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UA GenChem
General Chemistry IIGeneral Chemistry IICHEM 152 Unit 2CHEM 152 Unit 2
Week 6
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Week 6 Reading Assignment
Chapter 14 – Sections 14.6 through 14.9 (calculations - Le Châtelier)
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What to Expect
• Introduce Q
• Solve Equilibrium Problems– ICE Tables
• Discuss Le Châtelier’s Principle
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Q...the Reaction Quotient
• Quotient...what?quotient (n.) number resulting from the
division of one number by another
• At EQUILIBRIUM, Q = K
• For A + B 2 C
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Reaction Quotient Example
Write the reaction quotients for the following reactions:
2 N2O5(g) 4 NO2(g) + O2(g)
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
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Reaction Quotient Example
• Write the reaction quotient for the following reactions:
2 N2O5(g) 4 NO2(g) + O2(g)
Q = [NO2]4[O2]
[N2O5]2
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
Q = [CO2]3[H2O]4
[C3H8][O2]5
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Q vs. K
• At equilibrium, Q = K
So what happens when Q K?
Q < K or Q > K
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Problem
For the reaction N2O4(g) 2 NO2(g), the equilibrium constant is 0.21 at 100C.
At the point in the reaction where [N2O4] = 0.12 M and [NO2] = 0.55, is the reaction at equilibrium?
Q = [NO2]2 = [0.55]2 = 2.5
[N2O4] [0.12]
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ProblemFor the reaction N2O4(g) 2 NO2(g), the
equilibrium constant is 0.21 at 100C. At the point in the reaction where [N2O4]
= 0.12 M and [NO2] = 0.55, is the reaction at equilibrium?
Q = 2.5 0.21
In which direction will it proceed?
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ProblemFor the reaction N2O4(g) 2 NO2(g),
the equilibrium constant is 0.21 at 100C.
At the point in the reaction where [N2O4] = 0.12 M and [NO2] = 0.55, is the reaction at equilibrium?
In which direction will it proceed?
0 ∞1K 0.21
Q 2.5
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Shifting EquilibriumThe equilibrium state for a chemical process can be affected by changes in the concentration of reactants or
products, or by varying the temperature and pressure of the
system.The direction in which the
equilibrium shifts (increasing the
concentration of reactants or products) can be
predicted.
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Butane-Butane-Isobutane Isobutane
EquilibriumEquilibrium
K = [isobutane]
[butane] 2.5K =
[isobutane][butane]
2.5
butanebutane
isobutaneisobutane
Let us consider the following equilibrium
Shifting Equilibrium
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ButaneButane IsobutaneIsobutane
Assume you are at equilibrium with Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M.[iso] = 1.25 M and [butane] = 0.50 M.
Now add 1.50 M butane.Now add 1.50 M butane.
What are the concentrations of [iso] and What are the concentrations of [iso] and [butane] when the equilibrium is [butane] when the equilibrium is
re-established?re-established? K = 2.5 K = 2.5
Shifting Equilibrium
Q = [isobutane]
[butane]
1.250.50 + 1.50
= 0.63Q = [isobutane]
[butane]
1.250.50 + 1.50
= 0.63
To answer this question we can To answer this question we can calculate the “reaction quotient” calculate the “reaction quotient” QQ
and compare it to K:and compare it to K:
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Butane Butane IsobutaneIsobutane
Given that Given that Q= 0.63Q= 0.63 and and K =2.5K =2.5. . How can the equilibrium be re-established?How can the equilibrium be re-established?
The system must shift to increase Q until Q = K
This is done by increasing the concentration of ISOBUTANE and DECREASING that of BUTANE.
Adding reactants to an equilibrium shifts the equilibrium toward products
Shifting Equilibrium
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Le Châtelier’s Le Châtelier’s PrinciplePrinciple
The outcome is governed by the:The outcome is governed by the:
Le CHÂTELIER’S PRINCIPLELe CHÂTELIER’S PRINCIPLE
““...if a system at equilibrium is ...if a system at equilibrium is disturbed, the system tends to shift disturbed, the system tends to shift
its equilibrium position to counter the its equilibrium position to counter the effect of the disturbance.”effect of the disturbance.”
Changes in concentration, pressure, and temperature affect the position of
chemical equilibrium.
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Let’s Analyze ItN2O4 2NO2
What would you expect to happen if we double
the concentratio
n of NO2?
How does
K=[NO2]2
eq / [N2O4]eq
change?
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Consider the reaction
CaCO3(s) + CO2(aq) + H2O(l) Ca2+ (aq) + 2 HCO3
-
(aq)
Predict the effect on the equilibrium of:
-Removing CO2 -Adding more CaCO3
to the system -Adding CaCl2 to the
solution-Adding more water
-HCO3- is added
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Changing Volume or PressureN2O4 2NO2
Predict the effect of:• Reducing the
volume of the container to one half its initial value;• Increasing the volume of the container to two times its initial value.
K does not change
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Another ReactionH2 + I2 2HI
What would
happen in this case if
similar “stresses”
are applied to
this equilibriu
m?
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V (P) – Reaction shifts to the side with fewer moles of gas
V (P) – Reaction shifts to the side with greater moles of gas
The value of K DOES NOT CHANGE.
Summary
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Temperature Effects2NO2 N2O4
Let’s now analyze
the effect of
changing temperatu
re.
Evaluate the value
of K before and after
the change.
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Based on the shifts in the equilibrium
with changing
temperature, decide
whether this is an
endothermic or an
exothermic reaction.
PCl5(g) PCl3(g) + Cl2(g)
Temperature Effects
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Increase T More reactants (K decreases)
H = - (exothermic)
Temperature Effects
Decrease T More products (K increases)
2NO2 N2O4 + Heat
Treat the Heat as if it were a product
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Increase T More products (K increases)
H = + (endothermic)
Temperature Effects
Decrease T More reactants (K decreases)
Treat the Heat as if it were a reactant
Heat + PCl5(g) PCl3(g) + Cl2(g)
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Le Châtelier’s Le Châtelier’s PrinciplePrinciple
• Add or take away reactant or product; change volume or pressure: – K does not change– Reaction adjusts to new equilibrium
“position”• Change T:
– Change in K (effect depends on the sign of H0
rxn)– Therefore change in concentrations at
equilibrium• Use a catalyst:
- Reaction comes more quickly to equilibrium. K not changed; final concentrations not changed.
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Equilibrium Calculations
Let’s now learn how to calculate
equilibrium constants or the concentrations of reactant and
products at equilibrium.
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All [Products]eq known:H2(g) + I2(g) 2HI(g) @ 445°C
Initial EquilibriumEquilibrium
Constant
[H2] [I2] [HI] [H2] [I2] [HI]
0.50 0.50 0.0 0.11 0.11 0.78
0.0 0.0 0.50 0.055 0.055 0.39
]][I[H
[HI]
22
2
eq K
501][0.11][0.1
[0.78]2
50055][0.055][0.
[0.39]2
0.50 0.50 0.50 0.165 0.165 1.17 50165][0.165][0.
[1.17]2
1.0 0.5 0.0 0.53 0.033 0.934 5033][0.53][0.0
[0.934]2
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One [Product]eq KnownYou have the reaction 2 A(aq) + B(aq) 4 C(aq)
with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the
equilibrium concentration of C as [C]eq =0.50 M. What are the equilibrium concentrations? K?
2A B 4Cinitial molarity 1.00 1.00 0change in concentration
equilibrium molarity 0.50
I
C
E
+
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Find the value of K for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4]
= 0.115 M and the equilibrium [C2H2]eq = 0.035
MConstruct an ICE table for the reaction
For the substance
whose equilibrium
concentration is known,
calculate the change in
concentration
2CH4 C2H2 3H2
initial 0.115
0.000
0.000
change
equilibrium
0.035
Your Turn
+
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Find the value of K for the reaction
2 CH4(g) C2H2(g) + 3 H2(g) at 1700°C if the initial [CH4]
= 0.115 M and the equilibrium [C2H2]eq = 0.035
M
2CH4 C2H2 3H2
initial 0.115
0.000
0.000
change
equilibrium
0.035
Answer
+3x
+x-2x
0.1050.045
K = (0.035*0.1053)/(0.0452) = 0.020
K = [C2H2][H2]3
[CH4]2 +
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None [Product]None [Product]eqeq Known KnownImagine that we place 1.00 mol each of H2
and I2 in a 1.00 L flask, and they react according with the equation:
Kc = [HI]2
[H2 ][I2 ] = 55.3Kc =
[HI]2
[H2 ][I2 ] = 55.3
HH22(g) + I(g) + I22(g) (g) 2 HI(g) 2 HI(g)
If:
What are the values for the equilibrium concentrations of all the species in the system?
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Step 1.Step 1. Set up Set up a a table to define equilibrium table to define equilibrium concentrations. concentrations.
[H[H22]] [I[I22]] 2[HI]2[HI]
Initial Initial
ChangeChange
EquilibEquilib
HH22(g) + I(g) + I22(g) (g) 2 HI(g) 2 HI(g)K = 55.3K = 55.3
1.001.00 1.001.00 00
-x-x -x-x +2x+2x
1.00-x1.00-x 1.00-x1.00-x 2x2x
where where xx is defined as the amount of H is defined as the amount of H22 and and
II22 consumed on approaching equilibrium. consumed on approaching equilibrium.
Equilibrium Concentrations
+
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Step 2.Step 2. Put equilibrium concentrations into Put equilibrium concentrations into KKcc expression (expression (KK =[HI]=[HI]22/([H/([H22][I][I22]]).).
Kc = [2x]2
[1.00 - x][1.00 - x] = 55.3Kc =
[2x]2
[1.00 - x][1.00 - x] = 55.3
HH22(g) + I(g) + I22(g) (g) 2 HI(g) 2 HI(g)K = 55.3K = 55.3
Equilibrium Concentrations
[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M[HI] = 2x = 1.58 M
[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M[HI] = 2x = 1.58 M
Step 3.Step 3. Solve K expression – In this case, Solve K expression – In this case, we canwe can take square root of both sides. take square root of both sides.
x = 0.788x = 0.7887.44 = 2x
1.00 - x7.44 =
2x1.00 - x
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Nitrogen Dioxide Nitrogen Dioxide EquilibriumEquilibrium
NN22OO44(g) (g) 2 NO 2 NO22(g)(g)
Your Turn
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 KKc =
[NO2 ]2
[N2O4 ] = 0.0059 at 298 K
Find the equilibrium concentrations for
NO2 and N2O4 at 298 K if [N2O4]o=0.50 M.
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Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) 2 NO 2 NO22(g)(g)
If initial concentration of NIf initial concentration of N22OO44 is 0.50 M, what is 0.50 M, what are the equilibrium concentrations?are the equilibrium concentrations?
Step 1.Step 1. Set up a concentration table Set up a concentration table
[N[N22OO44]] 2[NO 2[NO22]]
InitialInitial 0.500.50 00
ChangeChange -x-x +2x+2x
EquilibEquilib 0.50 - x0.50 - x 2x2x
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 KKc =
[NO2 ]2
[N2O4 ] = 0.0059 at 298 K
Equilibrium Concentrations
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Step 2.Step 2. Substitute into K Substitute into Kcc expression and solve. expression and solve.
Kc = 0.0059 = [NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x) Kc = 0.0059 =
[NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x)
Rearrange: Rearrange: 0.0059 (0.50 - x) = 4x0.0059 (0.50 - x) = 4x22
0.0029 - 0.0059x = 4x0.0029 - 0.0059x = 4x22
4x4x22 + 0.0059x -0.0029= 0 + 0.0059x -0.0029= 0
This is a This is a QUADRATIC EQUATIONQUADRATIC EQUATION axax22 + bx + c = 0 + bx + c = 0
a = 4a = 4 b = 0.0059b = 0.0059 c = -0.0029c = -0.0029
Equilibrium Concentrations
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Equilibrium Concentrations
Solve the quadratic equation for x.Solve the quadratic equation for x.
axax22 + bx + c = 0 + bx + c = 0
a = 4a = 4 b = 0.0059b = 0.0059 c = -0.0029c = -0.0029
x = -b b2 - 4ac
2ax =
-b b2 - 4ac2a
x = -0.0059 (0.0059)2 - 4(4)(-0.0029)
2(4)x =
-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)
x = x = 0.0260.026 or or -0.028-0.028But a negative value is not reasonable.But a negative value is not reasonable.
Conclusion: Conclusion: x = 0.026 Mx = 0.026 M
[N[N22OO44] = 0.50 - x = 0.47 M] = 0.50 - x = 0.47 M
[NO[NO22] = 2x = 0.052 M] = 2x = 0.052 M
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Approximations to Simplify the Math
– If there is a large enough difference between the concentration and the K value – SIMPLIFY!
– Use approximation ONLY when the concentration and the K value are more than a factor of 100 or more apart
K = ax2
Conc - bx
K = ax2
Conc
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For the reaction
I2(g) 2 I(g)
the value of K = 3.76 x 10-5 at 1000 K.
If 1.00 moles of I2 is placed into a 2.00 L
flask and heated, what will be the
equilibrium concentrations of [I2] and
[I]?
Equilibrium Concentrations
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For the reaction I2(g) 2 I(g) the value of K = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] 2[I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
25
225
2
2
4500.01076.3
500.0
2
500.0
21076.3
I
I
x
x
x
x
K
Equilibrium Concentrations
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[I2] 2[I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
35
25
25
1017.24
1088.1
41088.1
4500.01076.3
x
x
x%1%434.0%100
500.0
1017.2 3
The approximation is valid!!
Equilibrium Concentrations
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For the reaction I2(g) 2 I(g) the value of K = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is
placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
[I2] 2[I]
initial 0.500 0
change -x +2x
equilibrium 0.500- x 2x
x = 0.00217
0.500 0.00217 = 0.498[I2] = 0.498 M
2(0.00217) = 0.00434[I] = 0.00434 M
52
2
2
c 1078.3498.0
00434.0
I
I K
.
Approximation is valid
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Summary Activity
1. Consider the reduction of carbon dioxide by hydrogen to give water vapor and carbon monoxide:
H2(g) + CO2(g) H2O(g) + CO(g) Kc=0.10 (at 420 oC)
Suppose the initial concentrations of CO2 and H2 are the same: 0.050 M. What are the
equilibrium concentrations of all the species at 420 oC?
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Summary Activity
2. If the same reaction as we looked at in Summary Activity 1 was endothermic, what would be the effect on the equilibrium by each of the following:
Decrease the Volume
Decrease the Temperature
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Summary Activity
3. Suppose that a mixture of 1.00 mol of HI(g) and 1.00 mol of H2(g) is sealed into a 10.0 L flask at 745 K. What will be the concentration of all species at equilibrium? 2HI(g) H2(g) + I2(g) Kc=0.0200