Lect w6 152_abbrev_ le chatelier and calculations_1_alg

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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


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


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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


[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


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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


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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]?


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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


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[I2] 2[I]

initial 0.500 0

change -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!!


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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


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

Lect w6 152_abbrev_ le chatelier and calculations_1_alg

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