Slide 2: corrected spelling of LeChatelier

Slide 32: reformatted

Slide 33: reformatted

Slide 40: what is this??

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.)

Outline

• Equilibrium defined• Equilibrium constant• ICE• Shifting equilibrium

– Equilibrium Law– LeChâtelier’s principle

• Practice problems

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.

Question 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 your limiting reagent?– How many grams of silver chloride will you

make?

Equilibrium

What does it mean?

That a reaction is going forward and backward at the same rate. (All reactants and products are present and actively interchanging)

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.)

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

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?

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

What does Keq tell you?

• A B Keq =

• For high 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

Keq for more complex Reactions

Writing Equilibrium Constants1) NO(g) + Cl2(g) ↔ NOCl(g) First, balance the equation.

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

2) H2(g) + I2(g) ↔ HI(g)

3) CaCO3(s) ↔ CaO(s) + CO2(g)

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 Keq at

407 K?

How would you determine the equilibrium constant?

Is this enough information to solve the problem?

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

.11 mol/L – 2x = 0.10 mol/L

x = 0.005 mol/L

0.005 mol/L

Temperature

Does temperature matter?

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

(red-brown) (colorless) 

Keq(407K) = 0.5

ICE in action

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?

You know how to find Keq. (What do you need?)

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

H2(g) I2(g) ↔ 2 HI(g)

Initial 2mol / 4L 2 mol / 4L

Change

Equilibrium

Keq = 25

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).

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?

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.

Can the Equilibrium Constant be Changed?

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

(red-brown) (colorless) 

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.

Does this method help explain the demonstrations that you just saw?

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:

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

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

• Keq = • Q, the reaction quotient, is used for this ratio under any

conditions at any point in time, not just equilibrium. • Q = • 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 + bB ↔ pP + qQ

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 ↔ C6H5CO2- + H3O+

(aq)

2.00 moles of C6H5CO2-, 1.00 moles of H3O

+ and

3.00 moles of C6H5CO2H are placed in 1 liter of

water.

What is Q under these conditions?Keq = 6.42 x10-5

Will the reaction proceed to form more C6H5CO2-(aq) and

H3O+(aq) or more C6H5CO2H(aq)?

H2O(l) + C6H5CO2H ↔ C6H5CO2- + H3O+

(aq)

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?

Q vs LeChâtelier

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?

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 vs LeChâtelier

Question 2a

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.

Question 2b 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?  

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?

Question 2cC(graphite) + S2(g)   ↔ CS2(g)

Question 3

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.

Question 4CO(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?

ICP- One Minute Paper

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 Chatelier'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.

Proposals

• Proposal Meetings During lab on Friday 11/3– 10 minutes– 1 hard copy of proposal for me to keep (in

template format)– No more than 5 minutes to present proposal

orally– Questions and discussion