Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6...

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Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682-694); 14.6 Changing Concentrations of Reactants or Product (p694-696); Changing Temperature (p698-701.)

Transcript of Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6...

Page 1: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Intersection 9: Equilibrium

10/31/06Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682-694); 14.6 Changing Concentrations of Reactants

or Product (p694-696); Changing Temperature

(p698-701.)

Page 2: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

A Gateway to Scientific EthicsScientists Employ the Scientific Method to Explore the Physical World.

• What are the ethical considerations associated with being a scientist?• How do they relate to the use of the Scientific Method?• Are there ethical considerations beyond the Scientific Method?

We will employ a case-study approach to examining these questions including a prominent case of fraud from Lucent Technologies as wellas the classic Millikan-Ehrenhaft debate.

6 – 8 PM Thursday Nov. 30th

Please RSVP by email to Prof Banaszak Holl([email protected])

Gateway evenings are optionaland will not affect your course grade

Gateway Chemistry 130/125/126Section 600

Dinner Provided

Page 3: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

MSF • Thank you!• What we can’t change:

– Exam time– Class lecture time

• Will be you prepared for your next chemistry class:– Chem 130 concepts– No answer keys

• How we hope to improve the course– Grading rubrics for remaining projects (ex. Proposal)– Try to preview topics and do more practice/homework

problems in lecture– Give GSIs some more time for problem solving in

studio

Page 4: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Outline

• Equilibrium defined• Equilibrium constant• ICE• Shifting equilibrium

– Equilibrium Law– LeChâtelier’s principle

• Practice problems

Page 5: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Reactions that Don’t Go to Completion

• Generally, we assume that reactions “go to completion”…as much of the reactants are used up as possible; there may be a limiting reagent and thus reactants may be left over

• We assume that the reaction can only go in the forward direction.

• A rate of (forward) reaction can be measured.

A

Page 6: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

What's a rate of reaction?

For a simple reactions A B, rate = k[A].

Most reactions slow down as they proceed and as the concentration/s of the reactant/s decreases (the rate approaches zero.)

A

Page 7: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Problem 1

2L of a 0.1M solution of magnesium chloride and 1 L of a 0.5 M solution of silver nitrate are mixed together.

– Write out the balanced net ionic equation– What is the limiting reagent?– How many grams of silver chloride will you

make?

A

Page 8: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

A New Scenario: Equilibrium

A reaction in equilibrium is going forward and backward at the same rate.

All reactants and products are present and actively interchanging

A

Page 9: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Forward and Reverse RatesReactions in equilibrium have both a

significant forward and reverse rate of reaction. A B

When equilibrium is reached, the rates of the forward and reverse reaction are equal AND are NOT equal to 0.

Forward reaction:  A → B

Reverse reaction:  B → A

A

Page 10: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

An Equilibrium Model

http://www.chm.davidson.edu/ronutt/che115/EquKin/EquKin.htm

2A B B 2A

When has the reaction reached equilibrium? (How can you tell?)

Is the forward reaction (2A B) still taking place?Is the reverse reaction (B 2A) still taking place?

A

Page 11: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);
Page 12: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Equilibrium Constant

• A B Forward Rate = kfwd [A]

• B A Reverse Rate = krev[B]

At equilibrium we have the following equality: kfwd[A] = krev[B]

forward rate = reverse rate

Rearranging this equation yields:

Keq = kfwd/krev = [B]/[A]

Keq is the equilibrium constant

M

Page 13: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

What does Keq tell you?

• A ↔ B Keq =

• For low values of Keq, do you expect there to be a higher concentration of products or reactants at equilibrium?

Concentration vs time

0

0.5

1

1.5

2

0 10 20 30

Time (seconds)

Co

nce

ntr

atio

n [

M]

A

B

Rate vs. Time

0

0.05

0.1

0.15

0.2

0.25

0 10 20 30

Time (seconds)

Rat

e [M

]/s

rate forw ard

rate backw ard

M

Page 14: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Keq for more complex reactions

M

aA + xX ↔ bB + yY

[B]b[Y]y

[A]a[X]xKeq =

Page 15: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Writing Equilibrium Constants

1) NO(g) + Cl2(g) ↔ NOCl(g)

# In any equilibrium expression, the concentration of a pure liquid (e.g water) or pure solid is considered a constant (1).

2) H2(g) + I2(g) ↔ HI(g)

3)# CaCO3(s) ↔ CaO(s) + CO2(g)

M

*First, balance the equation.

Page 16: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Equilibrium Constant Family

Keq -a generic equilibrium constant

Kc -an equilibrium constant calculated using equilibrium concentrations in M (mol/L)

Kp - associated with gaseous equilibria; found using equilibrium pressures (atm)Pressure is directly proportional to concentration (PV = nRT or P = (n/V)RT).

M

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Page 18: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Determining Keq

 2NO2 (g)  ↔   N2O4 (g)    ΔH = -24.02 KJ/mol

(red-brown) (colorless) 

Suppose that 0.55 moles of NO2 are placed in an empty 5.00 L flask

which is subsequently heated to 407 K.  By measuring the intensity of the color of red-brown NO2, it can be determined that

its concentration at equilibrium is 0.10 mol/L.   What is the expression for Keq? What is the value of Keq at 407 K?

How would you determine the equilibrium constant?

A

Keq = [N2O4]/[NO2]2

Page 19: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Put the Reaction on ICE2NO2 (g) ↔ N2O4 (g)

Initial Concentration (M)

Change in Concentration (M)

Equilibrium Concentration (M)

Keq = [N2O4]/[NO2]2 = 0.5

0.55 moles/5 L

0.10 mol/L

- 2x + x

0 moles/L

0.11 mol/L – 2x = 0.10 mol/L

x = 0.005 mol/L

0.005 mol/L

A

Page 20: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Temperature

Does temperature matter?

2NO2 (g)  ↔   N2O4 (g)    ΔH = -24.02 KJ/mol

(red-brown) (colorless) 

Keq(407K) = 0.5

A

Page 21: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

ICE applied

What if you want to know equilibrium concentrations?

H2 (g)  +  I2 (g) ↔ 2 HI(g)   Keq = 2.5 x 101

Two moles of hydrogen and 2 moles of iodine are added to a 4 L container; what are the concentrations of all reactants and products at equilibrium?

In the future, you can look up the equilibrium constants in Table 14.1 p685 as well as the Appendices) *

M

Page 22: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

H2(g) I2(g) ↔ 2 HI(g)

Initial 2mol / 4L 2 mol / 4L 0

Change

Equilibrium

Keq = 25 = [HI]2 = [2x]2

[H2][I2] [0.5-x][0.5-x]

M

-x

+2x

-x

0.5 - x0.5 - x

+2x

x = 0.8, 0.4

Keq = 2.5 x 101

Page 23: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Problem 2: Making Ammonia

Desired for fertilizing (belief that without ammonia, wouldn’t be able to feed world.)N2(g) + 3 H2(g) ↔ 2 NH3(g) Kc = 3.5 x108 (25oC)

– What is the Kc if the reaction were written for the production of 1 mole of ammonia?

1/2 N2(g) + 3/2 H2(g) ↔ NH3(g)

– If 10 moles of nitrogen and 10 moles of hydrogen are placed in a 1 L flask, how many moles of ammonia can you make? How many moles of starting material would be left over?

M

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Page 25: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Disturbing Equilibrium

• Sir Isaac Newton claimed that a ball at rest would remain at rest unless disturbed.  You might be tempted to apply this logic to equilibrium and get:   A reaction at equilibrium will remain at equilibrium unless disturbed; consequently, the reaction will shift so as to come back to equilibrium.

A

Page 26: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Can the Equilibrium Constant be Changed?

 2NO2 (g)  ↔   N2O4 (g)    ΔH = -24.02 KJ/mol

(red-brown) (colorless) 

A

Page 27: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Evaluating Changes in Equilibrium

Method 1: LeChâtelier's Principle

 if a system at equilibrium is disturbed or stressed by a change in temperature, pressure or concentration of one of the components, it will shift its equilibrium so as to oppose the stress.

How does LeChâtelier's Principle explain the demonstration that you just saw?

A

Page 28: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Can Equilibrium be Changed?

Fe(NO3)3 + KSCN ↔ Fe(SCN)+2 + KNO3 H < 0 red

Use LeChâtelier's Principle to predict what you will see:

A

Page 29: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Evaluating Changes in Equilibrium Method 2: Equilibrium Law (Q)

• Keq is used at equilibrium to represent the ratio of reactants to products for a give reaction.

Keq = [B]b[Y]y

[A]a[X]x

• Q, the reaction quotient, is used for this ratio under any conditions at any point in time, not just equilibrium.

Q = [B]b[Y]y

[A]a[X]x

• At equilibrium, Keq and Q are EQUAL. • According to the equilibrium law, the system will proceed

to bring Q and Keq equal to each other.

aA + xX ↔ bB + yY

M

Page 30: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Q vs. Keq • In general, how would the reaction proceed

to result in a decreased Q?

• What if an increased Q were the desired result?

M

Page 31: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Applying the Equilibrium Law

• What is the equilibrium expression for this reaction?

• Keq was determined to be 6.42x10-5 at 25oC.

• At equilibrium is this reaction product favored or reactant favored?

H2O(l) + C6H5CO2H (aq) ↔ C6H5CO2-(aq) + H3O+

(aq)

M

Page 32: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

2.00 moles of C6H5CO2-, 1.00 mole of H3O

+ and 3.00

moles of C6H5CO2H are placed in 1 liter of water.

What is the value of Q under these conditions?

Compare Q with Keq (6.42 x10-5). Will the reaction proceed to form more C6H5CO2

-(aq) and H3O+

(aq) or more C6H5CO2H(aq) or not change at all?

H2O(l) + C6H5CO2H ↔ C6H5CO2- + H3O+

(aq)

M

Page 33: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

LeChâtelier trumps Q

One instance where Le Châtelier's principle provides us with information that the equilibrium law cannot is in the case of changing temperature.  

Suppose we have the following reaction,

CaCO3(s) ↔ CaO(s) + CO2(g)     ΔH > 0

What happens if you increase the temperature?

M

Page 34: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

CaCO3(s) ↔ CaO(s) + CO2(g)     ΔH > 0

Using each method, explain what will happen to the concentration of CO2 if solid lime (CaCO3) is added to the system?

Q trumps LeChâtelier

M

Page 35: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);
Page 36: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Problem 3a

S2(g) and C(graphite) when placed together in a closed system form an equilibrium with CS2(g).  

C(graphite) + S2(g)   ↔ CS2(g)

Suppose that the equilibrium constant for this reaction is 4.0.  

Draw a qualitative graph that shows how the concentration of each gas changes with time if the system initially consists of pure S2(g) and graphite.

A

Page 37: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Problem 3b C(graphite) + S2(g)   ↔ CS2(g)

Draw a picture representing the molecules present under initial conditions and when the reaction reaches equilibrium.

Will the amount of graphite in the system be the same, more, or less at equilibrium than it was initially?  Why?  

A

Page 38: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Draw a second graph showing what happens if the system initially contains pure CS2(g) and graphite.  

Draw a picture representing the molecules present under initial conditions and when the reaction reaches equilibrium.

Will the amount of graphite have changed in this scenario? If so, how?

Problem 3cC(graphite) + S2(g)   ↔ CS2(g)

A

Page 39: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Problem 4

At room temperature, the equilibrium constant for the reaction:

2NO(g) ↔ N2(g) + O2(g) is 1.4 x1030

Is this reaction product-favored or reactant-favored?

In the atmosphere at room temperature, the concentration of N2 is 0.33 mol/L, and the concentration of O2 is about 25% of that value. Calculate the equilibrium concentration of NO in the atmosphere.

M

Page 40: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Problem 5CO(g) + H2O(g) ↔ CO2(g) + H2(g)

Kc = 4.00 at 500 K.

A mixture of 1.00 mol CO and 1.00 mol H2O is allowed to come to equilibrium in a flask of volume 0.5 L at 500K,

Calculate the final concentrations of all four species: CO, H2O, CO2 and H2

What would be the equilibrium concentrations if an additional 1.00 mol of each CO and H2O were added?

M

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Page 42: Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696);

Equilibrium Representation (Friday 11/10)

Your group will create a visual representation of a dynamic equilibrium. The medium is completely up to you (animation, skit, artwork, song, etc.), and creativity is encouraged. – The representative system that is in a stable dynamic

equilibrium. – A stress to the system and how it would respond

according to Le Châtelier's principle.

You will tell the class the system what species (chemical or otherwise) that are present etc., but the class will have to infer the stress placed on the system by its response to that stress.

A