Chemistry 102 - SEAS
Q, Keq, and Equilibrium
• At a given temperature and set of conditions (pressures or concentrations), we can tell if a
reaction is already at equilibrium, or which way it will approach equilibrium, by comparing
the reaction quotient (Q) to the equilibrium constant (Keq).
• Note: when comparing Q to Keq, you must compare the “same version” of Q and Keq.
That is, if you determine Q in terms of pressures (QP ), then you must compare to KP ; if you
determine Q in terms of concentrations, (QC), then you must compare to KC. Be careful !
• The first table (Table 1) below relates Q, its corresponding value of ∆G, Keq and the
subsequent approach to equilibrium. Remember the important relationships from which this
table derives:
Q
∆G = ∆Go + RTln(Q) = RTln K  ; where ∆Go = -RTlnKeq .
eq
• Remember, the approach to equilibrium is such that Q will either increase or decrease
(depending on the situation) until Q = Keq . In other words, the equilibrium constant
(Keq) is the “parameter” to which Q “must adjust” in order for the reaction to establish
equilibrium.
"products"
• The second table (Table 2) below relates the magnitude of Keq ∝ "reactants" .
Table 1: Q versus Keq and the “direction of approach” to equilibrium
Relationship of Q to Keq
Sign of ∆G Comments
Q < Keq
Rxn will proceed to products to establish
equilibrium.
Q > Keq
+
Rxn will proceed to reactants to establish
equilibrium.
Q = Keq
0
Rxn is already at equilibrium at the
specified conditions.
Table 2: Keq and the Composition of the equilibrium reaction mixture
Magnitude of Keq
Sign of ∆Go Composition of equilibrium rxn mixture
Keq > 1
Greater proportion of products.
Keq < 1
+
Greater proportion of reactants.
Keq = 1
0
Rxn is at equilibrium at standard conditions.
Each reactant and product gas is at 1.0 atm
Each reactant and product (aq) species is
at 1.0 M.
• Thus, Tables 1 & 2 allow us to look at Q and Keq, instead of ∆G and ∆Go, in order to
investigate the approach to and composition of the equilibrium system. Remember however,
Q and Keq give us the same information as ∆G and ∆Go - just in a different “form”.
1
Chemistry 102 - SEAS
Van't Hoff equation - Keq as a function of Temperature
• Equations:
∆Go = -RTlnKeq
and ∆Go = ∆Ho - T∆So .
• Assumptions:
∆Ho and ∆So are independent of temperature for chemical reactions (as discussed before).
• Derivation:
Setting the two equations above equal to each other, since they each equal ∆Go, we have:
-RTlnKeq = ∆Ho - T∆So .
• Consider two temperatures, T1 and T2 (which must be expressed in kelvin) and their
corresponding equilibrium constants Keq,1 and Keq,2 , respectively. [Note: for a gas
phase reaction, Keq must be KP ; and for an aqueous reaction, Keq must be KC.] Thus:
-RT1 lnKeq,1 = ∆Ho - T1 ∆So
and
-RT2 lnKeq,2 = ∆Ho - T2 ∆So
Dividing each equation through by R x temperature, we have:
∆Ho
-lnKeq,1 = RT
1
-
∆So
R
and
∆Ho
-lnKeq,2 = RT
2
-
∆So
R
Subtracting equation “2” (at temperature T2) from equation “1” (at equation T1), the
∆So
terms cancel. we have:
R
-lnKeq,1 + lnKeq,2
=
Rearranging:
∆Ho
= RT
1
∆Ho
RT1
-
∆Ho
RT2
-
∆So
R
∆Ho
RT2
 ∆Ho   1
 ∆Ho   T2 - T1 
 Keq,2
1
ln K
=  R   T - T  =  R   T T 

eq,1
1
2
2 1
- van't Hoff eqn.
An example calculation follows.
2
∆So
+ R
Consider the equilibrium between NO2(g) and N2O4(g):
2 NO2(g)
N2O4(g) ; ∆Go298 = _6 kJ and ∆Ho298 = _58 kJ.
Determine KP and KC at 25oC (298 K) and at 227oC (500 K).
Solution:
Let's find KP at each temperature, first. [Remember, for a gas phase reaction (or a heterogeneous
reaction where there are only gases, pure solids and pure liquids), ∆Go gives KP and the
van't Hoff equation will use KP . For aqueous reactions (homogeneous or heterogeneous)
∆Go gives KC and the van't Hoff equation will use KC. Be careful! ]
-∆Go298
At 298 K: ln(KP,298) =
RT
-(-6 x 103 J)
= 
 8.314 J  (298 K)

K
= 2.42
Thus, KP , 2 9 8 = antiln(2.42) or e2.42 = 11.24 . Using van't Hoff, we now find KP,500 .
Let T2 = 500 K and T1 = 298 K. Thus,
 ∆Ho  1
 ∆Ho  T2 - T1
 KP,500
1 

ln K
=


- =
 R   T T 
. Plugging in:
R  T1
T2
P,298
2 1
 KP,500
ln 11.42 
KP,500
11.42
 -58 x 103 J  500 K - 298 K 


= 
J   (500 K)(298 K)
 8.314 K 
= -9.46 .
= antiln(-9.46) or e-9.46 = 7.79 x 10-5 . Finally,
KP , 5 0 0 = (11.42)( 7.79 x 10-5) = 8.90 x 10- 4 .
Now, to get KC at each temperature, we use: KC = KP (RT)_ ∆ ng .
Here, ∆ng = [1 - 2] = -1. R must be the 0.0821 version, not the 8.314 version.
First, we find KC,298 .
KC , 2 9 8 = KP,298(RT)-∆ng = (11.42)[(0.0821)(298)]-(-1) = 279.4.
Now we'll find KC,500.
KC , 5 0 0 = KP,500(RT)-∆ng = (8.90 x 10_4)[(0.0821)(500)]-(-1) = 3.65 x 10- 2.
Note: KP (or KC) decreases as temperature increases. This is expected for an exothermic
(∆Ho < 0) reaction. For an endothermic (∆Ho > 0) reaction, KP (or KC) will increase
as temperature increases - as expected. Check it out for yourself!
3
Chemistry 102 - SEAS
Solving a “Typical” Equilibrium Problem - An Example
Equilibrium problems, that is, determining information about and/or from the composition of
an equilibrated chemical reaction, are very important in chemistry. The “typical” equilibrium
problem makes use of the reaction quotient (Q) and its relationship to the equilibrium constant
(Keq) of a chemical reaction. As discussed previously, Q and Keq are really just convenient
parameters which help us relate ∆G (and hence the Second Law of Thermodynamics) to the
solution of equilibrium problems - in terms of “measurable” quantities such as pressure or
concentration. We will solve such a “typical” sample problem below, keeping careful attention
to the “details”.
Example: Consider a 1.00 L bulb filled only with 10 torr of SO2(g) and 10 torr of Cl2(g), at
25oC (298 K).
10 torr SO 2
10 torr Cl 2
V = 1.00 L
T = 298 K
Maintaining constant volume and constant temperature, reaction is initiated (via a catalyst), and the
following equilibrium is established - in which SO2Cl2(g) (Sulfuryl chloride) is produced:
SO2 (g) + Cl2 (g)
SO2 Cl2 (g) .
At 298 K, KP = 2840 for the above equilibrium. Determine the equilibrium partial pressure
of each gas, when equilibrium is established.
Solution:
The “10” steps are listed below.
(1) Write down the equilibrium reaction and begin the construction of a “table” beneath the
reaction.
(2) In the “first row” of the table, write down the initial amount of each species - in the
appropriate units. If you are working with KP , all entries (throughout the table) must be in
atmospheres. If you are working with KC, all entries (throughout the table) must be in
Molarity (M) units. If we use the appropriate units for all entries, we will not only ensure
that the answers “come out right”, but we will ensure that KP (KC) is unitless.
Initial (atm):
SO2 (g) + Cl2 (g)
10
10
760
760
SO2 Cl2 (g) .
0
4
(3) Calculate Q, the reaction quotient, based on your listed initial amounts in the first row of
the table. Compare Q with the appropriate Keq value. If Q is in terms of concentrations
(QC) - compare with KC. If Q is in terms of atmospheres (QP ) - compare with KP .
If Q < Keq - reaction will proceed toward products to establish equilibrium
If Q > Keq - reaction will proceed toward reactants to establish equilibrium
If Q = Keq - reaction is already at equilibrium
Note: If Q = 0 (one or more products not present initially, but all reactants are present initially),
Q has to be less than Keq and thus, reaction will proceed toward products to establish
equilibrium. This is the case in this problem.
Q = rxn quotient (in terms of “initial” pressures, Po) =
PSO2Cl2,o
PSO2,o PCl2,o
.
0
Q =  10   10 
= 0
Q < KP - Proceeds to products to establish equilibrium.



 760  760
(4) Define a parameter - “x” - to reflect the changes that occur to establish equilibrium.
Define “x” in such a way so that it reflects positive change in the direction that the
reaction “wants” to proceed to establish equilibrium, i.e., the direction indicated by Q (see (3)).
Let x = increase in atmospheres of SO2 Cl2 (g), in order to reach equilibrium.
(5) Then, the changes are not independent ! They are related by the stoichiometry of the
reaction. [This is true because, at constant volume and constant temperature (for an ideal gas),
P ∝ n and concentration ∝ n.] Hence, the changes can all be expressed in terms of
the single parameter - “x”. Write down these changes, as a function of “x”, under each
species, in the second row of the table. Note: make sure “x” reflects correctly whether the
species will increase in amount (“+x”) or decrease in amount (“-x”) as equilibrium is
established (see (3)).
Initial (atm):
Change (atm)
(6)
SO2 (g) + Cl2 (g)
10
10
760
760
-x
-x
SO2 Cl2 (g)
0
+x
The third row is the algebraic sum of the initial and change (first and second rows of the
table) of each species. Thus, the third row clearly expresses the equilibrium amounts.
Initial (atm):
Change (atm)
Equil. (atm)
SO2 (g) + Cl2 (g)
10
10
760
760
-x
-x
10
10
-x
-x
760
760
5
SO2 Cl2 (g)
0
+x
+x
(7)
These equilibrium amounts are then plugged into the algebraic expression for Keq.
( “ Peq” denotes equilibrium pressure)
PS O2 C l2 ,eq
x
KP = 2840 = P
=  10
P
 2 S O2 ,eq C l2 ,eq

 760 - x
(8) This then provides you with one equation in one unknown (“x”). This equation must be
solved for “x”, using “suitable” algebraic techniques.
In principle, although sometimes mathematically difficult, the equation can be solved for “x”.
Here, we have a quadratic equation in “x”. By suitable re-arrangement we can put the
above expression in the form: ax2 + bx + c = 0 ; and then apply the quadratic formula:
-b ± b2 - 4ac
x± =
. [Note: you should become “fluent” with use of the quadratic
2a
formula, since many equilibrium expressions will give rise to quadratic equations!]
Putting the equilibrium expression in “x” in “suitable” form, we have:
2840 x2 - 75.74 x + 0.4917 = 0
(a
x2
b
x
c
. (Check this for yourself!)
)
Thus, by correspondence, we have:
a = +2840 ; b = -75.74 ; c = 0.4917
quadratic formula, we have:
x± =
. Substituting into the
-(-75.74) ± (-75.74)2 - (4)(2840)(0.4917)
2(2840)
= (0.01333 ± 0.002162) atm
We have two (2) roots (as usual) when we solve a quadratic equation; thus:
 760 torr
x+ = (0.01333 + 0.002162) atm = 0.01550 atm = (0.01550 atm) 1 atm. 
 760 torr
x- = (0.01333 - 0.002162) atm = 0.01117 atm = (0.01117 atm) 1 atm. 
= 11.78 torr;
= 8.49 torr.
Since “x” was defined as a positive quantity (See step (4)), it should be easy to choose the
“meaningful” solution. But wait! Both solutions are positive!! This situation is not
uncommon. Which solution is correct? Are they both correct? There is only one (1)
meaningful solution. In this case, we must go back to the table.
Remembering that P α n here, we realize that if the reaction went to completion, the maximum
pressure of SO2Cl2(g) that could be produced is only 10 torr (Why?). Also, according to the
table, “x” represents the (total) atmospheres of SO2Cl2(g) produced. Thus, 11.78 torr is
meaningless, since it indicates a greater pressure of SO2Cl2(g) than is stoichiometrically
possible! See how important it is to set up the table clearly!!
x- = 0.01117 atm. = 8.49 torr
is the proper solution.
6
(9) The parameter “x” is then used to determine the equilibrium amounts (atm or M) of each
species in the table. This can be achieved by utilizing the third row of the table (see (6)).
According to the third row of the table:
PS O2 C l2 ,eq
= x = 0.01117 atm = 8.49 torr
10
PS O2 ,eq
= 760 - x
= 0.00199 atm = 1.51 torr
10
PC l2 ,eq
= 760 - x
= 0.00199 atm = 1.51 torr.
(10) This step is optional, but useful. Plug the determined equilibrium amounts into the Keq
expression and evaluate it. If the result equals Keq (it should!) you can be reasonably
assured that you probably have not made a mistake. If the result does not equal Keq,
(“approximately”) you have made a mistake!! Go back and check!
PS O2 C l2 ,eq
0.01117
KP = P
=
= 2821 (??)
S O2 ,eq PC l2 ,eq
(0.00199)2
But KP = 2840!! Have we made a mistake? No! Due to “round-off” errors during the course
of the calculation, our equilibrium pressures do not give KP exactly. But it is close enough!
The above steps can be followed for any equilibrium problem (the specifics will, of course, be
different). “Skipping steps” or taking careless shortcuts will invariably lead to disaster!
Note: According to Dalton's Law, the total equilibrium pressure (see third row of the table):
 10

 10

Ptotal,eq = PSO2 Cl2,eq + PSO2,eq + PCl2,eq = x +  760 - x
+  760 - x
20
Ptotal,eq = 760 - x = 0.015145 atm = 11.51 torr .
And, the total initial pressure (see first row of the table):
10
10
Ptotal,o = PSO2 Cl2,o + PSO2,o + PCl2,o = 0 + 760
+ 760
Ptotal,o = 0.02632 atm = 20 torr .
Even though we have a closed system at constant volume and constant temperature,
Ptotal,o ≠ Ptotal,eq . Why?
Addendum: Consider what happens if the above equilibrated system is connected to a
previously empty bulb of equal volume (1.00 L) and the equilibrated gaseous mixture is allowed
to fill both bulbs (with a total volume = 2.00 L). After some time, equilibrium is re-established
at 25oC (298 K). See diagrams.
7
Before valve is opened.
1.51 torr Cl 2
Evacuated
1.51 torr SO 2
8.49 torr SO 2C l 2
Equilibrium
298 K
V = 1.00 L
V = 1.00 L
After valve is opened.
C l2
C l2
298 K
298 K
S O2
S O2
S O2C l2
S O2C l2
Vtotal = 2V = 2.00 L
Questions:
(a) What is the partial pressure of each gas, after the gases fill both bulbs, but
before equilibrium is established?
(b)Is the system at equilibrium after the expansion? Calculate Q to verify your
answer.
(c) What is the equilibrium partial pressure of each gas, after the expansion?
Solution:
(a) In order to find out if the system is at equilibrium (or not) after the expansion occurs, we
must calculate the pressures after the gases expand. These pressures will serve as the
initial pressures (Po) - which will allow us to determine if the system is now at equilibrium.
Since each gas is expanding at constant temperature (assumed) and constant amount (no gas
“leaks” out); we can use Boyle's Law. Thus, since each gas expands from V to 2V, the
“new” pressure of each gas will be half of its equilibrated pressure before expansion.
8
PSO2Cl2,eq
8.49 torr
PS O2 C l2 , o =
=
= 4.245 torr = 5.586 x 10_ 3 atm
2
2
PSO2,eq
1.51 torr
PS O2 , o =
=
= 0.755 torr = 9.934 x 10_ 4 atm
2
2
PCl2,eq
1.51 torr
PC l2 , o =
=
= 0.755 torr = 9.934 x 10_ 4 atm
2
2
(b) Following the suggestion, we calculate Q and compare to KP (= 2840).
PSO2Cl2,o
Q = rxn quotient (in terms of “initial” pressures, Po) = P
.
SO2,o PCl2,o
Q =
5.586 x 10-3
= 5660 - Q > KP
(9.934 x 10-4)(9.934 x 10-4)
The system is not at equilibrium and will proceed to reactants to establish
equilibrium.
(c) We will follow the same procedure as before, but this time we'll just give the “skeleton”
version. You should check through it yourself to make sure you understand!
Let x = increase in atmospheres of SO2 (g) (or Cl2 (g)), in order to reach
equilibrium. Remember, the reaction proceeds to the “left”.
Constructing the table, we have:
_
SO2(g)
+
Cl2(g)
SO2Cl2(g)
Initial (atm):
9.934 x 10-4
9.934 x 10-4
5.586 x 10-3
Change (atm)
+x
+x
-x
Equil. (atm)
9.934 x 10-4 + x
9.934 x 10-4 + x
5.586 x 10-3 - x
KP = 2840 =
PSO2Cl2,eq
PSO2,eq PCl2,eq
=
5.586 x 10-3 - x
2
(9.934 x 10-4 + x)
.
Again, we have a quadratic equation in “x”. By suitable re-arrangement we can put the
above expression in the form: ax2 + bx + c = 0 ; and then apply the quadratic formula:
x± =
-b ± b2 - 4ac
2a
.
Putting the equilibrium expression in “x” in “suitable” form, we have:
2840 x2 + 6.6425 x - 2.7834 x 10-3 = 0 (Check this for yourself!)
Thus, by correspondence, we have:
a = +2840 ; b = +6.6425 ; c = -2.7834 x 10-3
.
Substituting into the quadratic formula, we have:
x± =
-(+ 6.6425) ± (+ 6.6425)2 - (4)(2840)(-2.7834 x 10-3)
2(2840)
9
x± = (-1.1695 x 10-3 ± 1.5322 x 10-3) atm
We have two (2) roots, thus:
x+ = (-1.1695 x 10-3 + 1.5322 x 10-3) atm = +3.6276 x 10- 4 atm
x- = (-1.1695 x 10-3 - 1.5322 x 10-3) atm = -2.7017 x 10- 3 atm
Since “x” was defined as a positive quantity it is easy, this time, to choose the “meaningful”
solution. Only x+ is positive. Thus, x = +3.6276 x 10- 4 atm is the correct choice.
According to the third row of the table:
PS O2 C l2 ,eq = 5.586 x 10-3 - x = 5.223 x 10- 3 atm = 3.97 torr
PS O2 ,eq
= 9.934 x 10-4 + x = 1.356 x 10- 3 atm = 1.03 torr
PC l2 ,eq
= 9.934 x 10-4 + x = 1.356 x 10- 3 atm = 1.03 torr.
Now, plugging into the KP expression to check (Remember, it should equal 2840):
PS O2 C l2 ,eq
KP = P
S O2 ,eq PC l2 ,eq
=
5.223 x 10- 3
(1.356 x 10- 3) 2
= 2840 (Perfect!).
Note: When we expanded the equilibrated system from V to 2V (at constant temperature),
we initially decreased the pressure (via Boyle's Law) by increasing the volume.
The system re-equilibrated by “going backwards”, i.e., toward reactants. In other
words, it shifted to the side with the greater moles of gas (2 moles of gas on the
reactant side and 1 mole of gas on the product side). We will see that this is exactly what is
predicted qualitatively by a principle due to LeChâtelier, i.e., LeChâtelier's Principle!
More about that later!
Also: Comparing the equilibrium pressures after the volume change with the equilibrium
pressures before the volume change, we see that the “original” equilibrium pressures were
not attained! The equilibrium pressures after the volume change are all lower than the
equilibrium pressures before the volume change. In other words, “compensation is not
complete”! Why?
10
Chemistry 102 - SEAS
Approximate Solution of Equilibrium Problems - “Neglecting x”
Sometimes the algebraic equilibrium expression, i.e., the equation which relates the equilibrium
constant (KP or KC) to the equilibrium amounts (partial pressures or concentrations) - in terms
of “x”, is “intractable”. In other words, the exact algebraic solution of the problem is either
not possible or else (most times) the algebraic process of solution is complicated and/or tedious.
Thus, we wish (or must) seek approximate methods to find “x”. There are several approaches
- we will consider only one here; the technique of “neglecting x”. We will actually choose a
problem that can be solved exactly, in very easy fashion, via the quadratic formula. The reason
for this is so we can better assess our approximation(s). The best way to demonstrate this
method is by example.
Consider the following reaction, which can equilibrate at a certain temperature, T:
N2F4(g)
2 NF2(g) ;
KC = 8.10 at this temperature. Determine the equilibrium concentration (in M) of NF2 (g)
and of N2 F4 (g) if 0.050 mole of N2 F4 (g) and 0.500 mole of NF2 (g), at temperature T, are
initially placed in a previously empty 500.0 mL sturdy container. We set up our table, in
moles/L (M), remembering to calculate Q.
N2F4(g)
2 NF2(g)
0.050 moles
0.500 moles
[initial]
=
0.10
M
= 1.00 M
0.500 L
0.500 L
2
[NF2]o
Q = [N F ]
2 4 o
(1.00)2
= (0.10)
= 10
Q > KC - reaction proceeds toward reactants
(to the “left”) to reach equilibrium.
Let x = increase in concentration (in M) of N2 F4 (g) in order to establish
equilibrium. {Thus, decrease in concentration (in M) of NF2(g) in order to
establish equilibrium is “_2x”.} Setting up the complete table, we have:
[initial]
[change]
[equil]
N2F4(g)
0.10 M
+x
0.10 + x
2
[NF2]eq
KC = 8.10 = [N F ]
2 4 eq
=
2 NF2(g)
1.00 M
-2x
1.00 - 2x
(1.00 - 2x)2
(0.10 + x)
. This is a quadratic equation in “x”,
which, of course, we can solve exactly. However, for purposes of demonstration, we will (first)
solve this expression by “neglecting x”.
The algebraic equation is:
(1.00 - 2x)2
8.10 = (0.10 + x)
.
11
The method of “neglecting x” consists of assuming (initially) that “x” is “small” when
compared to the “initial” amount from which it is added/subtracted. In other words, we can
see from the above algebraic expression, that we have factors of the form: (Ao ± ax) , where
“ Ao” is the “initial” amount and “a” is the factor that multiplies x (in the above algebraic
equation, “a” = 2 in the numerator and “a” = 1 in the denominator. Thus, for a term of the form
( Ao ± ax), if we want to “neglect x”, we are approximating this term as: (Ao ± ax) ≈ Ao .
[Obviously, we can't neglect “x” in every term of our algebraic equilibrium expression - if we
did this, then we would eliminate “x” from the problem entirely! ] Now, since we assume that
“x” was small, that is, that we could neglect the term “ax” with respect to Ao; we must check to
make sure that the term “ax” is “small enough”. Criteria vary - the one we will use is the
following. If we approximate (Ao ± ax) as Ao , then the approximate value of “x”, xapprox
that we subsequently find must “pass” the following test - termed the 5 % rule:
If
axapprox
≤ 0.05 ; (Ao ± ax) ≈ Ao is valid.
Ao
In other words: “axapprox” (not just “xapprox”) must be less than or equal to 5 %
of Ao , in order to warrant the neglecting of “ax” with respect to Ao . If this criterion is
not met, our “neglecting of x” is not warranted, and we must go back to the original form of
the algebraic equilibrium equation. Let's apply this method to the sample problem above.
Looking at the algebraic expression once again, we have:
(1.00 - 2x)2
8.10 = (0.10 + x)
. We can “neglect x” in either of two places (not both);
Neglect “2x” with respect to 1.00 in the numerator, or neglect “x” with respect to 0.10 in the
denominator. Clearly, since 1.00 is ten times larger than 0.10, it should be a better approximation
to neglect “2x” with respect to 1.00 than to neglect “x” with respect to 0.10. So, let's neglect
“2x” with respect to 1.00 and solve.
(1.00 - 2x)2
(0.10 + x)
8.10 =
x ≈
(1.00)2
≈ (0.10 + x)
1.00 - (0.10)(8.10)
8.10
; where our “x” is now xapprox . Solving, we have:
= 0.0235 M
. Now we apply the 5 % rule, to check.
We must compare “2xapprox” to 1.00, i.e.,
2xapprox
(2)(0.0235 M)
=
= 0.047 which is ≤ 0.05. Thus, the approximation
1.00 M
1.00 M
is warranted.
The equilibrium concentrations (approximately) are:
[ N2 F4 ] eq = 0.10 + x = (0.10 + 0.0235) M = 0.1235 M
[NF2 ] eq = 1.00 - 2x = (1.00 - (2)(0.0235)) M = 0.953 M
and
.
[Just to see, let's, instead, neglect “x” with respect to 0.10. Thus, the algebraic
equilibrium expression becomes:
8.10 =
(1.00 - 2x)2
(0.10 + x)
_
(1.00 - 2x)2
(0.10)
; where our “x” is now xapprox . Solving:
12
x =
1.00 - (0.10) 8.10
2
= 0.358 M
. Now we apply the 5 % rule, to check.
We must compare “xapprox” to 0.10, i.e.,
xapprox
0.10 M
=
0.358 M
0.10 M
= 3.58 which is not ≤ 0.05. Thus, the approximation is
not warranted.]
[If the problem were solved exactly, you should verify the following. The algebraic
equilibrium expression, in quadratic form (i.e., ax2 + bx + c = 0) becomes:
4 x2 - 12.10 x + 0.19 = 0. Also, by use of the quadratic formula, you should verify that:
x± =
-b ± b2 - 4ac
2a
=
-(-12.10) ± (-12.10)2 - (4)(4)(0.19)
2(4)
x± = (1.5125 ± 1.4967) M. Verify that x+ = 3.0092 M and x- = 0.0158 M.
Finally, prove that x- = 0.0158 M is the only meaningful solution.
Thus, the exact equilibrium concentrations are:
[ N2 F4 ] eq = 0.10 + x = (0.10 + 0.0158) M = 0.1158 M
[NF2 ] eq = 1.00 - 2x = (1.00 - (2)(0.0158)) M = 0.968 M
and
.]
As you can see, the approximate and exact solutions are “close”, thanks to the 5 % rule!
13
Chemistry 102 - SEAS
Heterogeneous Equilibria - Examples
Although (pure) solids and (pure) liquids “do not enter” (i.e., enter as “1”) into the Keq
expressions, they do play a role in chemical equilibrium. Remember, all species in a chemical
reaction are “connected” by the stoichiometric chemical equation. Thus, pure solids and pure
liquids are affected by the species that are contained in the Keq expression (aqueous and gaseous
species) - albeit indirectly. An example or two may help.
Example #1: Consider the following equilibrium, between liquid methanol (CH3OH(l)),
H2(g) and CO(g):
CH3OH(l)
2 H2(g) + CO(g)
KC = 0.136 at 500 K. When 2.00 moles of CH3OH(l), 6.00 moles of H2(g), and
3.00 moles of CO(g) are placed in a 5.00 L container at 500 K, how many moles of H2(g),
CO(g), and CH3OH(l) are present in the container at equilibrium? [Assume that the liquid
methanol occupies a negligible volume of the container.]
Solution: We will work the problem in terms of KC (since it is given) and then convert the
(gaseous) concentrations to moles via: moles = concentration x volume. [We could equally
well have stated the problem in terms of KP and solved the entire problem with partial pressures.
Then, we could convert to (gaseous) moles via: n = PV/(RT).] The initial concentrations are:
6 moles
3 mole
[H2]o =
= 1.20 M
and [CO]o = 5 L
= 0.60 M . Thus, we calculate
5 L
2
Q (i.e., QC). Q = [H2]o [CO]o = (1.20)2(0.60) = 0.864 . So, Q > KC and the reaction
proceeds toward reactants to establish equilibrium. We set up our table, let x = the increase in
[CO] to establish equilibrium, the change in [CO] is “-x”. The completed table is below.
[initial]
[change]
[equil]
CH3OH(l)
“2.00 moles”*
------**-------------**--------
2 H2(g)
1.20 M
-2x
1.20 - 2x
+
CO(g)
0.60 M
-x
0.60 - x
The “*” means that only the gaseous species (i.e., only those that appear in the Keq
expression) should enter the table; and then, only in units of moles/L. Obviously, the moles of
liquid that are produced are related stoichiometrically to the “x” changes of the gaseous species.
More about that later. Let's find the equilibrium concentration of H2(g) and of CO(g).
2
KC = 0.136 = [H2]eq
[CO]eq = (1.20 - 2x)2(0.60 - x).
We can solve the above algebraic equation exactly, by suitable factoring. We have:
0.136 = (2)2(0.60 - x)3 . Thus,
3 0.136
= (0.60 - x) = x = 0.276 M .
4
[H2 ] eq = 1.20 - 2x = 0.648 M
(moles H2 ) eq = (0.648 M)(5 L) = 3.240 moles
14
; and
and [CO]eq = 0.60 - x = 0.324 M
(moles CO)eq = (0.324 M)(5 L) = 1.620 moles
Now:from the table, moles of CO(g) reacted = “x” x 5 L = (0.276 M)(5 L) =
1.380 moles of CO(g) reacted . The stoichiometry of the reaction “tells us” that the
• moles of CH3 OH(l) produced = moles of CO(g) reacted = 1.380 moles
.
• At equilibrium: moles CH3 OH(l) in container = (2 + 1.380)moles
= 3.380 moles .
Example # 2: Consider the reaction for the dehydration of CuSO4• 5H2O(s) (Blue Vitriol) at
500 K: CuSO4• 5H2O(s)
CuSO4(s) + 5 H2O(g) ; KP = 3.60 x 10-5.
(a) At 500 K, calculate the equilibrium pressure of H2 O(g) (in torr) and the minimum
number of grams of CuSO4 •
5H2 O(s) (M.W. = 249.75 g/mole) that must be placed in
a previously empty 5.00 L container, in order to achieve this H2O(g) pressure. [As usual,
the volume of both solids can be assumed to be negligible - when compared to the volume
of the 5.00 L container.]
(b) If the volume of the container is increased gradually, at constant temperature, what will
happen to the established equilibrium pressure of H2O(g)? Does this agree with LeChâtelier?
Solution: If we write down the equilibrium expression, we have:
5
KP = PH2O,eq .
(a) Since the law of chemical equilibrium depends on only one partial pressure, we can solve
directly: Thus:
PH2
O,eq
=
5
KP
=
5
3.60 x 10- 5
= 0.1292 atm.
= 98.19 torr
.
The moles of H2O(g) that must be produced to achieve equilibrium under these conditions is
moles of H2 O(g) at equilibrium =
PH2O,eqV
RT
0.0157 moles of H2 O(g) at equilibrium
=
(0.1292 atm.)(5.00 L)

L atm. 
 0.0821 mole K (500 K)
=
.
grams of Blue Vitriol needed =
 1 mole Blue Vitriol  249.75 g Blue Vitriol
(0.01157 moles H2O) 5 moles H O   1 mole Blue Vitriol 
2
= 0.784 g
.
Thus, a little bit more that 0.784 g of Blue Vitriol must initially be placed in the container, at
these conditions; since, at equilibrium, there must be some Blue Vitriol in the container also.
The actual amount of Blue Vitriol that must remain is not “critical” since solids “do not
appear” in the Keq expression.
15
5
(b) At constant temperature, KP does not change. Thus, since PH2 O,eq = KP , the
equilibrium pressure of H2O(g) will not change. In other words, as long as the
volume increase is gradual, the adjustment is “instantaneous” and the pressure will
never vary much from its previously established value. Compensation, in this case,
is complete! This does agree with LeChâtelier.
According to LeChâtelier, as the volume is increased (at constant temperature) the reaction
will shift to the side with more moles of gas, to re-establish equilibrium. In this case, more
moles of products (i.e., more H2O(g)) will be produced. Why doesn't the pressure of
H2O(g) increase? Because the mole increase exactly compensates the volume increase.
nH2ORT
Thus, since PH2O =
= [H2O]eqRT. nH2O and V increase by the same
V
amount - thus PH2 O and [H2O] does not change. What will happen as V is increased
gradually, at constant temperature, from an initial volume, Vo (5 L, here) and becomes
“very large”? The pressure of H2O(g) will stay constant until the Blue Vitriol runs out!
Now, the equilibrium is destroyed! Then, since no more moles of H2O(g) can be
produced, we have volume increase at constant temperature and (now) constant moles of
H2O(g). This is Boyle's Law - pressure will decrease as volume increases, i.e.,
1
P α V, a hyperbolic dependence. The graph below, where we plot PH2O versus Volume
(V). Vo denotes the initial volume of equilibration (5 L, here) and V* denotes the volume at
which the Blue Vitriol runs out and hence the onset of Boyle's Law.
Pwater
0.1292 atm.
Boyle's
Law
V*
V0
16
V
Chemistry 102 - SEAS
LeChâtelier's Principle - “Return to Equilibrium”
LeChâtelier: when a small disturbance or “stress” is applied to an equilibrated system, the
system will attempt to restore equilibrium in a fashion to minimize the effect of the stress.
“Stress”
Direction of Approach to Equilibrium
“Reason”
A. Temperature
change at
constant
pressure
Increase in temperature favors the endothermic reaction. Decrease in temperature favors the
exothermic reaction.
Van't Hoff
B. Pressure
change at
constant
temperature, by
increasing/
decreasing
volume
Decrease in pressure favors the side of
the reaction with the most moles of gas.
Increase in pressure favors the side of
the reaction with the least moles of gas.
Q versus Keq
C. Pressure
change, at
constant volume
and constant
temperature, by
adding an inert
gas
“No Effect”
“Ideal gas”;
Q versus Keq
D. Addition of an
inert gas, at
constant total
pressure and
constant
temperature
This “stress” is “equivalent to an increase in
volume, at constant temperature. The side of
the reaction with the most moles of gas is favored,
in order to restore equilibrium
E. Pressure or
Concentration
change, at
constant
temperature
and constant
volume
by adding or
removing
reactant/product
gaseous species
Removing product - “restoration” is toward products. Q versus Keq
Removing reactant - “restoration” is toward reactants.
Adding reactant - “restoration” is toward products.
Adding product - “restoration” is toward reactants.
17
Q versus Keq
“Stress”
Direction of Approach to Equilibrium
“Reason”
F. Adding or removing
pure solid or pure
liquid, that is a
reactant or product,
at constant
temperature and
constant volume.
“No effect” - only if
“enough” is present.
Q versus Keq
G. Adding a catalyst
“No Effect”
“Kinetics”
18