Transcript Equilibrium Constant and ICE Tables

7.3 The Equilibrium Constant:
Measuring Equilibrium Concentrations
Equilibrium Calculations (ICE Charts)
Qualitatively Interpreting the Equilibrium Constant
Meaning of Small Equilibrium Constant
SCH4U – Grade 12 Chemistry, University Preparation
Mr. Dvorsky
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SELF QUIZ: Equilibrium Expressions
• Write the equilibrium expression for the
following chemical equations:
1. CO(g) + 1/2 O2(g)
2. 2 HBr(g)
CO2(g)
H2(g) + Br2(g)
3. 4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
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We learned in previous lessons that…
• LAW OF CHEMICAL EQUILIBRIUM:
At equilibrium, there is a constant ratio between the
concentration of reactants and products in any change.
• Equilibrium constant (Keq or Kc) is the ratio of the
forward rate constant and reverse rate constant.
kf = Keq
kr
• The equilibrium expression is
aA + bB
cC+ dD
so:
Kc = [C]c[D]d
[A]a[B]b
• Equilibrium constant is effected by temperature.
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What if you don’t know the
equilibrium constant?
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Measuring Equilibrium Concentrations
• You can determine some equilibrium
concentrations if you know:
▫ the initial concentration(s) of the reactant(s) &
▫ the concentration of one product at equilibrium
[which will tell you the change in concentration(s)].
• Once you have the equilibrium concentrations,
you can find the equilibrium constant!!
• Use an ICE table to perform these calculations:
▫ I = initial concentration
▫ C = change in concentration
▫ E = equilibrium concentration
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Fe3+(aq) + SCN-(aq)
colourless
Fe(SCN)2+(aq)
red
• Fe3+(aq) + SCN-(aq) are both colourless.
• Fe(SCN)2+(aq) is red.
• Because this reaction involves a colour change,
you can determine the equilibrium
concentration of Fe(SCN)2+(aq) by measuring the
change in the intensity of colour (absorbance).
• Knowing the initial concentrations of Fe3+(aq)
and SCN-(aq), you can calculate the equilibrium
concentration of each ion using the chemical
equation.
• Using the equilibrium concentrations, you can
calculate the equilibrium constant!!
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Fe3+(aq) + SCN-(aq)
Concentration
Fe3+(aq)
(mol/L)
Initial
Change
Equilibrium
Fe(SCN)2+(aq)
SCN-(aq)
Fe(SCN)2+(aq)
0.0064
0.0010
0
-4.5 x 10-4
-4.5 x 10-4
4.5 x 10-4
0.006
5.5 x 10-4
4.5 x 10-4
Kc = [Fe(SCN)2+] =
[4.5x10-4]
= 136.36 ≈ 140
[Fe3+][SCN-] [0.006][5.5x10-4]
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What if you don’t know the
equilibrium concentrations?
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Measuring Equilibrium Concentrations
• You can determine some equilibrium
concentrations if you know:
▫ the initial concentration(s) of the reactant(s) &
▫ the equilibrium constant.
• Using this information and an ICE table, you can
calculate the equilibrium concentrations of all
the chemical compounds.
• Use an ICE table to perform these calculations:
▫ I = initial concentration
▫ C = change in concentration
▫ E = equilibrium concentration
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Sample Problem
  H + CO
CO (g) + H 2 O (g) 
2(g)
2(g)
At 700 K, the equilibrium constant is 8.3. Suppose
you start with 1.0 mol of CO(g) and 1.0 mol of H2O(g)
in a 5.0 L container. What amount of each
substance will be present in the container when the
gases are at equilibrium at 700 K?
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  H + CO
CO (g) + H 2 O (g) 
2(g)
2(g)
Reactants
Concentration
(mol/L)
Initial
Change
Equilibrium
Products
CO(g)
H2O(g)
H2(g)
CO2(g)
1/5 = 0.20
1/5 = 0.20
0
0
-x
-x
+x
+x
0.20 - x
0.20 -x
x
x
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  H + CO
CO (g) + H 2 O (g) 
2(g)
2(g)
Kc 
8 .3 
8 .3 
 2 . 88 
 2 . 88 
[H 2 ][CO
[CO][H
2
2
]
O]
( x )( x )
( 0 . 20  x )( 0 . 20  x )
( x)
2
( 0 . 20  x )
( x)
2
( 0 . 20  x )
(x)
( 0 . 20  x )
2 . 88 ( 0 . 20  x )  x
0 . 576  2 . 88 x  x
0 . 576  3 . 88 x
0 . 148  x
 2 . 88 
Doesn’t
make
sense
( x)
( 0 . 20  x )
 2 . 88 ( 0 . 20  x )  x
 0 . 576  2 . 88 x  x
 0 . 576   1 . 88 x
0 . 306  x
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  H + CO
CO (g) + H 2 O (g) 
2(g)
2(g)
So…
Concentration
(mol/L)
Reactants
Products
CO(g)
H2O(g)
H2(g)
CO2(g)
1/5 = 0.20
1/5 = 0.20
0
0
Change
-0.15
-0.15
+0.15
+0.15
Equilibrium
0.05
0.05
0.15
0.15
0.15*5 =
0.75 mol
0.15*5 =
0.75 mol
Initial
So, in a 5.0 L container:
Amount (n) at
Equilibrium
0.05*5 =
0.25 mol
0.05*5 =
0.25 mol
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The previous problem had perfect squares.
It was easy to find the square root both sides and
simplify for x.
Many problems do not involve perfect squares.
You will be required to use the quadratic equation
to find the value of x.
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Sample Problem
The above reaction has an equilibrium constant of 25.0 at
1100 K. In a 1.00 L reaction vessel, 2.00 mol H2(g) and 3.00
mol I2(g) react. What is the equilibrium concentration of
each gas?
• You need to find [H2], [I2], and [HI].
•You need to ensure you have a balanced equation.
•Set up an ICE table, and calculate the initial concentrations
of the gases.
•Let x equal all the change in the concentration of all
chemical species.
•Write the equilibrium expression.
Substitute and rearrange to solve for x.
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Reactants
Products
Concentration
(mol/L)
H2(g)
I2(g)
HI(g)
Initial
2.00
3.00
0
-x
-x
+2x
2.00 - x
3.00 -x
2x
Change
Equilibrium
K
c

[H I]
2
[H 2 ][I 2 ]
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K
c

2 5 .0 
2 5 .0 
[H I]
(2 x )
21x
2
2
x 
2
(2 .0 0  x )(3 .0 0  x )
4x
x 
2
(6 .0 0  2 .0 0 x  3 .0 0 x  x )
 4x
x 
2
2
x 
 125x  150  0
x 
x 
x 
125  55
180
x  4 .3
-b 
2
b
- 4ac
2a
2
125 
(  1 2 5) - 4 (2 1)(1 5 0 )
2(2 1)
125 
15625  12600
42
125 
3025
42
125  55
42
x 
42
42
 bx  c  0
2
2 5 .0 (6 .0 0  5 .0 0 x  x )  4 x
150  125x  25x
ax
[H 2 ][I 2 ]
2
2
2
Doesn’t
make
sense
x 
125  55
42
70
42
x  1 .7
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So…
Reactants
Products
Concentration
(mol/L)
H2(g)
I2(g)
HI(g)
Initial
2.00
3.00
0
Change
-1.7
-1.7
+2(1.7)
2.00 – 1.7
= 0.3
3.00 -1.7
= 1.3
3.4
Equilibrium
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So what does K mean??
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Why is the equilibrium constant important?
• K describes the extent of a reaction!
• A large
K >1 means: [products] > [reactants]
▫  position of equilibrium lies to the right
▫  equilibrium favours the products
• A Kc = 1 means: [products] = [reactants]
• A small Kc <1 means: [products] < [reactants]
▫  position of equilibrium lies to the left
▫  equilibrium favours the reactants
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Sample Problem
  COCl
CO (g) + Cl2 (g) 
2(g)
Consider the reaction of carbon monoxide and
chlorine gas to make phosgene. At 870K, the value
of Kc is 0.20 and at 370 K, the value of Kc is 4.6 x
107. Based only on the values of Kc, is the
production of COCl2(g) more favourable at the
higher or lower temperature?
▫Kc = 0.20 at 870 K
▫Kc = 4.6 x 107 at 370 K
•Is COCl2(g) a product or reactant?
•Which Kc favours COCl2(g )?
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So what does a really small Kc mean??
• When Kc is really small compared to the initial
concentration, the initial value minus x is
approximately equal to the initial value, so you can
ignore x!
• How do you know when you can do this?
Kc = 4.2 x 1o-8 & [NO] = 0.085 mol/L
• Divide the initial concentration by the value of Kc.
▫ If the answer is > 500, the approximation is justified.
▫ If the answer is < 500 but > 100, it may be justified.
▫ If the answer is < 100, you can’t ignore x and need to
solve the equilibrium expression in full.