Chapter 15. Chemical Equilibrium Always MC & Free …

Chapter 15. Chemical Equilibrium – Always MC & Free

Response Questions

(Most important topic – connected to Ch 14, 16, 17, 5, & 19)

Common Student Misconceptions • Many students need to see how the numerical problems in this chapter are solved. • Students confuse the arrows used for resonance and equilibrium. • Students often have problems distinguishing between K and Q.

15.1 The Concept of Equilibrium

Read p. 628-632.

Learning Objective Essential Knowledge

TRA-6.A Explain the relationship

between the occurrence of a

reversible chemical or physical

process, and the establishment of

equilibrium, to experimental

observations.

TRA-6.A.1 Many observable processes are reversible. Examples include

evaporation and condensation of water, absorption and desorption of a

gas, or dissolution and precipitation of a salt. Some important reversible

chemical processes include the transfer of protons in acid-base reactions

and the transfer of electrons in redox reactions.

TRA-6.A.2 When equilibrium is reached, no observable changes occur

in the system. Reactants and products are simultaneously present, and the

concentrations or partial pressures of all species remain constant.

TRA-6.A.3 The equilibrium state is dynamic. The forward and reverse

processes continue to occur at equal rates, resulting in no net observable

change.

TRA-6.A.4 Graphs of concentration, partial pressure, or rate of reaction

versus time for simple chemical reactions can be used to understand the

establishment of chemical equilibrium.

Consider colorless N2O4.

• At room temperature, it decomposes to brown NO2. N2O4(g) ⇄ 2NO2(g)

• At some time, the color stops changing and we have a mixture of N2O4 and NO2.

Chemical EQUILIBRIUM is the point at which the CONCENTRATIONS of all species are CONSTANT.

• Assume that both the forward and reverse reactions are elementary processes. • We can write rate expressions for each reaction.

• Forward reaction: N2O4(g) → 2 NO2 (g) Ratef = kf[N2O4] kf = rate constant (forward reaction)

• Reverse reaction: 2NO2(g) →N2O4(g) Rater = kr[NO2]

2 kr = rate constant (reverse reaction) Eventually we get to equilibrium where the forward and reverse rates are equal.

• At equilibrium: kc = kf = [NO2]

2 kr [N2O4]

• Rearranging, we get: kf/kr = kc

At equilibrium the concentrations of N2O4 and NO2 do not change. (This doesn’t mean that they don’t stop reacting)

Consider the following reaction as it approaches equilibrium in a closed container

N2O4(g) ⇄ 2 NO2(g)

❏ AP Chemistry Misconception:

❏ Students think that when we say

“Equilibrium” that we have

equal amounts of reactants and

products. (FALSE)

❏ Equilibrium, for this reaction,

starts at about 60 minutes- the

concentrations of the reactants

and products DO NOT

CHANGE (CONSTANT)

❏ Reactants decrease in

concentration and Products

increase in concentration

3

15.2 The Equilibrium Constant Read p. 632-637. Let’s Refresh our memory…remember Ch 14 disappearance of REACTANTS vs appearance of PRODUCTS EQUILIBRIUM DIAGRAMS – shows the disappearance of reactants vs appearance of products over time and when all species have achieved equilibrium

2 SO2(g) + O2(g) ⇄ 2 SO3(g)

Forward Rxn Reverse Rxn

Start with all reactants, no products and still end up

achieving equilibrium.

Start with all products, no reactants and still end up

achieving

Draw Equilibrium diagrams for the Haber process: (one of the most important reactions in chemistry = fertilizer!!)

N2(g) + 3H2(g) 2NH3(g)

Equilibrium (Make sure you understand this well!)

❏ When equilibrium is reached:

❏ Reactants and products are simultaneously present

❏ The concentrations or partial pressures of all species remain constant

❏ No observable changes occur in the system (remember they still collide/react)

❏ This is an example of a dynamic equilibrium. (important)

❏ A dynamic equilibrium exists when the rates of the forward and

reverse reactions are equal.

❏ No further change in reactant or product concentration occurs.

Examples

5


TRA-7.A Represent the reaction quotient Qc or Qp, for a reversible reaction, and the corresponding equilibrium expressions Kc = Qc or Kp = Qp.

TRA-7.A.1 The reaction quotient Qc describes the relative concentrations of reaction species at any time. For gas phase reactions, the reaction quotient may instead be written in terms of pressures as Qp. The reaction quotient tends toward the equilibrium constant such that at equilibrium Kc = Qc and Kp = Qp. As examples, for the reaction

TRA-7.A.2 The reaction quotient does not include substances whose concentrations (or partial pressures) are independent of the amount, such as for solids and pure liquids.

Equilibrium can be reached from either direction.

• We can write an expression for the relationship between the concentration of the reactants and products at equilibrium.

• This expression is based on the LAW of MASS ACTION For a general reaction, aA + bB cC + dD

• The equilibrium EXPRESSION is given by:

• Where Kc or Keq is the EQUILIBRIUM CONSTANT. • The subscript “c” indicates molar concentrations • Note that the equilibrium constant expression has products in the numerator and reactants

in the denominator.

❏ NOTE: SOLIDS & LIQUIDS ARE NOT WRITTEN IN EQUILIBRIUM

EXPRESSIONS* (Their concentrations remain constant)

❏ The reaction quotient does NOT include substances whose concentrations (or

partial pressures) are independent of the amount (like solids & pure liquids).

Kc & Kp ONLY - Only for gases and aqueous!!

Equilibrium Constants in Terms of Pressure, Kp

• When the reactants and products are gases, we can write an equilibrium expression using partial pressures rather than molar concentrations.

• The equilibrium constant is Kp where “p” stands for pressure. For the reaction:

aA + bB dD + eE

❏ No [ ] for pressure – only concentration uses brackets. ❏ Pressure values – use ( ) parenthesis.

Consider the reaction: N2O4(g) 2NO2(g)

• The equilibrium constant is given by:

• The equilibrium expression depends ONLY on stoichiometry.

o It does not depend on the reaction mechanism. (pathway) • REMEMBER: K ONLY DEPENDS ON TEMP !!! So, K will only change if temp

changes! • We generally omit the units of the equilibrium constant, Kc, Keq, or Kp only.

o Finally a chapter where you don’t have to write units!! ☺ IMPORTANT: Kc, Kp, and Keq use EQUILIBRIUM concentration/pressures!

o The value of K does not depend on initial concentrations/pressures of products or reactants…. More coming in 15.5 and 15.6.

][

][

42

22

ON

NOKc =

bB

aA

eE

dD

pPP

PPK

)()(

)()(=

7

Write equilibrium expressions for the following reactions. (Remember, we do not include solids or liquids)

1. CH4(g) + 2 O2(g) ⇄ CO2(g) + 2 H2O(l) Kc =

2. 2 Pb(NO3)2(s) ⇄ 2 PbO(s) + 4 NO2(g) + O2(g) Kp =

3. C8H10N4O2(aq) + H2O(l) ⇄ C8H10N4O2H+(aq) + OH－(aq) Kb =

4. Al(NO3)3(s) ⇄ Al3+(aq) + 3 NO3－(aq) Ksp

=

5. 2NO (g) + Cl2 (g) 2NOCl (g) Kp =

Examples

The numerical values of Kc and Kp will differ. • You can calculate one from the other using the ideal gas equation:

PV = nRT so P = (n/V)RT

• The quantity (n/V) = molarity of X .

P = [X]RT

• Since [X] is concentration, we can use relate it to Kc to solve for Kp:

Kp = Kc(RT)∆n

• where ∆n = (moles of gaseous products) – (moles of gaseous reactants).

Kc = Kp ∆n = 0

(Only when the same number of moles of gas appear on both

the reactant and product side of the equation)

6. N2(g) + 3H2(g) 2NH3(g)

Kc = 9.60 at 300 oC. Calculate Kp for this reaction at this temperature

7. 2 SO3(g) 2SO2(g) + O2(g)

Kc = 4.08 x 10-3 at 1000 K. Calculate Kp for this reaction at this temperature

Examples

9

15.3 Interpreting and Working with Equilibrium Constants Read p. 637-641.

Learning Objective Essential Knowledge TRA-6.B Explain the relationship between the direction in which a reversible reaction proceeds and the relative rates of the forward and reverse reactions.

TRA-6.B.1 If the rate of the forward reaction is greater than the reverse reaction, then there is a net conversion of reactants to products. If the rate of the reverse reaction is greater than that of the forward reaction, then there is a net conversion of products to reactants. An equilibrium state is reached when these rates are equal.

TRA-7.C Explain the relationship between very large or very small values of K and the relative concentrations of chemical species at equilibrium.

RA-7.C.1 Some equilibrium reactions have very large K values and proceed essentially

to completion. Others have very small K values and barely proceed at all.

The Magnitude of Equilibrium Constants

The larger the K:

• the more products (and less reactants) are present at equilibrium. • If K >> 1, then products are favored at equilibrium. • Equilibrium lies to the right.

The smaller the K:

• the more reactants (and less products) are present at equilibrium. • If K << 1, then reactants are favored at equilibrium. • Equilibrium lies to the left.

1. Consider the following reaction:

N2O4(g) 2NO2(g) The equilibrium constant for this reaction is:

Does the reaction favor the products or reactants?

Does it lie to the left or to the right?

212.0][

][

42

22 ==ON

NOKc

Examples

***The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

2. When we write the equilibrium expression for the reverse reaction, 2NO2(g) N2O4(g)

Write the equilibrium constant expression (Kc) for this reaction



AP Chemistry: 7.1-7.6 Equilibrium, Reversible Reactions, and the Equilibrium Constant WATCH THIS VIDEO TO ASSIST IN THIS QUESTION (~31min)

2 A(g) + B(g) ⇄ 2 C (g) A(g) and B(g) react to form C(g), according to the balanced equation above. In an experiment, a previously evacuated rigid vessel is charged with A(g), B(g), and C(g), each with a concentration of 0.0100 M. The following table shows the concentrations of the gases at equilibrium at a particular temperature.

[A]eq [B]eq [C]eq

0.018 0.014 0.0020

1. Calculate the value of Kc.

Examples

https://www.youtube.com/watch?v=iPWe-jHRLMw&list=PLoGgviqq4845Sy3UfnNh_PljzAptMR7MQ&index=15

11

2. If the experiment is repeated at a higher temperature, Kc, is found to have a larger value.

Describe the effect of the temperature change on the concentrations of the gases at

equilibrium.


3. Magnitude of Equilibrium Constant

Reaction Equilibrium Reactions at 298 K Keq

1 Br2(g) + Cl2(g) ⇄ 2 BrCl(g) 10.

2 N2(g) + O2(g) ⇄ 2 NO(g) 4.2 × 10－31

The table above shows data for two reactions carried out in two separate experiments. The students started with 2 evacuated 1.0-Liter rigid containers at a constant temperature of 298 K. To each container 0.50 mol of the appropriate reactants was added, and the reaction was allowed to reach equilibrium. Based on this information, how do the relative concentrations of BrCl and NO present inside their respective containers at equilibrium compare to one another?

Examples



TRA-7.D Represent a multistep process with an overall equilibrium expression, using the constituent K expressions for each individual reaction.

TRA-7.D.1 When a reaction is reversed, K is inverted.

TRA-7.D.2 When the stoichiometric coefficients of a reaction are multiplied by a

factor c, K is raised to the power c.

TRA-7.D.3 When reactions are added together, the K of the resulting overall reaction

is the product of the K’s for the reactions that were summed.

TRA-7.D.4 Since the expressions for K and Q have identical mathematical forms, all

valid algebraic manipulations of K also apply to Q.

Relating Chemical Equations and Equilibrium Constants – RULES!! (read carefully)

❏ If a chemical equation is modified in some way, then the equilibrium constant for the equation

changes because of the modification. There are three common modifications:

1. If you reverse the equation, invert the equilibrium constant.

2. If you multiply the coefficients in the equations by a factor, raise the equilibrium

constant to the same factor.

3. If you add two or more individual chemical equations to obtain an overall

equation, multiply the corresponding equilibrium constants by each other to

obtain the overall equilibrium constant.

See example of how to apply the rules above:

Given the following information:

HF (aq) H+ (aq) + F- (aq) Kc =6.8 x 10-4

H2C2O4 (aq) 2H+ (aq) + C2O4 -2 (aq) Kc =3.8 x 10-6

Determine the value of Kc for the reaction:

2HF (aq) + C2O4 -2 (aq) 2 F- (aq) + H2C2O4 (aq) Kc = ??

Answer

If you multiply the first elementary reaction by 2, the Kc is raised to the power of 2.

2HF (aq) 2H+ (aq) + 2F- (aq) Kc =(6.8 x 10-4 )2 = 4.6 x 10 -7

If you reverse the second elementary reaction, the Kc is now the reciprocal

2H+ (aq) + C2O4 -2 (aq) H2C2O4 (aq) Kc = 1 = 2.6 x 105

3.8 x 10-6

13

The two equations can now be summed up to give the net overall equation and we can multiply the individual Kc values to get the overall equilibrium constant, Kc

2HF (aq) 2H+ (aq) + 2F- (aq)

2H+ (aq) + C2O4 -2 (aq) H2C2O4 (aq)

2HF (aq) + C2O4 -2 (aq) 2 F- (aq) + H2C2O4 (aq) Kc = (4.6 x 10 -7)(2.6 x 105)= 0.12



4. Given the following information:

H2 (g) + I2 (g) 2HI (g) Kp =54.0

N2 (g) + 3H2 (g) 2NH3 (g) Kp =1.04 x 10-4

Determine the value of Kp for the reaction:

2NH3 (g) + 3I2 (g) 6HI (g) + N2 (g) Kp = ???

Examples


5. Properties of the Equilibrium Constant

The chemical equations and equilibrium expressions for two reactions at the same temperature

are given above. Based on the information, how can they be used to calculate the value of K3 for

Reaction 3 at the same temperature?

Examples


15


The chemical equations and equilibrium expressions for two reactions at the same temperature

are given above. Based on the information, how can you manipulate the K1 to get the K2 value

for Reaction 2 at the same temperature?


Based on the equilibrium constants given above, what is the correct expression for the equilibrium constant for Reaction 3?

Examples

15.4 Heterogeneous Equilibria Read p. 641-644.

• Equilibrium in which all reactants and products are present in the same phase are called HOMOGENEOUS equilibrium.

• Equilibrium in which one or more reactants or products are present in a different phase are called HETERGENEOUS equilibrium.

CaCO3(s) CaO(s) + CO2(g)

Kc & Kp expressions: Only for gases and aqueous!! (No solids or liquids are written in kc equation since their concentrations are constant) PbCl2(s) Pb2+(aq) + 2Cl–(aq) Kc = [Pb2+][Cl–]2

H2O(l) + CO32–(aq) OH–(aq) + HCO3

–(aq)

Write the Kc equilibrium expression for the following.

(a) 2Ag (s) + Zn +2 2Ag +2 (aq) + Zn (s)

Write the Kp equilibrium expression for the following.

(b) 3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4H2(g)

][ 2COK c =

][

]][[2

3

3

−

−−

=CO

HCOOHKc

2COp PK =

Examples

17

15.5 Calculating Equilibrium Constants

Read p. 644-646.

Learning Objective Essential Knowledge TRA-7.B Calculate Kc or Kp based on experimental observations of concentrations or pressures at equilibrium.

TRA-7.B.1 Equilibrium constants can be determined from experimental measurements of the concentrations or partial pressures of the reactants and products at equilibrium.

TRA-7.E Identify the concentrations or partial pressures of chemical species at equilibrium based on the initial conditions and the equilibrium constant.

TRA-7.E.1 The concentrations or partial pressures of species at equilibrium can be

predicted given the balanced reaction, initial concentrations, and the appropriate K.

TRA-7.F Represent a system undergoing a reversible reaction with a particulate model.

TRA-7.F.1 Particulate representations can be used to describe the relative numbers of

reactant and product particles present prior to and at equilibrium, and the value of the

equilibrium constant

H2(g) + I2(g) ⇄ 2 HI(g)

1. At Equilibrium, [H2] = 0.100 M , [I2] = 0.100 M and [HI] = 0.714 M. Calculate the value of Kc.

Examples

2. A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472oC. The equilibrium mixture of the gases were found to contain 7.38atm H2, 2.46atm N2, and 0.166 atm NH3. From the data, write a balanced chemical equation, write the Kp equilibrium expression, and calculate Kp for the reaction.

3. For the Haber process, the Kp = 1.45 x 10 -5 at 500 oC.

N2(g) + 3H2(g) 2NH3(g)

In an equilibrium mixture of the three gases at 500 oC, the partial pressures are 0.928atm H2 and 0.432atm N2. What is the partial pressure of NH3 in this equilibrium mixture?(p.647)

Examples

19

***Sometimes we don’t know the equilibrium concentrations of all chemical species in an equilibrium mixture. If we know the equilibrium concentration of at least one species, we can use stoichiometry to deduce the equilibrium concentrations of the others. We use a method called ICE (Initial, Change, Equilibrium) Charts ICE Chart Rules: • If an initial and an equilibrium concentration (or partial pressures) is given for a species,

calculate the change in concentration. • Use the coefficients in the balanced chemical equation to calculate the changes in

concentration of all species. • Determine the equilibrium concentrations of all species. • Use equilibrium concentrations these to calculate the value of the equilibrium constant.

See example of step-by-step instructions using ICE charts

A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448oC is allowed to reach equilibrium. The equilibrium mixture for HI is 1.87 x 10-3M. Calculate Kc for the reaction at 448oC.

H2 (g) + I2 (g) 2HI (g)

First, we create an ICE Chart and put in the data that we are given into the chart:

H2 (g) + I2 (g) 2HI (g)

Initial 1.000 x 10-3 M 2.000 x 10-3 M 0M

Change

Equilibrium 1.87 x 10-3M.

Second, we calculate the change in concentration of HI, which is the difference between the equilibrium values and the initial values:

Change in [HI] = 1.87 x 10-3M – 0M = +1.87 x 10-3M ( + = gained products)

H2 (g) + I2 (g) 2HI (g)

Initial 1.000 x 10-3 M 2.000 x 10-3 M 0M

Change + 1.87 x 10-3M.


Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H2] and [ I2]

H2 (g) + I2 (g) 2HI (g)

Initial 1.000 x 10-3 M 2.000 x 10-3 M 0M

Change -0.935 x 10-3M -0.935 x 10-3M. + 1.87 x 10-3M.


Fourth, we calculate the equilibrium concentrations of H2 and I2 by taking the difference between the initial concentrations and the change

H2 (g) + I2 (g) 2HI (g)

Initial 1.000 x 10-3 M 2.000 x 10-3 M 0M

Change -0.935 x 10-3M -0.935 x 10-3M. + 1.87 x 10-3M.

Equilibrium 0.065 x 10-3 M 1.065 x 10-3 M 1.87 x 10-3M.

Fifth, we use the concentrations values for the equilibrium concentrations to calculate Kc

Kc = [HI]2 = [1.87 x 10-3M]2 = 51 [H2 ][ I2 ] [0.065 x 10-3 M ][1.065 x 10-3 M ]

- = Lost reactants + = Gained Products

21

4. Calculating Equilibrium Concentrations from Initial Concentration (p. 648)

A 1.000L flask is filled with 1.00mol of H2 and 2.00mol of I2 at 448 oC. The value of the equilibrium constant Kc for the reaction is 50.5. What are the equilibrium concentrations of I2, H2, and HI.

I2(g) + H2(g) 2HI (g)

Examples

AP Chemistry: 7.7-7.10 Calculating Equilibrium Concentrations and Le Châtelier's Principle

WATCH THIS VIDEO TO ASSIST IN THIS QUESTION (~12min)

Calculating Equilibrium Concentrations - Guided Practice # 2

H2(g) + I2(g) ⇄ 2 HI(g) Kc = 51.0

A sample of H2(g) and I2(g) is added to a previously evacuated reaction vessel. The initial concentrations are the following.

[H2] = [I2] = 2.0 M

Calculate the concentrations of all three gases in the reaction at equilibrium. Use the ICE Table to help you compute your answer.

H2(g) + I2(g) 2 HI(g)

Initial 2.0 M 2.0 M 0

Change － x － x + 2 x

Equilibrium 2 － x 2 － x 2 x

(NOTE: YOU WILL NOT HAVE TO USE QUADRATICS ON THE AP EXAM!)

https://www.youtube.com/watch?v=IIeDMFWf7nI&list=PLoGgviqq4845Sy3UfnNh_PljzAptMR7MQ&index=17

23

Graphical representation of the answer to Guided Practice # 2 Where is equilibrium achieved for all chemicals? Draw a line


H2(g) + I2(g) ⇄ 2 HI(g) Kc = 51.0

A sample of HI(g) is added to a previously evacuated reaction vessel. The initial concentration is the following.

[HI] = 3.0 M


(NOTE: in order to reach equilibrium, the reaction must proceed from right to left ∴ + reactants, － products)

H2(g) + I2(g) 2 HI(g)

Initial

Change

Equilibrium

Graphical representation of the answer to Guided Practice # 3

AP TIPS: CALCULATING EQUILIBRIUM CONCENTRATIONS *IMPORTANT NOTES*

❏ FOR ANY REACTION, THE EQUILIBRIUM CONCENTRATIONS OF THE

REACTANTS AND PRODUCTS DEPEND ON THE INITIAL

CONCENTRATIONS

❏ HOWEVER, THE EQUILIBRIUM CONSTANT IS ALWAYS THE SAME AT A

GIVEN TEMPERATURE, REGARDLESS OF THE INITIAL

CONCENTRATIONS.

❏ THERE IS NO SUCH THING AS THE LAW OF CONSERVATION OF MOLES.

(MOLES CAN CHANGE….GRAMS CANNOT)

25

15.6 Applications of Equilibrium Constants Read p. 646-650.


TRA-7.A Represent the reaction quotient Qc or Qp, for a reversible reaction, and the corresponding equilibrium expressions Kc = Qc or Kp = Qp.

TRA-7.A.1 The reaction quotient Qc describes the relative concentrations of reaction species at any time. For gas phase reactions, the reaction quotient may instead be written in terms of pressures as Qp. The reaction quotient tends toward the equilibrium constant such that at equilibrium Kc = Qc and Kp = Qp. As examples, for the reaction

TRA-7.A.2 The reaction quotient does not include substances whose concentrations (or partial pressures) are independent of the amount, such as for solids and pure liquids.

Predicting the Direction of Reaction

• For a general reaction: aA + bB dD + eE

• We define Q, the REACTION QUOTIENT, (at any point in time besides equilibrium):

We can compare Qc with Kc or Qp with Kp:

• If Q = K, then the system is at equilibrium. • If K > Q, concentration of products are too small and the reactants too large.

o It will achieve equilibrium when more products are formed. o It moves from left to right.

• If K < Q, concentration of products are too large and the reactants too small. o It will achieve equilibrium when more reactants are formed. o It moves from right to left

Q vs. K

• It is calculated the same way as Kc or Kp • Q is used to determine

1. which way the reaction is moving (left → right or right → left) 2. whose concentrations/pressures need to increase/decrease to achieve equilibrium.

ba

ed

QBA

ED=

27

1. Predicting the Direction to Achieve Equilibrium (p.646)

At 448oC , the equilibrium constant Kc for the reaction is 50.5.

H2 (g) + I2 (g) 2HI (g)

Predict which direction the reaction will proceed to reach equilibrium if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2 , and 3.0 x 10-2 mol of I2 in a 2.00L container.

Examples


2. 7.3 Reaction Quotient and Equilibrium Constants & 7.4 Calculating the Equilibrium

Constant

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in with each system will proceed to reach equilibrium.

(a) 2 2 NH3(g) ⇄ N2(g) + 3 H2(g) Kc = 17 [NH3] = 0.20 M [N2] = 1.00 M [H2] = 1.00 M

(b) 2 2 NH3(g) ⇄ N2(g) + 3 H2(g) KP = 6.8 × 104 NH3= 3.0 atm N2 = 2.0 atm H2 = 1.0 atm

(c) 2 2 SO3(g) ⇄ 2 SO2(g) + O2(g) Kc = 0.230 [SO3] = 0.00 M [SO2] = 1.00 M [O2] = 1.00 M

Hint: Zero on Top = 0. Zero on Bottom = ∞

(d) 2 2 SO3(g) ⇄ 2 SO2(g) + O2(g) KP = 16.5 SO3= 1.00 atm SO2 = 1.00 atm O2 = 1.00 atm

Examples


29


WATCH THIS VIDEO TO ASSIST IN THIS QUESTION (~18min)


H2(g) + I2(g) ⇄ 2 HI(g) Kc = 51.0

A sample of H2(g) and I2(g) and HI(g) is added to a previously evacuated reaction vessel. The initial concentration are the following.

[H2] = [I2] = 1.0 M

[HI] = 3.0 M


PROBLEM: WHICH DIRECTION WILL YOU GO? TOWARDS REACTANTS OR PRODUCTS? NEED TO CALCULATE Q FIRST!

(a) What is the reaction quotient for the given initial conditions? Which direction will

the reaction move towards?

❏ If the reaction proceeds towards products, － reactants, +

products

❏ If the reaction proceeds towards reactants, + reactants, －

products


(b) Now that you know which direction, complete the ICE table to identify the

equilibrium concentrations of all species.

H2(g) + I2(g) 2 HI(g)

Initial

Change

Equilibrium

Graphical representation of the answer to Guided Practice # 4

31

An easier way to remember which direction for K and Q

7.8 Representations of Equilibrium : SUMMARY OF GUIDED PRACTICE 2, 3 & 4

For H2(g) + I2(g) ⇄ 2 HI(g) All have the SAME Kc!!!! BECAUSE TEMP didn’t change!!!

AP EXAM FRQ PRACTICE - 7.1-7.6 - HOMEWORK (Answers shown at the beginning of 7.7-7.10 video)

1. In an experiment, X(g) and Y(g) were combined in a rigid container at constant temperature

and allowed to react as shown in the equation below. The graph provides the data collected

during the experiment.

X(g) + Y(g) ⇄ XY(g)

Indicate whether you agree or disagree with the statement below. Support your answer with a

short explanation.

“The reaction was about to reach equilibrium at time =A because the

concentrations of X and XY were almost the same.”

Examples

33

AP EXAM FRQ PRACTICE - 7.1-7.6 - HOMEWORK (CONTINUED)

2. An equimolar mixture of X(g) and Y(g) is placed inside a rigid container at constant

temperature. Each particle diagram below represents the changes that occur over

time.

Draw a particle diagram in the box for t = 400 s that is consistent with the conclusion that

equilibrium was reached for this reaction after 300 seconds, but before 400 seconds.

N2(g) + O2(g) ⇄ 2 NO(g)

3. At high temperatures, N2(g) and O2(g) can react to produce nitrogen monoxide, NO(g), as

represented by the equation above.

(a) Write the expression for the equilibrium constant, KP, for the forward reaction.

(b) A student injects N2(g) and O2(g) into a previously evacuated, rigid vessel and raises the

temperature of the vessel to 2000oC. At this temperature the initial partial pressures of

N2(g) and O2(g) are 6.01 atm and 1.61 atm, respectively. The system is allowed to reach

equilibrium. The partial pressure of NO(g) at equilibrium is 0.122 atm. Calculate the

value of KP.

35

15.7 Le Châtelier’s Principle (VERY IMPORTANT) Read p. 650-659.


TRA-8.A Identify the response of a system at equilibrium to an external stress, using Le Chatelier’s principle.

TRA-8.A.1 Le Chatelier’s principle can be used to predict the response of a system to

stresses such as addition or removal of a chemical species, change in temperature,

change in volume/ pressure of a gas-phase system, or dilution of a reaction system.

TRA-8.A.2 Le Chatelier’s principle can be used to predict the effect that a stress will

have on experimentally measurable properties such as pH, temperature, and color of a

solution.

TRA-8.B Explain the relationships between Q, K, and the direction in which a reversible reaction will proceed to reach equilibrium.

TRA-8.B.1 A disturbance to a system at equilibrium causes Q to differ from K, thereby

taking the system out of equilibrium. The system responds by bringing Q back into

agreement with K, thereby establishing a new equilibrium state.

TRA-8.B.2 Some stresses, such as changes in concentration, cause a change in Q only.

A change in temperature causes a change in K. In either case, the concentrations or

partial pressures of species redistribute to bring Q and K back into equality.

Supposed that a chemical reaction has already reached equilibrium.

❏ The rate of the forward reaction = the rate of the reverse reaction.

❏ The concentration (or partial pressures) of each reactant and product are no longer

changing over time

What would happen if a system at equilibrium experiences some sort of external stress?

❏ Le Châtelier’s Principle - if a system at equilibrium is disturbed (or experiences an

external stress), the position of the equilibrium shifts in the direction that relieves the

stress.

❏ What is a stress?

Change in Concentration Change in Pressure/Volume Change in Temperature

Recall that when Q = K, the system is at equilibrium.

❏ If a change occurs that causes Q to become less than K, the equilibrium position will shift

from left to right until Q = K.

❏ If a change occurs that causes Q to become greater than K, the equilibrium position will

shift from right to left until Q = K.

If you need additional help, watch the video by Farabaugh to help you explain! (~24min)


Stresses on Equilibrium # 1 - Change in Concentration

2 NO2(g) ⇄ N2O4(g) brown colorless

❏ Adding More Reactants

❏ Suppose that the system shown above is at equilibrium and additional NO2(g) is

added to the reaction vessel (at constant temperature). (Concentration of NO2

increases - denoted by larger brackets)

Explanation:

Q is now LESS than K.

The equilibrium position shifts to the right

until Q = K.

❏ Adding More Products

❏ Suppose that the system shown above is at equilibrium and additional N2O4(g) is

added to the reaction vessel (at constant temperature). (Concentration of N2O4

increases - denoted by larger brackets)

Explanation:

Q is now GREATER than K.

The equilibrium position shifts to the left

until Q = K.

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❏ Removing Some Reactants

❏ Suppose that the system shown above is at equilibrium and some NO2(g) is

removed from the reaction vessel (at constant temperature). (Concentration of

NO2 decreases - denoted by larger brackets)

Explanation:

Q is now GREATER than K.

The equilibrium position shifts to the left

until Q = K.

❏ Removing Some Products

❏ Suppose that the system shown above is at equilibrium and some N2O4(g) is

removed from the reaction vessel (at constant temperature). (Concentration of

N2O4 decreases - denoted by larger brackets)

Explanation:

Q is now LESS than K.

The equilibrium position shifts to the right

until Q = K.

AP CHEMISTRY EXAM TIP:

❏ DO NOT SAY “BECAUSE OF LE CHÂTELIER’S PRINCIPLE”.

❏ INSTEAD SAY “AS A RESULT OF THIS CHANGE, Q WILL BE < OR > K, AND

THE EQUILIBRIUM WILL SHIFT TOWARD RIGHT/LEFT UNTIL Q=K.

Stresses on Equilibrium # 2 - Change in Pressure/Volume


❏ Adding an Inert Gas

❏ Suppose that the system shown above is at equilibrium and a sample of He(g) (or

Ne or Ar) is added to the reaction vessel (at constant temperature).

Q = K.

There is no shift in the equilibrium

position.


❏ Decreasing the Volume of the Reaction Vessel

❏ Suppose that the system shown above is at equilibrium and the volume of the

reaction vessel is reduced by half(at constant temperature).

Let’s say [NO2]eq = 1.0 M and [N2O4]eq = 5.0 M. If I halve the volume, I double the

concentration.

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❏ Increasing the Volume of the Reaction Vessel

❏ Suppose that the system shown above is at equilibrium and the volume of the

reaction vessel is doubled(at constant temperature).

Let’s say [NO2]eq = 1.0 M and [N2O4]eq = 5.0 M … if I Double the Volume, I Halve the

Concentration.

AP Tips: SUMMARY OF VOLUME CHANGES:

❏ DECREASING THE VOLUME OF THE REACTION VESSEL CAUSES A SHIFT

TOWARD THE SIDE OF THE EQUATION WITH FEWER MOLES OF GAS

❏ INCREASING THE VOLUME OF THE REACTION VESSEL CAUSES A SHIFT

TOWARD THE SIDE WITH MORE MOLES OF GAS

EQUILIBRIUM PARTICLE DIAGRAMS


Decreasing the Volume of the Reaction Vessel (Increasing the Pressure) AKA Boyle’s Law: inverse relationship to P vs. V. Le Chatelier’s principle predicts that if pressure is increased, the system will shift to counteract

the increase. • That is, the system shifts to remove gases and decrease pressure. • An increase in pressure favors the direction that has fewer moles of gas.

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Increasing the Volume of the Reaction Vessel (Decreasing the Pressure)

AP Tips: SUMMARY OF PRESSURE/VOLUME CHANGES:

❏ DECREASING THE VOLUME (INCREASE PRESSURE) OF THE REACTION

VESSEL CAUSES A SHIFT TOWARD THE SIDE OF THE EQUATION WITH

FEWER MOLES OF GAS

❏ INCREASING THE VOLUME (DECREASE PRESSURE) OF THE REACTION

VESSEL CAUSES A SHIFT TOWARD THE SIDE WITH MORE MOLES OF

GAS

❏ IF BOTH SIDES OF THE EQUATION HAVE EQUAL MOLES, THERE WILL

BE NO SHIFT (Q WILL ALWAYS EQUAL K)

❏ THIS ALSO WORKS FOR AQUEOUS EQUILIBRIA AS WELL WITH

DISSOLVING IONS

❏ CHANGING P OR V DOES NOT AFFECT Kc OR Kp

Stresses on Equilibrium # 3 - Change in Temperature


DARKER BROWN LIGHTER BROWN

These observations can tell us information about this reaction. Is this reaction, as written, endothermic or exothermic?

The equilibrium constant (Kc or Kp) is temperature dependent!!!!! Changing the temperature will also change the value of K. The information in the data table above can tell us about this reaction. Is this reaction, as written, endothermic or exothermic?

2 NO2(g) ⇄ N2O4(g)

Temperature Kc

15oC 160

100oC 5.0

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ANSWER: When you increase temperature, it is like adding heat. When you did this, your value of Kc decreased. (When you removed heat, your value of Kc increases) THIS MEANS HEAT IS A PRODUCT.

2 NO2(g) ⇄ N2O4(g) + HEAT

THEREFORE IT IS AN EXOTHERMIC REACTION.

SUMMARY OF TEMPERATURE CHANGES

HEAT + A ⇄ B ∆H > 0

❏ For an ENDOTHERMIC reaction…

❏ Increasing temperature (T) causes a shift toward the right, and K increases.

❏ Decreasing temperature (T) causes a shift toward the left, and K decreases.

A ⇄ B + HEAT ∆H < 0

❏ For an EXOTHERMIC reaction…

❏ Increasing temperature (T) causes a shift toward the left, and K decreases

❏ Decreasing temperature (T) causes a shift toward the right, and K increases.

Conclusion….. A change in temperature causes a change in the value of K.

Type of reaction Temperature Favors K

Endo ↑ products ↑

Endo ↓ reactants ↓

Exo ↑ reactants ↓

Exo ↓ products ↑

Stresses on Equilibrium # 4 – Effect of Catalysts

A catalyst lowers the activation energy barrier for the reaction AND provides an alternate pathway

• Therefore, a catalyst will decrease the time taken to reach equilibrium. • A catalyst DOES NOT affect the composition of the equilibrium mixture.

1.Using Le Chatelier’s Principle to predict shifts in equilibrium

PCl5(g) PCl3(g) + Cl2(g) ∆H = 87.9 kJ In which direction will the equilibrium shift when the following occur: Change Shifts Reason

(a) Cl2 (g) is removed

(b) Temperature is decreased

(c) The volume of the reaction system is increased

(d) PCl3 is added

(e) Increase in Temperature

Examples

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2. Using Le Chatelier’s Principle to predict shifts in equilibrium

N2(g) + 3H2(g) 2NH3(g) ∆H = -92.0 kJ In which direction will the equilibrium shift when the following occur: Change Shifts Reason

(a) Increase in pressure

(b) Increase in temperature

(c) Increase in concentration of reactants

(d) Increase in contraction of the product

(e) Adding catalyst


WATCH THIS VIDEO TO ASSIST IN THIS QUESTION (~42min) Reaction Quotient and Le Châtelier’s Principle - Guided Practice

PCl3(g) + Cl2(g) ⇄ PCl5(g)

Using the chemical equilibrium shown above, convey how the equilibrium will shift by completing the particle diagram drawn in the box labeled “New Equilibrium.”

Examples


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AP EXAM FRQ PRACTICE 7.7 - 7.10 - HOMEWORK (Answers shown at the beginning of 7.11-7.13 video)

CaCO3(s) ⇄ CaO(s) + CO2(g)

When heated, calcium carbonate decomposes according to the equation above. In a study of the decomposition of calcium carbonate, a student added a 50.0 g sample of powdered CaCO3(s) to a 1.00 L rigid container. The student sealed the container, pumped out all of the gases, then heated the container in an oven at 1100 K. As the container was heated, the total pressure of the CO2(g) in the container was measured over time. The data are graphed and plotted in graph below.

The student repeated the experiment, but this time the student uses a 100.0 g sample of powdered CaCO3(s). In this experiment, the final pressure in the container was 1.04 atm, which was the same final pressure as in the first experiment.

(a) Calculate the number of moles of CO2(g) present in the container after 20 minutes of

heating.

(b) The student claimed that the final pressure in the container in each experiment became

constant because all of the CaCO3(s) had decomposed. Based on the data in the

experiments, do you agree with this claim? Explain.

(c) After 20 minutes some CO2(g) was injected into the container, initially raising the

pressure to 1.5 atm. Would the final pressure inside the container be less than, greater

than, or equal to 1.04 atm. Explain your reasoning.

(d) Are there sufficient data obtainer in the experiments to determine the value of the

equilibrium constant, KP, for the decomposition of CaCO3(s) at 1100 K? Justify your

answer.

Examples

AP EXAM FRQ PRACTICE 7.7 - 7.10 - HOMEWORK (CONTINUED)

Ba2+(aq) + EDTA4－ (aq) ⇄ Ba(EDTA)2－ (aq) K = 7.7 × 107

The polyatomic ion C10H12N2O84－ is commonly abbreviated as EDTA4－. The ion can form

complexes with metal ions in aqueous solutions. A complex of EDTA4－ with Ba2+ ion forms according to the equation above. A 50.0 mL volume of a solution that has an EDTA4－(aq) concentration of 0.30 M is mixed with 50.0 mL of 0.20 M Ba(NO3)2 to produce 100.0 mL of solution.

a. Considering the value of K for the reaction, determine the concentration of Ba(EDTA)2－

(aq) in the 100.0 mL of solution. Justify your answer.

b. The solution is diluted with distilled water to a total volume of 1.00 L. After equilibrium

has been reestablished, is the number of moles Ba2+(aq) of present in the solution greater

than, less than, or equal to the number of moles of Ba2+(aq) present in the original

solution before it was diluted? Justify your answer.

Examples

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