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Entry Task: Jan
th
8
Tuesday
At 1100 K, Kp = 0.25 atm1 for the reaction:
2 SO2(g) + O2(g) ↔ 2 SO3(g)
What is the value of Kc at this temperature?
Equilibrium
Agenda:
• Discuss Equilibrium constant Kp and Kc
ws
• In-Class ICE table practice
• HW: Pre-Lab Determining Keq Constant
Equilibrium
Equilibrium
1. Calculate Kp and Kc for the following
reaction…..
CH3OH (g)  CO (g) + 2H2 (g)
Given the following equilibrium pressures at
25oC:
P CH3OH=6.10 x 10-4 atm
2
[ PCO ] [ PH 2]
PCO = 0.387 atm
Kp 
PH2= 1.34 atm
[ PCH 3OH ]
[0.387] [1.34]
Kc 
[6.10 x10  4]
2
Kc= 1.14 x 10-3
Equilibrium
1. Calculate Kp and Kc for the following
reaction…..
CH3OH (g)  CO (g) + 2H2 (g)
Given the following equilibrium pressures at
25oC:
2
P CH3OH=6.10 x 10-4 atm
[0.387] [1.34]
Kc 
PCO = 0.387 atm
[6.10 x10  4]
PH2= 1.34 atm
Kp= Kc(RT)∆n
∆n = 3-1 =2
Kp= (1.14 x 10-3)((0.0821)(298))2
Kp= 0.68
Equilibrium
2. A 127°C, Kc = 2.6 × 105 mol2/L2 for the
reaction: 2 NH3(g)  N2(g) + 3H2(g)
Calculate Kp at this temperature.
Kp= Kc(RT)∆n
∆n = 4-2 =2
Kp= (2.6 × 105)((0.0821)(400))2
Kp= 2.8x10-2
Equilibrium
3. At 1100 K, Kp = 0.25 atm1 for the reaction:
2 SO2(g) + O2(g)  2 SO3(g)
What is the value of Kc at this temperature?
Kp= Kc(RT)∆n
∆n = 2-3 =-1
0.25= (X)((0.0821)(1100))-1
Kc= 23
Equilibrium
4. Phosphorus pentachloride dissociates on
heating: PCl5(g)  PCl3(g) + Cl2(g) If Kc equals
3.28x10-2 at 191°C, what is Kp at this
temperature?
Kp= Kc(RT)∆n
∆n = 2-1 =1
Kp= (3.28x10-2)((0.0821)(464))1
Kp= 1.25
Equilibrium
5. Nitrogen dioxide dimerizes to form dinitrogen
tetraoxide: 2 NO2(g) N2O4(g)
Calculate the value of Kc, given that the gas
phase equilibrium constant, Kp, for the reaction is
1.3 × 103 at 273 K. (R = 0.08206 L·atm/mol·K)
Kp= Kc(RT)∆n
∆n = 1-2 =-1
1.3 x10-2= (X)((0.08206)(273))-1
Kc= 0.29
Equilibrium
6. An equilibrium mixture of SO3, SO2, and O2 at
1000 K contains the gases at the following
concentrations: [SO3] = 0.41 M, [SO2] = 0.032 M,
and [O2] = 0.59 M. What is the equilibrium
constant (Kc) and the Kp value for the
decomposition of SO3?
2 SO3(g) 2 SO2(g) + O2(g)
2
2
[0.032] [0.59]
Kp 
2
[0.41]
[ PSO 2] [ PO 2]
Kp 
2
[ PSO 3]
Kc= 3.59 x 10-3
Equilibrium
6. An equilibrium mixture of SO3, SO2, and O2 at
1000 K contains the gases at the following
concentrations: [SO3] = 0.41 M, [SO2] = 0.032 M,
and [O2] = 0.59 M. What is the equilibrium
constant (Kc) and the Kp value for the
decomposition of SO3?
2 SO3(g) 2 SO2(g) + O2(g)
Kp= Kc(RT)∆n
∆n = 3-2 =1
Kp= (3.59 x 10-3)((0.0821)(1000))1
Kp= 2.95 x 10-1
Equilibrium
Equilibrium
15.4- Calculating Equilibrium Constant
When a reaction has reached equilibrium,
we often don’t know HOW the initial
concentrations of the species have changed
from the equilibrium concentrations.
Or we were given JUST the initial
concentrations, how can we determine the
equilibrium concentrations?
Equilibrium
Equilibrium
To look at all these variables at the same
time we need to create an ICE table.
I is initial concentration,
C is the change in concentration
E is the concentration at equilibrium.
Equilibrium
There are two types of ICE tables
Type 1
A. The initial or equilibrium concentration of some
substances must be determined.
B. Initial or equilibrium concentrations of some
substances are given, but not both. Change is
therefore treated as an unknown (x)
C. The equilibrium constant is given.
Equilibrium
There are two types of ICE tables
Type 2
A. The equilibrium constant or concentration
must be determined.
B. Initial and equilibrium concentrations of at
least one substance are given so that
change can be calculated directly.
C. All other initial and equilibrium concentrations
of substances are determined directly from
the table.
Equilibrium
Walkthrough-(See pg. 571-2 Sample Exercise)
A closed system initially containing
1.000 x 10−3 M H2 and 2.000 x 10−3 M I2
At 448C is allowed to reach equilibrium.
Analysis of the equilibrium mixture shows that the
concentration of HI is 1.87 x 10−3 M. Calculate
Kc at 448C for the reaction taking place, which is
H2 (g) + I2 (g) 2 HI (g)
Equilibrium
STEPS TO ICE
① Make Sure that all concentrations are in
M- molarity!! (Done for you)
② Set up table- ICE (as you see it) and
then species at top.
③Place the known concentrations provided
in the question into table.
④Put in the CHANGE for HI (subtract
equilibrium from initial)
Equilibrium
② Set up table- ICE (as you see it) and
then species at top.
[H2], M
[I2], M
[HI], M
Initially
Change
At
equilibrium
Equilibrium
③Place the known concentrations provided
in the question into table.
[H2], M
Initially
[I2], M
1.000 x 10-3 2.000 x 10-3
[HI], M
0
Change
At
equilibrium
1.87 x 10-3
Equilibrium
④Put in the CHANGE for HI (subtract
equilibrium from initial)
[H2], M
Initially
[I2], M
1.000 x 10-3 2.000 x 10-3
[HI], M
0
Change
+1.87 x 10-3
At
equilibrium
1.87 x 10-3
Equilibrium
⑤NOW we have to use the stoichiometry of the
reaction to get the change of H2 and I2. Put a
negative sign in front because they are reactants.
Initially
[H2], M
[I2], M
[HI], M
1.000 x 10-3
2.000 x 10-3
0
Change
+1.87 x 10-3
At
equilibrium
1.87 x 10-3
H2 (g) + I2 (g) 2 HI (g)
1.87 x 10-3mol
L
1 mol H2
2 mol of HI
= 0.935 x 10-3
Same goes for iodine
Equilibrium
⑤ Stoichiometry tells us [H2] and [I2]
decrease by half as much
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
0
Change
-9.35 x 10-4
-9.35 x 10-4
+1.87 x 10-3
At
equilibrium
1.87 x 10-3
The change MUST be in the negative because
Equilibrium
they are reactants!!
⑥ Subtract the initial concentrations from
the change which will provide the
equilibrium value.
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3 2.000 x 10-3
Change
-9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At
equilibrium
6.5 x
10-5
1.065 x
10-3
0
1.87 x 10-3
1.000 x10-3 – 9.035 x10-4) =6.5 x 10-5
2.000 x10-3 – 9.035 x10-4) =1.065 x 10-3
Equilibrium
⑦ Finally, provide the equilibrium
expression for this reaction.
Substitute the equilibrium values from chart
into the expression to solve for Kc.
=
[HI]2
Kc =
[H2] [I2]
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
Kc = 51
Equilibrium
Let’s Try another
Sulfur trioxide decomposes at high temperature in a
sealed container: 2SO3(g)  2SO2(g) + O2(g). Initially,
the vessel is charged at 1000 K with SO3(g) at a
concentration of 6.09 x 10-3 M. At equilibrium the SO3
concentration is 2.44 x10-3M. Calculate the value of Kp
at 1000 K.
2SO3
INITIAL
CHANGE
EQUILIBRIUM

6.09 x 10-3 M
2SO2
O2
0M
0M
2.44 x10-3M
Equilibrium
Let’s Try another
2SO3
INITIAL
CHANGE
EQUILIBRIUM

6.09 x 10-3 M
2SO2
O2
0M
0M
-3.65 x 10-3 M
2.44 x10-3M
What information can we fill in with what we are given?
We can fill in the change of SO3.
6.09 x 10-3 M - 2.44 x10-3M = -3.65 x 10-3 M
Equilibrium
Let’s Try another
2SO3
INITIAL
CHANGE
EQUILIBRIUM

6.09 x 10-3 M
2SO2
O2
0M
0M
-3.65 x 10-3 M +3.65 x 10-3 M +1.83 x 10-3 M
2.44 x10-3M
If SO3 went down by -3.65 x 10-3 M we have to use
stoichiometry to find out the relationship the products
have with the reactant.
2
moles
SO
-3
2
= +3.65 x 10-3 M
3.65 x 10 M of SO3
2 mole of SO3
of SO
2
3.65 x 10-3 M of SO3 1 moles O2
2 mole of SO3
-3 M
= +1.83 x 10
Equilibrium
of O2
Let’s Try another
2SO3
INITIAL
CHANGE
EQUILIBRIUM

2SO2
O2
0M
0M
6.09 x 10-3 M
-3.65 x 10-3 M +3.65 x 10-3 M +1.83 x 10-3 M
2.44 x10-3M 3.65 x 10-3 M 1.83 x 10-3 M
Now subtract the initial from the change to get the
equilibrium.
0 - 3.65 x 10-3 M = 3.65 x 10-3 M
0 – 1.83 x 10-3 M = 1.83 x 10-3 M
Equilibrium
Let’s Try another
2SO3
INITIAL
CHANGE
EQUILIBRIUM

6.09 x 10-3 M
2SO2
O2
0M
0M
-3.65 x 10-3 M +3.65 x 10-3 M +1.83 x 10-3 M
2.44 x10-3M 3.65 x 10-3 M 1.83 x 10-3 M
Use the equilibrium concentrations to find Kcplug n chug
2
[ SO 2] [O 2]
Kc 
[ SO 3]2
-3 2
-3
[3.65 x 10 ] [1.83 x 10 ]
Kc 
-3 2
[2.44 x10 ]
(1.33225 x10-5)(1.83 x10-3) = 2.4380175 x10-8
-6
5.9536
x10
Kc= 4.11 x 10-3
Equilibrium
In many situations we will know the
value of the equilibrium constant
and the initial concentrations of all
species. We must then solve for
the equilibrium concentrations. We
have to treat them as variables or
“x”
Equilibrium
A chemist has a container of A2 and B2 and they
react as given: A2 (g) + B2 (g)  2 AB (g)
Kc = 9.0 at 100°C
If 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0
L container, what are the equilibrium
concentrations of A2, B2, and AB?
① Convert to molarity!!
1 mol = 0.50 M
2.0 L
Equilibrium
A chemist has a container of A2 and B2 and they
react as given: A2 (g) + B2 (g)  2 AB (g)
Kc = 9.0 at 100°C
If 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0
L container, what are the equilibrium
concentrations of A2, B2, and AB?
[A2], M
Initially
0.50 M
[B2], M
0.50 M

[AB], M
0.0 M
Change
At
equilibrium
Equilibrium
[A2], M
Initially
Change
At
equilibrium
0.50 M
-x
0.50-x
[B2], M

0.50 M
[AB], M
0.0 M
-x
0.50-x
+ 2x
2x
Fill in the chart with variables!!
A2(g) + B2(g)  2AB (g)
You need to (for this ICE table) to factor in the
stoich relationship)
Equilibrium
Initially
Change
At
equilibrium
[A2], M
[B2], M
0.50M
0.50 M
-x
0.50-x
-x
0.50-x
We have to work backwards!!
Provide the Kc expression
(2x)2
9.0=
(0.50-x)(0.50-x)

[AB], M
0M
+ 2x
2x
[AB]2
Kc =
[A2] [B2]
Equilibrium
9.0=
3.0=
(2x)2
Root both sides
(0.50-x)(0.50-x)
Or
(0.50-x)2
(2x)
(0.50-x)
Get 0.50-x out by
multiply on both
sides
Equilibrium
3.0=
(2x)
(0.50-x)
3.0(0.50-x)=
1.5-3x= 2x
2x
Get 0.50-x out by
multiply on both
sides
Multiply 3 through
Add 3x to both sides
to get rid of -3x
Equilibrium
1.5-3x= 2x
Add 3x to both sides
to get rid of -3x
1.5 = 5x
Divide both sides by
5 to get 5 out of
there
0.3 = x
SO back to the chart with
our x value of 0.3
Equilibrium
[A2], M
Initially
Change
At
equilibrium
0.50 M
-x
0.50-0.3 =
0.2M
[B2], M
0.50 M
-x

[AB], M
0.0 M
+ 2x
0.50-0.3 = 2(0.3) =
0.2M
0.6M
Now plug in our “x” value of 0.3
Equilibrium
Sample Exercise 15.11
A 1.000-L flask is filled with 1.000 mol of H2 and
2.000 mol of I2 at 448°C. The value of the
equilibrium constant (Kc) is 50.5.
H2(g) + I2(g)  2HI (g)
What are the concentrations of H2, I2, and HI in
the flask at equilibrium?
Equilibrium
Initially
Change
At
equilibrium
[H2], M
[I2], M
1.000M
2.000 M
-x
-x
1.000-x
2.000-x

[HI], M
0M
+ 2x
2x
Fill in the chart with variables!!
H2(g) + I2(g)  2HI (g)
You need to (for this ICE table) to factor in the
stoich relationship)
Equilibrium
Initially
Change
At
equilibrium
[H2], M
[I2], M
1.000M
2.000 M
-x
-x
1.000-x
2.000-x

[HI], M
0M
+ 2x
2x
Fill in the chart with variables!!
H2(g) + I2(g)  2HI (g)
You need to (for this ICE table) to factor in the
stoich relationship)
Equilibrium
(2x)2
50.5=
(1.000-x)(2.000-x)
50.5 =
4x2
2-3x+x2
50.5 (2-3x+x2)=4x2
Factor the
denominator
Get the
denominator out
of there multiply
both sides
Multiply 51
though
Equilibrium
50.5 (2-3x+x2)=
4x2
Multiply 51
though
101-151.5x+50.5x2= 4x2
Subtract 4x2
from both
sides
101-151.5x+46.5x2= 0
C - Bx
+ Ax2 = 0
Quadratic
equation
You have program your calculator OR do it
the LONG WAY
Equilibrium
101-151.5x+46.5x2= 0
C - Bx
+ Ax2 =
X= -(-151.5)± √151.52 – 4(46.5)(101)
2(46.5)
Equilibrium
101-151.5x+46.5x2= 0
C - Bx
+ Ax2 =
X= -(-151.5)± √22952.25 –18786
93
Equilibrium
101-151.5x+46.5x2= 0
C - Bx
+ Ax2 =
X= -(-151.5)± √4166.25
93
Equilibrium
101-151.5x+46.5x2= 0
C - Bx
+ Ax2 =
X= -(-151.5)+ 64.5
= 2.32
93
X= -(-151.5)- 64.5
93
= 0.935
Equilibrium
Which is the correct
X= -(-151.5)+ 64.5 = 2.32
93
X= -(-151.5)- 64.5
= 0.935
93
Substitute the value for X. If we get a
negative concentration that value would
NOT be the correct on.
Equilibrium
Initially
Change
At
equilibrium
[H2], M
[I2], M
1.000M
2.000 M
-x
-x
1.000-x
2.000-x
[H2] 1.000- 2.32= -1.32

[HI], M
0M
+ 2x
2x
NOT correct value!!
From Quadratic 1
Equilibrium
Initially
Change
At
equilibrium
[H2], M
[I2], M
1.000M
2.000 M
-0.935
0.065M

[HI], M
0M
-0.935
2(0.935)
1.065 M
1.87 M
[H2] 1.000- 0.935= 0.065M
Correct value!!
From Quadratic 2
Equilibrium
Initially
Change
At
equilibrium
[H2], M
[I2], M
1.000M
2.000 M
-0.935
0.065M

[HI], M
0M
-0.935
2(0.935)
1.065 M
1.87 M
Plug the equilibrium concentrations into Kc
expression see if its 50.5
[1.87] 2
3.4969
Kc 
Kc 
Kc=
50.5
[0.065] [1.065]
0.069225
Equilibrium
Equilibrium
Equilibrium
Equilibrium
HW:
PreLab Determining Kc
constant and ice tables
ws
Equilibrium
ICE TABLES
Equilibrium
Sample Exercise 1
• An equilibrium was established after 0.100
mol of hydrogen gas and 0.100 mol of
iodine gas were added to an empty 1.00 L
reaction vessel and heated to 700 K. The
color intensity of the mixture changed from
deep purple to a lighter purple color. At
equilibrium, concentration of iodine was
0.0213 mol/L.
• Calculate the equilibrium concentrations of
hydrogen gas and hydrogen iodide gas.
Equilibrium
Sample Exercise 1-Plan a Strategy
• You will need to write the balanced chemical equation
for the reaction, set up an ICE table, and enter the
given data.
• Then, using the given data and your stoichiometric
skills, you will find the change in the concentration of
each species. This will involve finding change in [I2] by
subtraction and changes in [H2] and [HI] by
stoichiometry using the change in [I2]. Update the ICE
table as values are calculated.
• Calculate the equilibrium concentrations by addition or
subtraction. Since H2 and I2 react to produce HI, their
concentrations should be lower at equilibrium.
• Communicate the answers.
Equilibrium
Sample Exercise 1• Step 1: Write the equation and build the
ICE table.
Equilibrium
Fill in information from question
Initial
0.100 M
0.100 M
0.000 M
Change
Equilibriu
m
0.0213 M
Equilibrium
Sample Exercise 1• Step 2: Calculate the changes in [I2], [H2], and [HI].
• Use the given change with the # moles
Equilibrium
Fill in information from question
Initial
0.100 M
0.100 M
Change
0.0787 M
Equilibriu
m
0.0213 M
0.000 M
Equilibrium
Sample Exercise 1• Step 2: Calculate the changes in [I2], [H2], and [HI].
• Use mole ratios with the CHANGE in I2
Equilibrium
Fill in information from question
Initial
Change
Equilibriu
m
0.100 M
0.100 M
0.0787 M
0.0787 M
0.000 M
0.0213 M
Equilibrium
Sample Exercise 1• Step 2: Calculate the changes in [I2], [H2], and [HI].
• Use mole ratios with the CHANGE in I2
Equilibrium
Fill in information from question
Initial
Change
Equilibriu
m
0.100 M
0.100 M
0.000 M
0.0787 M
0.0787 M
0.157 M
0.0213 M
Equilibrium
Sample Exercise 1• Step 3: Calculate the equilibrium concentrations.
• Subtract the initial from the change (reactants)
• Add the initial from the change (product)
[H2] Equilibrium = 0.100 M - 0.0787 M = 0.0213 M
Equilibrium
Fill in information from question
Initial
Change
Equilibriu
m
0.100 M
0.100 M
0.000 M
0.0787 M
0.0787 M
0.157 M
0.0213 M
0.0213 M
Equilibrium
Sample Exercise 1• Step 3: Calculate the equilibrium concentrations.
• Subtract the initial from the change (reactants)
• Add the initial from the change (product)
[HI] Equilibrium = 0.000 + 0.1574 = 0.1574 M
Equilibrium
Fill in information from question
Initial
Change
Equilibriu
m
0.100 M
0.100 M
0.000 M
0.0787 M
0.0787 M
0.157 M
0.0213 M
0.0213 M
0.157 M
Equilibrium
Step 4: ANSWER
At equilibrium the concentrations of hydrogen gas
and hydrogen iodide gas are 0.0213 M and 0.157
M respectively
Equilibrium
Things to note:
• You should see that the initial concentrations of the species placed in
the vessel are greater than their equilibrium concentrations. That's
because these species have been consumed to make the product.
• The product species have higher concentrations at equilibrium
because initially they were not present in the vessel.
• All aqueous/gas species have non-zero concentrations at equilibrium.
• It is always a good idea to review the final numbers in an ICE table to
ensure that the values make sense. For example, it would not make
sense in the above example for the concentration of hydrogen gas to
be higher at equilibrium than it was initially.
Equilibrium
Sample Exercise 2
• An equilibrium was established in a 1.00
L container when 0.120 mol of
phosphorus pentachloride gas was
decomposed at 500 K. At equilibrium,
the concentration of chlorine gas was
0.0540 mol/L.
• Calculate the equilibrium constant for
this system and state a conclusion
Equilibrium
about the position of the equilibrium
Sample Exercise 2• Step 1: Write the equation and build the
ICE table.
Equilibrium
Fill in information from question
Initial
0.120 M
0.000 M
0.000 M
Change
Equilibriu
m
0.0540 M
Equilibrium
Sample Exercise 2• Step 2: Calculate the changes in given species.
• Use the given change with the # moles
Equilibrium
Fill in information from question
Initial
Change
Equilibriu
m
0.120 M
0.000 M
0.000 M
0.0540 M
0.0540 M
Equilibrium
Sample Exercise 2• Step 2: Calculate the changes in given species.
• Now we have 1 for the change can use stoich to figure
out the other changes
Equilibrium
Fill in information from question
Initial
Change
Equilibriu
m
0.120 M
0.000 M
0.0540 M
0.000 M
0.0540 M
0.0540 M
Equilibrium
Sample Exercise 2• Step 2: Calculate the changes in given species.
• Now we have 1 for the change can use stoich to figure
out the other changes
Equilibrium
Fill in information from question
Initial
Change
Equilibriu
m
0.120 M
0.000 M
0.0540 M
0.0540 M
0.000 M
0.0540 M
0.0540 M
Equilibrium
Sample Exercise 2• Step 3: Calculate the equilibrium concentrations.
• Subtract the initial from the change (reactants)
• Add the initial from the change (product)
[PCl5] Equilibrium = 0.120 M - 0.0540 M = 0.066
M
Equilibrium
Fill in information from question
0.120 M
0.000 M
Change
0.0540 M
0.0540 M
Equilibriu
m
0.0066 M
Initial
0.000 M
0.0540 M
0.0540 M
Equilibrium
Sample Exercise 2• Step 3: Calculate the equilibrium concentrations.
• Subtract the initial from the change (reactants)
• Add the initial from the change (product)
[PCl3] Equilibrium = 0.000 + 0.0540 = 0.0540 M
Equilibrium
Fill in information from question
0.120 M
0.000 M
Change
0.0540 M
0.0540 M
0.0540 M
Equilibriu
m
0.0066 M
0.0540 M
0.0540 M
Initial
0.000 M
Equilibrium
Sample Exercise 2• Step 4: Write the K expression and solve.
Equilibrium
Sample Exercise 2• Calculate the equilibrium constant for this
system and state a conclusion about the
position of the equilibrium
The equilibrium constant for the decomposition of
phosphorus pentachloride at 500 K is 4.4 x 10-2.
The equilibrium position lies to the left, or in other
words, the reactant is favored.
Equilibrium