4/29/2014
Equilibrium Law Calculations
The purple colour of the I2 vapour was used to
monitor the reaction, and from the decreased
intensity of the purple colour it was determined that
at equilibrium, the I2 concentration had dropped to
0.020 mol/L. What is the value of Kc for this
reaction at this temperature?
(with RICE charts)
H2 + I2  2HI
H2
I2
HI
R
1
1
2
I
0.100
0.100
0
C
E
-0.08
-0.08
+0.16
0.02
0.02
0.16
Ratio, Initial, Change, Equilibrium
[HI]2
[.160]2
Kc=
= 64
=
[H2] [I2]
[.020] [.020]
R
PCl3 + Cl2  PCl5
PCl3
Cl2
1
1
I
0.2
0.1
0
C
E
-0.08
-0.08
+0.08
0.12
0.02
0.08
PCl5
1
Ratio, Initial, Change, Equilibrium
Kc =
[PCl5]
[PCl3] [Cl2]
Example 1:
At a certain temperature, a mixture of H2 and I2 was
prepared by placing 0.100 mol of each into a 1.00L
flask. After a period of time, the equilibrium was
established for the reaction.
H2 + I2  2HI
=
[.08]
[.12] [.02]
Example 2:
A student placed 0.20 mol of PCl3 (g) and 0.10 mol
of Cl2 (g) into a 1.00L flask at 250˚C. The reaction,
PCl3 + Cl2  PCl5
was allowed to come to equilibrium, at which time it
was found that the flask contained 0.12 mol of PCl3.
a)What were the initial concentrations of the reactants
and product?
b) What were the changes in concentration?
c) What were the equilibrium concentrations?
d) What is the value of Kc for this reaction?
Using Kc to Calculate Equilibrium
Concentrations.
(Perfect Squares)
• A more complex type of calculation
involves the use of initial concentrations
and Kc to compute equilibrium
concentrations.
= 33.3
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4/29/2014
Calculating Equilibrium Concentrations
Example 3.
CO + H2O  CO2 + H2
The reaction above has Kc = 4.06 at 500˚C.
If 0.100 mol of CO and 0.100 mol of H2O
(g) are placed in a 1.00L reaction vessel at
this temperature, what are the
concentrations of the reactants and products
when the system reaches equilibrium?
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, K, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
Kc =
4.06 =
[CO2][H2]
[CO][H2O]
CO + H2O  CO2 + H2
[CO2][H2]
[CO][H2O]
CO
• We must find values for the concentrations that
satisfy the Kc equation.
• Because we don’t know what these concentrations
are, we must represent them as unknowns,
algebraically.
• Allow x to be equal to the number of moles per
litre of CO that reacted, the change in
concentration of CO and H2O would then be –x
because of the decrease in concentration.
• The coefficients of x are the same as the
coefficients in the balanced equation.
H2O
CO2
H2
R
I
C
E
CO + H2O  CO2 + H2
R
CO
1
H2O
1
0.100
0.100
-x
-x
C
E 0.10 - x 0.10 - x
I
CO2
1
H2
1
0
0
+x
+x
x
x
[x]2
[CO2][H2]
=
= 4.06
[0.10 -x]2
[CO][H2O]
x/[0.10-x] = 2.01, x = 0.201-2.01x, 3.01x = 0.201
x=0.0668
Kc =
Example 4:
During an experiment, 0.200 mol of H2 and
0.200 mol of I2 were placed into a 1.00L
vessel. The reaction was allowed to come
to equilibrium. Find the equilibrium
concentrations of H2, I2 and HI? k = 49.5.
H2 + I2  2HI
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H2
H2 + I2  2HI
I2
HI
H2 + I2  2HI
I2
1
R
R
H2
1
I
I
0.200
0.200
0
C
E
C
E
-x
-x
+2x
0.200 - x
Kc =
[HI]2
0.200 - x
HI
2
2x
[2x]2
=
= 49.5
[H2][I2]
[0.200 -x]2
2x/[0.2-x] = 7.04, 2x = 1.408-7.04x, x=0.156
H2 (I2 also): 0.2 - 0.156 = 0.0443 M
HI: 2(0.156) = 0.312 M
Example 5.
The equilibrium constant Kc for the reaction
was found to be 0.500 at a certain
temperature. If 0.300 mol of SO3 and 0.300
mol of NO were placed in a 2.00L container
and allowed to react, what would be the
equilibrium concentrations of each gas?
SO3 + NO  NO2 + SO2
SO3
NO
NO2
SO2
R
I
C
E
SO3 + NO  NO2 + SO2
R
SO3 + NO  NO2 + SO2
SO3
NO
NO2
SO2
1
1
1
1
0.150
0.150
-x
-x
C
E 0.15 - x 0.15 - x
I
0
0
+x
+x
x
x
[x]2
[NO2][SO2]
=
= 0.50
[SO3][NO]
[0.15 -x]2
.707=x/[0.15-x], 0.106-0.71x=x, x=0.062
SO3, NO: 0.15 - 0.062 = 0.088 M
NO2, SO2: = 0.062 M
Kc =
Example 6.
At a certain temperature the reaction below
has a Kc=0.400. Exactly 1.00 mol of each
gas was placed in a 100L vessel and the
mixture was allowed to react. What was the
equilibrium concentration of each gas?
CO + H2O  CO2 + H2
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4/29/2014
CO + H2O  CO2 + H2
CO
H2O
CO2
CO + H2O  CO2 + H2
CO
H2O
CO2
1
1
1
H2
R
R
I
I
C
E
+x
+x
C
E 0.01+x 0.01+x
0.010
0.010
H2
1
0.010
0.010
-x
-x
0.01 - x 0.01 - x
[CO2][H2]
[0.01 - x]2
=
= 0.40
[CO][H2O]
[0.01+x]2
.6325=(0.01-x)/(0.01+x), x=0.00225
CO, H2O: 0.010 + 0.00225 = 0.0123 M
CO2, H2: = 0.010 - 0.00225 = 0.0078 M
Kc =
Br2 (g) 2 Br (g)
Example 7:
At 12800C the equilibrium constant (Kc) for the reaction
Br2 (g)
Br2
2Br (g)
Br
R
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M
and [Br] = 0.012 M, calculate the concentrations of these
species at equilibrium.
I
C
E
At 12800C the equilibrium constant (Kc) for the reaction
Br2 (g)
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M
and [Br] = 0.012 M, calculate the concentrations of these
species at equilibrium.
Let x be the change in concentration of Br2
Ratiol
Initial (M)
Change (M)
Equilibrium (M)
Kc =
[Br]2
[Br2]
Kc =
Br2 (g)
2Br (g)
1
0.063
2
0.012
-x
+2x
0.063 - x
0.012 + 2x
(0.012 + 2x)2
= 1.1 x 10-3
0.063 - x
(0.012 + 2x)2
= 1.1 x 10-3
0.063 - x
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
-b ± b2 – 4ac
x=
ax2 + bx + c =0
2a
Kc =
2Br (g)
x = -0.0105 x = -0.00178
Initial (M)
Change (M)
Equilibrium (M)
Solve for x
14.4
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.063 – x = 0.0648 M
14.4
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4/29/2014
Example 8:
The following reaction has an equilibrium constant of 25.0
at 1100K.
H2 (g) + I2 (g)
2HI (g)
If 2.00 mol of hydrogen gas and 3.00 mol of iodine gas are
placed in a 1.00L reaction vessel at 1100K, what is the
equilibrium concentration of each gas?
H2
H2 + I2  2HI
I2
HI
R
I
C
E
Equilibrium calculations when Kc is very small
• Thus far, problems have been designed so
that the solution for x is straightforward
• If the problems were not so carefully designed
we might have to use quadratic equation to
solve the problem.
• If Kc is very large or very small we can use a
simplification to make calculating x simple
• Setting up the RICE chart is the same, but the
calculation of Kc is now slightly different
Equilibrium calculations when Kc is small
Looking at the equilibrium law below:
4x3
= small Kc
[0.100 - 2x]2
For Kc to be small, top must be small, bottom
must be large (relative to top)
For top to be small, x must be small
If x is small, then 0.100 - 2x  0.100
Notice that we can only ignore x when it is in a
term that is added or subtracted.
500 Rule
• To test whether you can use a simplifying
assumption, divide the smallest initial
concentration by the Kc value.
• If you get a number > 500, your assumption
is justified and you can proceed with your
calculation ignoring x values that are added
or subtracted.
This is called using a
SIMPLIFYING ASSUMPTION
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Validating a Simplifying
Assumption
500 Rule
• If you get a number between 100 and 500,
you must validate your assumption at the
end of your calculation. (see below)
• If you get a number < 100, you cannot use a
simplifying assumption and must use the
solution to the quadratic equation to solve
for x.
• If the answer is <5% , your simplifying
assumption is justified and you are finished
• If the answer is > 5%, your simplifying
assumption is not justified and you must
redo your calculation using the solution to
the quadratic equation.
N2 + O2  2NO
Example 9.
N2
In air at 25˚C and 1 atm, the N2
concentration is 0.033M and the O2
concentration is0.00810M. The reaction
has Kc=4.8 x 10-31. Taking the N2 and O2
concentrations given above as initial values,
calculate the equilibrium NO concentration.
O2
NO
R
I
C
E
N2 + O2  2NO
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N2 + O2  2NO
N2 + O2  2NO
R
N2
1
O2
1
NO
2
I
0.033
0.00810
0
C
E
-x
-x
+2x
0.033-x
0.00810-x
2x
[NO]2
[2x]2
Kc =
=
= 4.8 x 10-31
[N2][O2] [0.033-x][0.0081-x]
[2x]2
[0.033-x][0.0081-x]
= 4.8 x 10-31
Small Kc: Thus numerator is small and x must be
small: x is negligible when adding or subtracting
Simplifying Assumption:
0.0081
4.8 x 10-31
= 1.7 x 1028
1.7 x 1028 > 500, therefore, the simplifying
assumption can be used. Since the value
is >> 500, no validation is required.
N2 + O2  2NO
0.033-x ≈ 0.033
0.0081-x ≈ 0.0081
[2x]2
[0.033][0.0081]
= 4.8 x 10-31
[2x]2 = 1.28 x 10-34
2x
Example 10:
Given the following reaction information at
a certain temperature, determine the
[equil], if [initial] are [HCl]=2.0 M, [H2]=1.0
M, [Cl2]=0 M
2HCl  H2 + Cl2 Kc= 3.2 x 10–34
= 1.13 x 10-17
[NO] = 2x = 1.1 x 10-17M
2HCl  H2 + Cl2 Kc= 3.2 x 10–34
HCl
H2
2HCl  H2 + Cl2 Kc= 3.2 x 10–34
Cl2
R
R
HCl
2
I
I
2.0
1.0
0
C
E
C
E
-2x
+x
+x
2.0-2x
1.0+x
x
Kc =
H2
1
Cl2
1
[H2] [Cl2] [1.0+x] [x] [1.0] [x]
= 3.2 x 10–34
=
=
[HCl]2
[2.0-2x]2 [2.0]2
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2HCl  H2 + Cl2 Kc= 3.2 x 10–34
Kc =
[1] [x]
[H2] [Cl2] [1+x] [x]
= 3.2 x 10–34
=
=
2
2
[HCl]
[2]2
[2-2x]
x = (3.2 x 10–34)(4) = 1.3 x 10–33
[HCl] = 2.0-2x = 2.0- 2(1.3 x 10-33) = 2.0 M
[H2]=1.0 – 2(1.3 x 10-33) = 1.0M
[Cl2]= x =1.3 x 10–33M
Therefore:
[HCl] = 2.0M, [H2]=1.0M and [Cl2]=1.3 x 10–33M
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