Worksheet 16 - Equilibrium
Chemical equilibrium is the state where the concentrations of all reactants and
products remain constant with time.
Consider the following reaction:
H2O + CO Æ H2 + CO2
Suppose you were to start the reaction with some amount of each reactant (and no H2
or CO2). As the reaction proceeds, the concentrations of CO and H2O will decrease as
they are converted into CO2 and H2. Eventually, the reaction reaches a point at which
the concentrations of reactants and products stop changing. Reactant is still present
(see the non-zero concentration of reactants in the illustration below?), but it looks like
the reaction has stopped.
In fact, the reaction has NOT stopped. Reactions occur when molecules collide with
one another. As the concentrations of CO and H2O decrease, it becomes less likely for
these two molecules to collide, even though they are both still present. As the
concentration of the products increases, the probability of a collision between H2 and
CO2 increases. In this case, the reaction can occur in both the forward (as written) and
reverse (products converting to reactants) directions. At equilibrium, the rate of the
forward reaction is equal to the rate of the reverse reaction.
The equilibrium constant, K, is used to determine the relative concentrations of products
and reactants at equilibrium. According to the law of mass action, if a chemical reaction
has the form
aA + bB ↔ cC + dD
then the equilibrium constant can be expressed as
c
d
C ] [D]
[
K=
[A]a [B]b
1. In an experiment conducted at 74°C, the equilibrium concentrations of reactants and
products for the equation shown below were [CO] = 1.2x10-2 M, [Cl2] = 0.054 M and
[COCl2] = 0.14 M.
CO( g ) + Cl2 ( g ) ↔ COCl2 ( g )
a) What is the equilibrium expression for this reaction?
K=
[COCl2 ]
[CO][Cl2 ]
b) Calculate the value of the equilibrium constant
K=
(0.14M )
[COCl2 ] =
= 216
[CO][Cl2 ] (1.2 ×10−2 M )(0.054M )
2. One student performed an experiment in which nitrogen and hydrogen gas were
mixed together to form ammonia. Equilibrium concentrations of the three species
were [NH3] = 0.157 M, [N2] = 0.921 M and [H2] = 0.763 M.
N 2 ( g ) + 3H 2 ( g ) ↔ 2 NH 3 ( g )
a) What is the equilibrium expression for this reaction?
2
[
NH 3 ]
K=
[N 2 ][H 2 ]3
b) Calculate the value of the equilibrium constant
2
2
(
[
0.157M )
NH 3 ]
K=
=
= 0.0603
3
3
[N 2 ][H 2 ] (0.921M )(0.763M )
3. Another student performed the same reaction (to form ammonia) with the same
equilibrium constants, but wrote the balanced equation as
1
3
N 2 ( g ) + H 2 ( g ) ↔ NH 3 ( g )
2
2
a) What is the equilibrium expression when the reaction is written this way?
K=
[NH 3 ]
[N 2 ]1 / 2 [H 2 ]3 / 2
b) Calculate the value of the equilibrium constant
K=
(0.157M )
[NH 3 ] =
1/ 2
3/ 2
(0.921M )1 / 2 (0.763M )3 / 2
[N 2 ] [H 2 ]
= 0.245
c) How are the equilibrium constants from questions 2b and 3b related?
The equilibrium constant from 3b is the square root of the equilibrium
constant from 2b. In other words (K2b)1/2 = K3b or (0.0603)1/2 = 0.245. When
all the coefficients in the balanced equation are multiplied by a constant
value (½), the equilibrium value is raised to that same power (½).
4. The equilibrium constant for the following reaction at 100°C is 7.10. The equilibrium
concentrations of the reactants were found to be [Br2] = 2.3x10-3 M and [Cl2] =
1.2x10-2 M.
Br2 ( g ) + Cl2 ( g ) ↔ 2 BrCl ( g )
a) What is the equilibrium concentration of BrCl?
2
2
[
[
BrCl ]
BrCl ]
K = 7.10 =
=
[Br2 ][Cl2 ] (2.3 × 10−3 M )(1.2 × 10−2 M )
[BrCl]2 = 1.96 × 10−4
[BrCl] = 1.4 × 10−2 M
b) If the reaction were performed in a 5.0 L container, how many moles of BrCl
would be produced?
1.4 × 10 −2 mol BrCl
5.0 L ×
= 7.0 × 10 − 2 mol BrCl
1L
5. Which of the above reactions had the highest value of K? What does this indicate
about the extent of the reaction?
The reaction in question #1 had the highest K value (K = 216). This indicates
that this reaction goes the furthest toward completion, or in other words that
at equilibrium the majority of the reactants have reacted to form product.
When using gases, the equilibrium expression can also be written using pressures
of the reactants and products rather than concentrations. Remember that
concentration has units of moles (n) per liter (V). The ideal gas law can be
rearranged to the form
P=
nRT ⎛ n ⎞
= ⎜ ⎟ RT = CRT
V
⎝V ⎠
where C represents concentration. Using the last example again, we can derive an
equilibrium expression involving pressures:
Br2 ( g ) + Cl2 ( g ) ↔ 2 BrCl ( g )
2
⎛ PBrCl ⎞
⎟
⎜
2
[
BrCl ]
RT ⎠
⎝
K=
=
[Br2 ][Cl2 ] ⎛⎜ PBr2 ⎞⎟⎛⎜ PCl2
⎜ RT ⎟⎜ RT
⎠⎝
⎝
P
(RT ) = PBrCl = K
= BrCl
p.
⎞ PBr2 PCl2 (RT )2 PBr2 PCl2
⎟⎟
⎠
2
2
2
This equilibrium expression is called Kp because it uses pressures instead of
concentrations. Kp has the same form as K, but with pressures.
6. The following equilibrium pressures at a certain temperature were observed for the
reaction shown below. Calculate Kp for this reaction.
2 NO2 ( g ) ↔ 2 NO( g ) + O2 ( g )
PNO = 6.5x10-5 atm
PNO2 = 0.55 atm
Kp =
(6.5 × 10
=
(PNO )2 PO
(P )
2
2
NO2
−5
)(
2
PO2 = 4.5x10-5 atm
)
atm 4.5 × 10 −5 atm
= 6.3 × 10 −13
2
(0.55atm)
In some cases, K = Kp. This is only true if the number of moles of gas on the product
side of the reaction are equal to the number of moles of gas on the reactant side.
In general,
K p = K (RT )
Δn
, where Δn = moles of product gas – moles of reactant gas
-1
and R = 0.08206 L atm mol K-1
7. Calculate K for the reaction in problem 6 at 100°C
K=
Kp
(RT )Δn
6.3 × 10 −13
=
= 2.05 × 10 −14
L ⋅ atm ⎞
⎛
3−2
⎜ 0.08206
⎟(373K )
mol ⋅ K ⎠
⎝
Note that this K value is very small (<<1). This reaction favors the reactant side (a
lot of reactant is left compared to the amount of product present at equilibrium)
All of the examples considered at this point have involved gases. If a chemical
reaction involves solids or pure liquids, they are not included in the equilibrium
expression. The fundamental reason for this is that the concentration of a solid or pure
liquid cannot change.
The equilibrium expression for the reaction CaCO3 ( s ) ↔ CaO( s ) + CO2 ( g ) would
be written simply as K = [CO2] because the CaCO3 and CaO are solids and must be
excluded.
8. Write the equilibrium expression for the reaction
CO2 ( g ) + C ( s) ↔ 2CO( g )
2
[
CO]
K=
[CO2 ]
The carbon is left out of the equilibrium expression because it is a solid
9. Write the expressions for K and Kp for the following process:
Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and
chlorine gas
PCl5 ( s) ↔ PCl3 (l ) + Cl2 ( g )
K = [Cl2 ]
K p = PCl2
The phosphorus pentachloride and phosphorus trichloride are left out
of the equilibrium expression because they are solid and pure liquid,
respectively.