x is the concentration of NH3 present at equilibrium:
N2(g) + 3 H2(g)
Concentrations at equilibrium:
0.500 – x
2 NH3(g)
1.50 – 3x
2x
Using an ICE table to show all of the above in a more compact format:
(M)
(I) Initial
(C) Change
(E)Equilibrium
N2
0.500
-x
0.500 - x
H2
1.50
- 3x
1.50 – 3x
NH3
0
+ 2x
2x
At equilibrium, we know that:
Kc  0.159 
[NH3 ]2
[N 2 ][H 2 ]3
so 0.159 
(2x)2
(0.500 - x)(1.50 - 3x)3
I apologize for choosing a problem that was much more annoying mathematically than I realized it would be
at first. To avoid getting a fourth degree polynomial expression on the right, you must realize that you can
factor out a 3 from the right-most expression in the denominator. This will allow you to see that the right
hand side is a perfect square and so you can take the square root of both sides:
0.159 
(2x) 2
(2x) 2
(2x) 2


3
3
3
(0.500 - x)[3(0.50 0 - x)]
(0.500 - x)3 (0.500 - x)
27(0.500 - x) 4
 0.159(27) 
2
(2x)

(0.500 - x) 4
0.159(27) 
2
(2x)
2x
 2.072 
4
(0.500 - x)
(0.500 - x) 2
This leads (finally) to a quadratic equation that you can solve for x:
0.5180 – 2.072x + 2.072x = 2x 
2
x
Technically, this should be 
2.072, but the negative root
will yield a negative value of
x, which is non-sensible in
this problem (where x was
defined as a positive
quantity), so I will omit that
calculation here.
2
2.072x - 4.072x + 0.5180 = 0
 (4.072)  (4.072) 2  4(2.072)(0.5180)
2(2.072)
= +1.829… or 0.1367…
The positive root yields a non-sensible value here ([N2]eq would be negative) and is not considered further.
Using x = 0.1367:
[N2]eq = 0.500 M – 0.1367 M = 0.363 M;
[H2]eq = 1.50 M – 3(0.1367 M) = 1.09 M
[NH3]eq = 2(0.1367 M) = 0.273 M (assume 3 SF here; hard to propagate here with SF rules [not worth it])
Check results by substituting back into the Law of Mass Action:
Kc 
[NH3 ] 2eq
3
2 eq
[N2 ] eq [H ]

(0273)2
0.3631.093
 0.15854...  0.159 Good! (agrees with given value of 0.159)
4)
A 1.00 L reaction vessel is filled with 1.00 mol of H2, 2.00 mol of I2. What will the concentrations of H2, I2, and HI be at
equilibrium (i.e., after equilibrium is reached)? The equilibrium constant Kc (at the T this reaction is carried out) is
50.5.
H2(g) + I2(g)
2 HI(g)
Answers: [H2]eq = 0.065 M ; [I2]eq = 1.065 M; [HI]eq = 1.87 M
Work: If x is defined to be the concentration of H2 that reacts, then you have:
[H2]eq = 1.00 M – x (Don’t forget to calculate concentrations! 1.0 mol/1.0 L = 1.0 M)
[I2]eq = 2.00 M - x
[HI]eq = 0 + 2x
(Note: 2.0 mol HI/1.0 L = 2.0 M)
Using an ICE table to show how the above was arrived at:
(M)
(I) Initial
(C) Change
(E)Equilibrium
H2
1.00
-x
1.00 - x
I2
2.00
-x
2.00 – x
If you substitute these values into: K c 
2
[HI]eq
[H2 ]eq [I2 ]eq
 2x 
 50.5 you will get:
2
(1.00  x )(2.00  x )

2
HI
0
+ 2x
2x
2
 50.5
2
2
4x = 50.5(2 – 2x – x + x ) = 50.5 (2 – 3x + x ) = 101 – 151.5x + 50.5x
 46.5x – 151.5x + 101 = 0
2


151.5 
x
  151.5 
 151.52  446.5101
246.5
4166.25 151.5  64.554..

 2.323...OR 0.93498...
93
93
So, rounding to 3 decimal places (for convenience), x turns out to be either 2.323 or 0.935. 2.323
yields unreasonable values for [H2]eq (it would be negative!). 0.935 yields reasonable values:
Substituting 0.935 back into the equations for [ ]eq’s above (see ICE table, also) yields
[H2]eq = 1.00 – 0.935 = 0.065 M = 0.07 M (technically, this would only have 1 SF; seems non-ideal…)
[I2]eq = 2.00 – 0.935 = 1.065 M = 1.07 M
[HI]eq = 0 + 2(0.935) = 1.87 M
Check results by substituting back into the Law of Mass Action (using unrounded values to minimize
rounding error in the check):
Kc 
[HI]2eq
[H2 ] eq [I2 ] eq

(1.87)2
 50.5149..  50.5 Good! (agrees with given value of 50.5)
0.0651.065
5) The partial pressures in an equilibrium mixture of NO, Cl2, and NOCl at 500 K are as follows:
PNO = 0.240 atm;
PCl
= 0.608 atm;
2
PNOCl
= 1.35 atm. What is Kp at 500 K for the reaction (equation!):
2 NO(g)
+ Cl2(g)
2 NOCl(g)
K=?
Answer: 52.0
Reasoning:
This system is AT EQUILIBRIUM at the “beginning” of the problem (with the partial pressures
given). So just set up the equilibrium constant expression and substitute in the values of
(equilibrium) partial pressures given in the problem.
P 

P  P
2
Kp
NOCl
eq
2
NO
eq
Cl2


(1.35)
2
2
(0.240) (0.608)
eq
 52.04..  52.0