Chem 1011 – Intersession 2011
Class #23
03-Jun-11
1
Class 23: Finding Equilibrium
Concentrations / LeChatelier’s
Principle
•
2
Announcements
• Monday, June 6th
▫ No Lab – Labs will resume Wed, June 8th
▫ Tutorial – Everyone go to the first tutorial session (2:00pm) for
that day only
Sec 14.8 – Finding Equilibrium Concentrations
▫
• Wednesday, June 8th
Given the equilibrium constant and all but one of the
equilibrium concentrations of the reactants and the
products.
Given the equilibrium constant and the initial
concentrations or pressures.
Simplifying approximations in working equilibrium
problems
▫
▫
▫ Class at 11am will outline material covered on the midterm
• Thursday, June 9th
▫ Review Session from 12:00pm – 2:00pm in C-3033
▫ Come with Questions!
• Friday, June 10th
▫ MIDTERM #2 @ 9:00am in C-4002
• Tuesday, June 21st
▫ First day of Summer
▫ Oh right... the final exam. 9:00am – 11:30am, Room TBD
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Finding Equilibrium Concentrations
4
Finding Equilibrium Concentrations
• Scenario 1
• The Plan:
▫ Given
1.
 The Equilibrium Constant
 All but one Equilibrium Concentration / Pressure
Identify Knowns and Unknowns


▫ Find
 The remaining equilibrium value
2.
• Example
▫ In an equilibrium mixture, the concentration of COF2 is
0.255 M and the concentration of CF4 is 0.118 M. What is
the equilibrium concentration of CO2?
[COF2] = 0.255 M
[CO2]
[CF4] = 0.118 M
Kc = 2.00
Rearrange

2 COF2(g) ⇌ CO2(g) + CF4(g) Kc = 2.00 @ 1000oC
Given
Find
Solve for the unknown ([CO2])
COF2 2  K  CO 2 CF4  COF2 2
c
CF4 
CF4 
COF2 2
2
K c COF2 
CO 2  
CF4 
5
Finding Equilibrium Concentrations
3.
Substitute and Solve
CO 2   K c COF2 
CF4 
CO 2   1.10 M
2


(2.00)(0.255) 2
(0.118)
Note: Since K is unitless, all units are omitted from this
calculation. However, since your are solving for
concentration, but sure to include the appropriate units.
6
Problem
•
Diatomic iodine (I2) decomposes at high
temperatures to form I atoms according to the
reaction:
I2(g) ⇌ 2 I(g)
Kc = 0.011 at 1200 oC
In an equilibrium mixture, the concentration
of I2 is 0.010 M. What is the equilibrium
concentration of I?
Finding Equilibrium Concentrations
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Chem 1011 – Intersession 2011
Class #23
03-Jun-11
7
Finding Equilibrium Concentrations
8
Finding Equilibrium Concentrations
• Scenario 2 (more common)
• Our ICE Table will look like:
▫ Given
 The Equilibrium Constant
 Initial Concentrations of Reactants and Products
▫ Find
 All Equilibrium Concentrations
•
Consider the reaction:
A(g) ⇌ 2 B(g)
Kc = 0.33
The initial concentration of A = 1.0 M. Solve for the
equilibrium concentrations of A and B.
Kc 
B2
A
Substitute
0.33 
(2 x) 2
(1.0  x)
Need to solve for x
This is a Quadratic!
Dun dun dun....
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Finding Equilibrium Constants
Finding Equilibrium Constants
3.
• Strategy:
1.
10
Substitute values into the expression for K, and solve for
x. Since these are quadratics, you may need the quadratic
formula:
Determine the direction of shift
If you start with an initial amount of 0 for any substance,
then all species on that side must gain
If you have non-zero initial amounts for all, then you must
compare Q to K to determine the direction of the shift




2.
If

Construct your ICE Table


Define all changes in terms of “x”
The magnitude of the change depends on the coefficient.
The coefficient in the equation will be the coefficient for x
Then:
x
 b  b 2  4ac
2a
Often times you may be able to simplify and avoid the
dreaded equation – be on the lookout for that!
The quadratic equation results in 2 values of x – we are only
interested in the one that leads to positive equilibrium
concentrations

If Q < K – Shift to the right (towards products)
If Q > K – Shift to the left (towards reactants)
ax 2  bx  c  0
4. After finding x, return to the ICE table and substitute the
value for x to find all equilibrium concentrations.
11
Example 14.9
Example 14.9
• Consider the following reaction:
N2(g) + O2(g) ⇌ 2 NO(g)
Kc = 0.10 @ 2000oC
A reaction mixture (at 2000oC) initially contains [N2] =
0.200 M and [O2] = 0.200 M. Find the equilibrium
concentrations of the reactants and products at this
temperature.
1. Determine the direction of the shift
▫
We have an initial [N2] and [O2] – but not [NO].
Therefore, without finding Qc, we know the equilibrium
must shift to the right


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Positive Change for NO
Negative Change for N2 and O2
Finding Equilibrium Concentrations
2. Construct an ICE table
▫
▫
Define all changes in terms of “x”
Remember that the coefficient in the equation is
the coefficient with x!
N2(g)
+
O2(g)
⇌
2 NO(g)
[I]
0.200 M
0.200 M
0
[C]
–x
–x
+2x
[E]
0.200 – x
0.200 – x
2x
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Chem 1011 – Intersession 2011
Class #23
03-Jun-11
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Example 14.9
Example 14.9
3. Substitute values into the expression for K and solve
for x (use the quadratic equation, if necessary)
NO2
Kc 
N 2 O 2 
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• Crisis averted! No quadratic needed here!
(2 x) 2
0.10 
(0.200  x)(0.200  x)
Substitute
(2 x) 2
 2x 


2
(0.200  x)  0.200  x 
2
Square Root
 2x 
0.10  

 0.200  x 
2x
0.200  x
2 x  0.316(0.200  x)
2 x  0.0632  0.316 x
2.316 x  0.0632
0.316 
The denominator can be simplified!
(0.200 – x)(0.200 – x) = (o.200 – x)2
0.10 
 2x 
0.10  

 0.200  x 
2
x
0.0632
2.316
x  0.027 M
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Example 14.9
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Problem
4. After finding x, substitute it in the ICE table and solve
for equilibrium concentrations.
N2(g)
+
O2(g)
⇌
2 NO(g)
•
[I]
0.200 M
0.200 M
0
[C]
–0.027 M
–0.027 M
+2 (0.027 M)
[E]
0.173 M
0.173 M
0.054 M
•
Consider the following reaction:
I2(g) + Cl2(g) ⇌ 2 ICl(g)
NO2  0.0542  0.097  0.10
N 2 O2  0.1730.173
Kp = 81.9 @ 25 oC.
The initial pressure of I2 is 0.150 atm, and Cl2 is 0.125
atm. Find the equilibrium partial pressures of all
three gases.
Check: Do these concentrations give you Kc?
Kc 

▫
This example will need the quadratic equation!
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The “Simplifying Assumption”
•
•
We know that when the equilibrium constant is very
small (negative exponent), the position of equilibrium
favors the reactants.
Therefore, in these cases:
If the initial concentration of reactant is relatively
large, it will not change significantly when it reaches
equilibrium.
▫
the [X]equilibrium = ([X]initial  ax)  [X]initial
▫

▫
2
The equilibrium concentration of reactant is
approximately the same as the initial concentration.
This assumes the reaction is proceeding forward
Finding Equilibrium Concentrations
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The “Simplifying Assumption”
• How do we know if we can do this?!?
▫ If K is small, go for it. But, it’s a catch-22 – you cannot
validate your assumption until after you’ve solved for x
▫ We check by comparing the approximate value of x to the
initial concentration.
▫ If the approximate value of x is less than 5% of the initial
concentration, the approximation is valid.
▫ If you make “the assumption,” but it is then proved invalid,
you need to re-do the calculation without making that
assumption
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Chem 1011 – Intersession 2011
Class #23
03-Jun-11
19
Example 14.12
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Example 14.12
• Consider the following reaction for the decomposition of
hydrogen disulfide:
2 H2S(g) ⇌ 2 H2(g) + S2(g)
2. Construct an ICE table
▫
▫
Kc = 1.67 x 10–7 @ 800oC
Define all changes in terms of “x”
Remember that the coefficient in the equation is the
coefficient with x!
A 0.500-L reaction vessel initially contains 0.0125 mol of H2S
at 800oC. Find the equilibrium concentrations of H2 and S2.
H 2S 
1. Determine the direction of the shift
▫
V

0.0125 mol
 0.0250 mol  L1
0.500 L
2 H2S(g)
We have an initial [H2S] – but not [H2] or [S2]. Therefore,
without finding Qc, we know the equilibrium must shift to the
right


n H 2S
Positive Change for H2 and S2
Negative Change for H2S
⇌
2 H2(g)
+
S2(g)
[I]
0.0250 M
0
0
[C]
–2x
+2x
+x
[E]
0.0250 – 2x
2x
x
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Example 14.12
Example 14.12
3. Substitute values into the expression for K and solve
for x (use the quadratic equation, if necessary)
Kc 
H 2 2 S2 
H 2S2
1.67 10-7 
Substitute
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1.67 10-7 
( 2 x) 2 ( x)
(0.0250  2 x) 2
1.67 10-7 
4 x3
4 x3

2
(0.0250  2 x)
0.02502
1.67 10-7 
4 x3
6.25 10 4


4 x3  1.67 10-7 6.25 104
4 x3
(0.0250  2 x) 2
x3 
Nasty! It is possible to solve for x here, but it is fairly difficult.
However, Kc is a very small number – so we can assume x is small.
Therefore:
0.0250 – 2x  0.0250
1.044 10
4

10
x  3 2.609 1011  2.97 104 M
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Example 14.12
• Before going any further, since you now know x, check
the validity of your assumption to make sure it’s good!
x
2.97 104 M
100% 
100%  1.19%
H 2S
0.0250 M
1.19%  5%
 Assumption is Good!
• Now that x has been validated, you can substitute it back
into the ICE table...
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Example 14.12
• At equilibrium:
[H2S] = 0.0250 M – 2x = 0.0250 M – 2(2.97 x 10–4 M)
= 0.0244 M
[H2] = 2x = 2(2.97 x 10–4 M)
= 5.94 x 10–4 M
[S2] = x = 2.97 x 10–4 M
▫ Check! Would you still get roughly the same value of K?
Finding Equilibrium Concentrations
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Chem 1011 – Intersession 2011
Class #23
03-Jun-11
25
Problem
•
Next Chapter
Consider the following reaction:
2 H2S(g) ⇌ 2 H2(g) + S2(g)
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•
Sec 14.9 – Le Chatelier’s Principle: How a
System at Equilibrium Responds to Disturbances
▫
Kc = 1.67 x 10-7 @ 800oC
▫
▫
Initially, [H2S] = 0.100 M and [H2] = 0.100 M. Find
the equilibrium concentration of S2.
•
Sec 15.2 – The Nature of Acids and Bases
•
Sec 15.3 – Definitions of Acids and Bases
▫
▫
Finding Equilibrium Concentrations
The effect of a concentrations change on
equilibrium
The effect of a volume or pressure change
The effect of a temperature change
The Arrhenius Definition
The Brønsted-Lowry Definition
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