Unit 2 (Part II): Chem 3202. Equilibrium
Equilibrium must have: 1) Reversible reactions (Rx). 2) A closed system
3) Constant Pressure and temperature 4) Observable unchanging
properties
An Example for Achieving Equilibrium:
ex:
Assume [CO] and [H2] are placed in a sealed flask. At time zero, the [CO]
and [H2] are at their maximums, so initially the reaction between is fast.
As the CO and H2 reactants react, their concentrations decrease. As time
goes on, they are less likely to collide and react, the forward Rx slows.
Now let’s go back to consider the product as it forms. At time zero, there
isn't any CH3OH. Its concentration increases as the forward reaction
proceeds. However, as you can see from the slope of the graph, the rate
of its formation decreases. That's because CH3OH decomposes to form
H2 and CO. Eventually, the rates of the
opposing reactions equalize and the amounts of all three substances
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become stable.
At this point there won't be any further change in the amounts of CO, H2,
and CH3OH in the flask, but the two reactions are still occurring! It's just
that as fast as one CO and two H2 combine to make CH3OH, one CH3OH
decomposes to replace them. The system is in dynamic equilibrium.
Notice that the concentrations of the 3 species are not the same. The
only things that are equal at equilibrium are the rates of change.
How to Solve Equilibrium Problems
Equilibrium problems are a common type of problem in Chemistry which
involve the equilibrium constant K.
For the generic reaction
mA + nB → xC + yD,
the equilibrium constant is
The double arrow indicates the reaction goes in both directions (and
never reaches completion). At equilibrium, the rates of the forward and
reverse Rx’s are equal, and there is no further change in concentration.
Kc is a constant for a given reaction at a given temperature. An ICE table
is a way of organizing the data. The three letters in the acronym stand for
initial, change, and equilibrium. Here come some ICE examples.
Exercise 1: An equilibrium was established after 0.100 mol/L of
hydrogen gas and 0.100 mol/L of iodine gas were added to a reaction
vessel and heated to 700 K. At equilibrium, concentration of iodine was
2
0.0213 mol/L. Calculate the equilibrium concentrations of hydrogen &
hydrogen iodide gas.
Step 1: Write the equation and build the ICE table. Check that AL initial
values are CONCENTRATION, if not, make it so.
[H2]
Initial
Change
[I2]
0.100
-X
Equilibrium
[HI]
0.100
0.000
-X
+2X
0.0213
Step 2: Calculate the changes in [I2], [H2], and [HI]
Update the ICE table:
[H2]
Initial
0.100 M
Change
0.0787 M
Equilibrium
[I2]
[HI]
0.100 M
0.000 M
0.0787 M 0.1574 M
0.0213 M
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Step 3: Calculate the equilibrium concentrations.
[H2] Equilibrium = 0.100 M - 0.0787 M = 0.0213 M
[HI] Equilibrium = 0.000 + 0.1574 = 0.1574 M
Example 2: Calculate the Kc at 25 oC for the reaction:
2 NOCl(g)  2 NO(g) + Cl2(g)
Using an initially placed 2.00 mol of NOCl in a 1.00 -L flask, and the
concentration of NO after equilibrium is achieved is 0.66 mol/L.
Solution:
Step 1: Write the equation and build the ICE table. Check that ALL initial
values are CONCENTRATION, if not, make it so.
The initial concentration [NOCl]0 =
I
C
E
2.0 mol
= 2.00 M
1.00 L
2 NOCl(g)
2.00
- 2x
(2.00 - 2x)
2 NO(g)
0
2x
2x = 0.66
Cl2(g)
0
x
x
From the table we have
[NO] = 2x = 0.66 M
x = 0.33 M
Thus [NOCl] = (2.00 - 2x) = (2.00 - 2(0.33)) = 1.34M
[Cl2] = x = 0.33 M
Step 3: We now have the information we need to calculate Kc:
2
æç [NO]2 [Cl 2 ]ö
(0.66) (0.33)
Kc =
=
= 0.080
è [NOCl]2 ø ( at equilibrium )
(1.34 )2
Calculating Equilibrium Concentrations from Percent Reaction
Exercise 3:
In a 1.00 L vessel, 0.200 mol of nitrogen and 0.200 mol of oxygen gas
were reacted at 773 K to produce nitrogen monoxide gas. The percent
reaction was found to be 12.5%. Calculate the equilibrium concentrations
of all three species.
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I
0.200
0.200
0.000
C
0.0250
0.0250
0.0500
E
Calculate the equilibrium concentrations.
[N2] Equilibrium = 0.200 M - 0.0250 M = 0.175 M
[O2] Equilibrium = 0.200 - 0.0250 = 0.175 M
[NO] Equilibrium = 0.000 + 0.0500 = 0.0500 M
THE FOLLOWING ARE MORE PROBLEMS, YOU SHOULD TRY THEM
YOURSELF, AND NOT LOOK AT THE SOLUTION UNTIL YOU HAVE
WORKED THEM OUT ON YOUR OWN.
Solving equilibrium problems using Kc
If you are given any chemical reaction at equilibrium and the
concentration of each species, you can calculate the value of the
equilibrium constant. For ex, if you are given the following equation:
N2(g) + 3 H2(g)
2 NH3(g) + 92 kJ
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Suppose at equilibrium the concentration of
N2 = 0.142 mol/L, the [H2] = 0.341 mol/L and the [NH3] = 0.220 mol/L.
You can calculate the value of the Keq by first writing the Keq expression:
Keq =
[NH3]2
[N2] [H2]3
If we fill in the numbers we get:
Keq =
[NH3]2 =
[N2] [H2]3
(0.220 mol/L)2
(0.142 mol/L)(0.341 mol/L)3
[NH3]2 =
[N2] [H2]3
(0.220 mol/L)2
(0.142 mol/L)(0.341 mol/L)3
Working this out gives
Keq =
= 8.60
This means that products are slightly favoured over reactants, since the
ratio is slightly over 1 and Products are favored.
More ICE box problems
This type of problem gives you information about the initial situation,
before equilibrium is established. Then you get 1 or more pieces of
information concerning the situation at equilibrium.
1) Consider the following reaction: 2 A(g) + B(g)
C(g) + 3 D(g)
0.100 mol/L each of A and B are placed in a flask and allowed to come to
equilibrium. At equilibrium 0.022 mol/L of C is present.
Calculate the value of the Keq.
Step 1. Write down what you know:
2 A(g)
[I]
+
0.100 mol/L
B(g)
C(g)
0.100 mol/L
0.0
+
3 D(g)
0.0
[C]
[Equilibrium]
0.022 mol/L
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Step 2. Determine the Change in Concentration:
2 A(g)
+
B(g)
[I] 0.100 mol/L
[C] -(2/1)0.022 M
C(g)
0.100
+
3 D(g)
0.0
-(1/1)0.022 M
= -0.044M
0.0
+0.022 M
= -0.022M
[Equilibrium]
+(3/1)0.022M
= +0.066M
0.022 mol/L
You know the change in concentration of C is +0.022 mol/L, since it
started at 0.0 mol/L and ended at 0.022 mol/L at equilibrium. The
changes of all other components is proportional to the change in [C].
This works exactly the same as mole ratio, mole bus, does in gravimetric
stoichiometry.
Step 3. Determine the Equilibrium Concentrations:
2 A(g)
+
B(g)
<->
[Initial] 0.100 mol/L
0.100 mol/L
[C] -(2/1)0.022M
or -0.044 mol/L
-(1/1)0.022 M
or -0.022 mol/L
[Equilibrium]
0.056
0.078
C(g)
+
3 D(g)
0.0
+0.022M
0.0
+(3/1)0.022M=
or +0.066 mol/L
0.022
0.066
To get each equilibrium concentration subtract the change in each case
from the initial concentration.
Step 4:
Write the Keq expression and calculate the Keq:
Keq =
[C][D]3 =
[A]2[B]
(0.022 mol/L)(0.066 mol/L)2
(0.056 mol/L)2(0.078 mol/L)
Keq = 0.026
======================================================
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2). Consider the following reaction:
A(g) + B(g)
2 C(g)
If the initial concentration of C is 0.350 mol/L, calculate the
equilibrium concentrations of all three species if the value of the Keq is
145.
Step 1. Write down what you know:
A(g)
[Initial]
+
0.0
B(g)
2 C(g)
0.0
0.350 mol/L
[Change]
[Equilibrium]
Step 2. Determine the Change in Concentration:
A(g)
+
B(g)
[Initial]
0.0
0.0
[Change]
+x
+x
2 C(g)
0.350 mol/L
- 2x
[Equilibrium]
When you don’t know something in math, you call it ‘x’. The change in
concentration of ‘C’ reflects the stoichiometry of the equation. The
reactants are positive x because since they start at zero, they have no
where to go but up.
Step 3. Determine the Equilibrium Concentrations:
A(g)
+
B(g)
[Initial]
0.0
0.0
[Change]
+x
+x
x
x
[Equilibrium]
2 C(g)
0.350 mol/L
- 2x
0.350 mol/L - 2x
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Step 4: Write the Keq expression and calculate the value of x:
[C]2
[A][B]
Keq =
(0.350 - 2x)2
(x)(x)
145 =
=
(0.350 - 2x)2
x2
Take the square root of both sides:
√145 =
√(0.350 - 2x)2
√x2
x = 0.0249 mol/L
Step 5.
Solve for the concentration of each species.
A(g)
[Initial]
+
0.0
[Change]
0.0
+x
[Equilibrium]
0.0249 mol/L
B(g)
x
2 C(g)
0.350 mol/L
+x
- 2x
x
0.350 mol/L - 2x
0.0249 mol/L
0.350 mol/L – 2(0.0249 mol/L)
= 0.300 mol/L
Ice Box Problems – Practice
o
1. At 430 C the Kc is 1.84 x 10-2for the following reaction:
2HI(g)  H2(g) + I2(g)
If 0.100 mol HI(g) is placed in a 1.00 L container and allowed to reach
equilibrium, find the equilibrium concentrations.
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The reaction CO(g) + H2O(g)  CO2(g) + H2(g) has a Keq of 4.06 at
2.
o
500 C. If 0.100 mol each of CO(g) and H2O(g) are placed in a
flask, what are the equilibrium [ ]'s of the products?
1.00 L
3.
At equilibrium the concentration of CO is 0.010 M, H2O is 0.020 M,
CO2 is 0.010 M, and H2 is 0.010 M. For the following
reaction, does
the equilibrium favour the formation of the reactants or the products?
How do you know this?
CO(g) + H2O(g)  CO2(g) + H2(g)
4.
If 0.054 moles of N2O4 is placed in a 2.0 L flask, which is then
sealed. At equilibrium, it is found that [NO2] = 0.050 M. Find the Keq for
the following reaction:
N2O4 (g)  2 NO2 (g)
5.
For the reaction H2 (g)+ F2 (g)  2HF (g), the equilibrium constant
is 15.0. If the initial [H2] and [F2] are each 0.200 M, what is
the
equilibrium concentration of HF?
o
6.
For the reaction following reaction the Keq = 2.00 at 1000 C:
2COF2(g)  CO2(g) + CF4(g)
If a 6.00L vessel contains 0.105 mol COF2, 0.220 mol CO2, 0.550
o
mol CF4at 1000 C, will the mixture be at equilibrium?
o
At 1530 C the value of Keq for the reaction below is 1.20 x 10-4
N2 (g)+ O2 (g) 2NO (g)
If 0.3366 mol of nitrogen gas and 0.3366 mol of oxygen gas are placed in
a 3.00L vessel and allowed to reach equilibrium, determine the [NO]
once equilibrium has been established.
7.
8.
At a particular temperature, Keq for the reaction
H2(g) + I2(g)  2HI(g)
has a value of 55.6. If the initial [H2] = 0.200M and the initial [I2] =
0.200M, what is the equilibrium [HI]?
9.
Consider the following equilibrium reaction
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H2(g) + Br2(g)  2HBr(g) + energy
o
The Keq for this reaction is at 25 C is 1.02. At equilibrium the
concentration of HBr is 0.50 M. Assuming H2 and Br2 are present in
equal amounts, calculate the concentration of H2 at equilibrium.
10.
Reactants W and X each at a concentration of 0.80 mol/L, react
slowly to produce U and V according to the following
equation:
W+XU+V
At equilibrium, the molarity of U is 0.60M. Calculate K?
11.
A 1.00L reaction container initially contains 9.28 x 10 mol of H2S(g)
-3
-3
At equilibrium the concentration of H2S = 7.06 x 10 M. Calculate the
equilibrium constant for this reaction.
2H2S(g)  2H2(g) + 2S(g)
12. Calculate Keq given 2A(g) + B(g)  3C(g) where there is an initial
of [A] is 0.95M, [B] is 0.79M and the final concentration of [B] is 0.61M.
Answers:
1. [HI] = 0.0787M [H2] = [I2]= 0.0107M
2. [CO2]=[H2]=0.0668M
3. Neither the reactants are favored because Keq is 0.50 or close to one,
to be “favoured” it should be a factor of 10 from zero.
4. 1.3
5. [HF] = 0.264M
6. No because the calculated Keq does not equal the accepted Keq
-3
7. 1.22 x 10
9. 0.50M
M
8. [HI] = 2x= 0.315M
10. 9.0
-7
11. 4.87 x 10
12. 0.74
Le Chatelier’s Principle
A reaction at equilibrium will proceed in a direction that relieves the
stress put upon it. The equilibrium position changes but the equilibrium
constant does not change unless temp changes: Possible stresses are:
A) Temperature : If exo, a temperature increase will make rxn
proceeds to the left (to reactants). If endothermic, proceed right .
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B) Volume: Look at the number of moles of gas, decreasing the
volume proceed towards the side with fewer moles of gas
C) Pressure: Look at the number of total moles of gas on the reactant
then product side. Pressure can be changed in 2 ways
1) Changing volume: pressure and volume inversely linked
2) Adding an inert gas: Changing pressure this way does NOT change
reaction rate, it does nothing.
D) Concentration
1)If a reactant’s concentration is increased, reaction shifts towards right
products (RUR, [reactant] up … go right)
2) If a product’s concentration is increased, reaction shifts to reactants
(PUL , [product] up…go left)
Note: A catalyst may be introduced to any system at equilibrium BUT
reactants nor products will increase in any amount. No side will be
favored.The rate of the forward reaction will increase BUT so will the rate
of the reverse rection. THUS, the exchange of reactants and products will
increase but the ultimate amounts will not.
AND, an INERT gases will not favor the reactants or prodiucts
Le Chatelier’s Examples:
2H2(g) + O2(g) + energy  2H2O(g)
1) If [H2] increases, the reaction will proceed to right to produce more
product and used up the excess [H2]
2) If [H2] is decreased, the reaction will proceed to left to make up some
of the H2 that was lost
3)If temperature is increased, because energy is a reactant, reaction will
proceed to right to use up the added energy.
4)If temperature is decreased, because energy is a reactant, reaction will
proceed to left to make up some of the energy that was lost
5) When pressure is increased (or Volume decreases ). Reaction will
proceed to make fewer moles of gas, will shift to right
6) When pressure in decreased (or Volume increased).Reaction will
proceed to make more moles of gas, will shift to left
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