Post on 19-Jan-2016
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Equilibrium
Agenda:
• Sign off/Discuss Equilibrium constant ws• Collect Beanium Equilibrium Activity• In-Class ICE table practice
• HW: Ice Tables ws
Equilibrium
I can…
• Determine the direction of the reaction by the magnitude of the Kc.
• Use the ICE tables to predict changes concentration and its effect on the Kc value.
Equilibrium
Flip NotesEquilibrium concentrations of each species in 0.2M iodic acid. HIO3, has a Ka = 0.17
HIO3 (aq) H+ (aq) + IO3- (aq)
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 tablesType 1A. 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 tablesType 2A. 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 448C 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 448C 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
Equilibrium
Equilibrium
③Place the known concentrations provided in the question into table.
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change
Equilibrium 1.87 x 10-3
Equilibrium
④Put in the CHANGE for HI (subtract equilibrium from initial)
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 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.
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change +1.87 x 10-3
At equilibrium
1.87 x 10-3
1.87 x 10-3mol L
1 mol H2
2 mol of HI
= 0.935 x 10-3
Same goes for iodine
H2 (g) + I2 (g) 2 HI (g)
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 they are reactants!!
Equilibrium
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 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At equilibrium
1.87 x 10-3
1.000 x10-3 – 9.35 x10-4 =6.5 x 10-5
6.5 x 10-5 1.065 x 10-3
2.000 x10-3 – 9.35 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.
=(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
Kc =[HI]2
[H2] [I2]
Kc = 51
Equilibrium
Let’s Try anotherSulfur 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.
INITIAL CHANGE EQUILIBRIUM
2SO3 2SO2 O2
6.09 x 10-3 M 0 M 0 M
2.44 x10-3M
Equilibrium
Let’s Try another
What information can we fill in with what we are given?
INITIAL CHANGE EQUILIBRIUM
2SO3 2SO2 O2
6.09 x 10-3 M 0 M 0 M
2.44 x10-3M
We can fill in the change of SO3.
6.09 x 10-3 M - 2.44 x10-3M = -3.65 x 10-3 M
-3.65 x 10-3 M
Equilibrium
Let’s Try another
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.
INITIAL CHANGE EQUILIBRIUM
2SO3 2SO2 O2
6.09 x 10-3 M 0 M 0 M
2.44 x10-3M
3.65 x 10-3 M of SO3= +3.65 x 10-3 M
of SO2
-3.65 x 10-3 M
2 moles SO2
2 mole of SO3
+3.65 x 10-3 M
3.65 x 10-3 M of SO3= +1.83 x 10-3 M
of O2
1 moles O2
2 mole of SO3
+1.83 x 10-3 M
Equilibrium
Let’s Try another
Now subtract the initial from the change to get the equilibrium.
INITIAL CHANGE EQUILIBRIUM
2SO3 2SO2 O2
6.09 x 10-3 M 0 M 0 M
2.44 x10-3M
0 - 3.65 x 10-3 M = 3.65 x 10-3 M
-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
1.83 x 10-3 M+1.83 x 10-3 M
3.65 x 10-3 M
Equilibrium
Let’s Try another
Use the equilibrium concentrations to find Kc- plug n chug
INITIAL CHANGE EQUILIBRIUM
2SO3 2SO2 O2
6.09 x 10-3 M 0 M 0 M
2.44 x10-3M-3.65 x 10-3 M +3.65 x 10-3 M
1.83 x 10-3 M+1.83 x 10-3 M
3.65 x 10-3 M
23
22
2
][
][][
SO
OSOKc
23-
-32-3
] x102.44[
]10 x 1.83[]10 x 3.65[cK
(1.33225 x10-5)(1.83 x10-3) = 2.4380175 x10-8
5.9536 x10-6
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°CIf 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 mol2.0 L
= 0.50 M
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°CIf 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 [B2], M [AB], M
Initially
Change
Equilibrium
0.50 M 0.50 M 0.0 M
Equilibrium
[A2], M [B2], M [AB], M
Initially
Change
Equilibrium
0.50 M 0.50 M 0.0 M
Fill in the chart with variables!!
You need to (for this ICE table) to factor in the stoich relationship)
A2(g) + B2(g) 2AB (g)
-x -x + 2x
0.50-x 0.50-x 2x
Equilibrium
We have to work backwards!!
Provide the Kc expression
9.0=(2x)2
(0.50-x)(0.50-x)
Kc =[AB]2
[A2] [B2]
[A2], M [B2], M [AB], M
Initially 0.50M 0.50 M 0 M
Change
At equilibrium
-x
-x + 2x
0.50-x 0.50-x 2x
Equilibrium
Get 0.50-x out by multiply on both sides
Root both sides9.0=
(2x)2
(0.50-x)(0.50-x)Or
(0.50-x)2
3.0=(2x)
(0.50-x)
Equilibrium
Get 0.50-x out by multiply on both sides
3.0=(2x)
(0.50-x)
3.0(0.50-x)=2x Multiply 3 through
1.5-3x= 2x 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 [B2], M [AB], M
Initially
Change
At equilibrium
0.50 M 0.50 M 0.0 M
Now plug in our “x” value of 0.3
-x -x + 2x 2(0.3) =0.6M
0.50-0.3 = 0.2M
0.50-0.3 = 0.2M
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
[H2], M [I2], M [HI], M
Initially 1.000M 2.000 M 0 M
Change
At equilibrium
Fill in the chart with variables!!
-x
-x + 2x
1.000-x 2.000-x 2x
You need to (for this ICE table) to factor in the stoich relationship)
H2(g) + I2(g) 2HI (g)
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000M 2.000 M 0 M
Change
At equilibrium
Fill in the chart with variables!!
-x
-x + 2x
1.000-x 2.000-x 2x
You need to (for this ICE table) to factor in the stoich relationship)
H2(g) + I2(g) 2HI (g)
Equilibrium
50.5=(2x)2
(1.000-x)(2.000-x)
50.5 =4x2
2-3x+x2
Factor the denominator
Get the denominator out of there multiply both sides
50.5 (2-3x+x2)=4x2 Multiply 51 though
Equilibrium
50.5 (2-3x+x2)= 4x2 Multiply 51 though
101-151.5x+50.5x2= 4x2Subtract 4x2
from both sides
101-151.5x+46.5x2= 0 Quadratic equation
C - Bx + Ax2 = 0
You have program your calculator OR do it the LONG WAY
Equilibrium
101-151.5x+46.5x2= 0C - Bx + Ax2 =
X= -(-151.5)± √151.52 – 4(46.5)(101)
2(46.5)
Equilibrium
101-151.5x+46.5x2= 0C - Bx + Ax2 =
X= -(-151.5)± √22952.25 –18786
93
Equilibrium
101-151.5x+46.5x2= 0C - Bx + Ax2 =
X= -(-151.5)± √4166.25
93
Equilibrium
101-151.5x+46.5x2= 0C - Bx + Ax2 =
X= -(-151.5)+ 64.5
93= 2.32
X= -(-151.5)- 64.5
93= 0.935
Equilibrium
Which is the correct
X= -(-151.5)+ 64.5
93= 2.32
X= -(-151.5)- 64.5
93= 0.935
Substitute the value for X. If we get a negative concentration that value would
NOT be the correct on.
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000M 2.000 M 0 M
Change
At equilibrium
-x
-x + 2x
1.000-x 2.000-x 2x
[H2] 1.000- 2.32= -1.32
From Quadratic 1
NOT correct value!!
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000M 2.000 M 0 M
Change
At equilibrium
-0.935
-0.935 2(0.935)
0.065M 1.065 M 1.87 M
[H2] 1.000- 0.935= 0.065M
From Quadratic 2
Correct value!!
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000M 2.000 M 0 M
Change
At equilibrium
-0.935
-0.935 2(0.935)
0.065M 1.065 M 1.87 M
Plug the equilibrium concentrations into Kc expression see if its 50.5
]065.1[]065.0[
]87.1[ 2
cK069225.0
4969.3cK Kc= 50.5
Equilibrium
Equilibrium
Equilibrium