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

SG Questions – Read p. 628-632. Answer the following questions

1. What is chemical equilibrium?

2. Explain what happens during equilibrium.

3. How does equilibrium relate to reaction rates?

4. How is equilibrium represented in symbols?

5. Explain what Is meant by the statement “equilibrium is dynamic”

Lecture Outline

• 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) →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:

kf = [NO2]2

kr [N2O4] • Rearranging, we get:

kf/kr = a constant

• At equilibrium the concentrations of N2O4 and NO2 do not change. (This doesn’t mean that they don’t stop reacting) • This mixture is called an equilibrium mixture. • 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. • The double arrow implies that the process is dynamic.

15.2 The Equilibrium Constant SG Questions – Read p. 632-637. Answer the following questions

1. What is the Haber process?

2. Explain the law of mass action.

3. What is the equilibrium constant?

4. What is the equilibrium expression?

5. Write the equilibrium expression for aA+bB=cC+dD

6. What is the equilibrium expression dependent on?

7. How do you solve for Kc

8. What is Kp

9. Compare and contrast Kc and Kp.

10. Do Kc and Kp have a relationship? Can they ever equal each other?

11. Why doesn’t K have units?

• Consider the Haber process: (one of the most important reactions in chemistry = fertilizer!!)

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

• The process is carried out at high temperature (500 oC) and pressure (200 atm) in the

presence of a catalyst.

3

Equilibrium Diagram – remember Ch 14 disappearance of REACTANTS vs appearance of PRODUCTS

Non-Catalyst rxn vs Catalyst rxn – Remember from Ch 14 (lowers Ea, provides alt route, speeds up rxn)

Equilibrium Diagrams Forward Rxn Reverse Rxn

Understand this importance of the following: • If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will

reach equilibrium with constant concentrations of nitrogen, hydrogen, and ammonia.

• However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2 and H2 will be produced until equilibrium is achieved.

• No matter what the starting compositions of reactants and products are, the equilibrium mixture contains the same relative concentrations of reactants and products.

• 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 dD + eE • The equilibrium expression is given by:

• Where Kc is the equilibrium constant. (REMEMBER: k ONLY DEPENDS ON TEMP !!! So, k will only change if temp changes) • The subscript “c” indicates that molar concentrations

• Note that the equilibrium constant expression has products in the numerator and reactants in the denominator.

• The value of Kc does not depend on initial concentrations of products or reactants. • Consider the reaction:

N2O4(g) 2NO2(g) • The equilibrium constant is given by:

• The equilibrium expression depends ONLY on stoichiometry. • It does not depend on the reaction mechanism. (pathway) • The value of Kc varies with temperature • We generally omit the units of the equilibrium constant, Kc or Kp only. (finally a chapter where you don’t have to write units!! )

ba

ed

cBA

EDK

][][

]][[

][

][

42

22

ON

NOKc

Kc & Kp ONLY - Only for gases and aqueous!! (No solids or liquids are written

in kc equation since their concentrations are constant)

5

Practice Problem #1 Write the equilibrium expression for Kc for the following reactions:

(a) 2O3 (g) 3O2 (g)

(b) 2NO (g) + Cl2 (g) 2NOCl (g)

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

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

Practice Problem #2 Write the equilibrium expression for Kp for the following reactions:

(a) 2O3 (g) 3O2 (g)

(b) 2NO (g) + Cl2 (g) 2NOCl (g)

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

bB

aA

eE

dD

pPP

PPK

)()(

)()(

• 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)

Practice Problem #3

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

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

(b) 2 SO3(g) 2SO2(g) + O2(g)

Kc = 4.08 x 10-3

at 1000 K. Calculate Kp for this reaction at this temperature

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15.3 Interpreting and Working with Equilibrium Constants SG Questions – Read p. 637-641. Answer the following questions

1. What does the magnitude of equilibrium tell you?

2. Copy figure 15.6

3. Fill in the statement

- If K>>1(large K): ___________ lies to the ________, __________ predominate.

- If K<<1(small K): ___________ lies to the ________, __________ predominate.

4. How does the value of K change?

5. Explain the difference between a large value of K and a smaller value of K.

6. How is the magnitude of an equilibrium constant related to the relative amounts of reactants and

products present in equilibrium mixture?

7. How is K manipulated to reflect changes in the chemical equation?

8. How are the equilibrium expressions related for the forward and reverse reaction?

9. Is it possible to calculate the equilibrium constant for a rxn if we know the equilibrium constants

for other reactions that add up?

The Magnitude of Equilibrium Constants

• The equilibrium constant, K, is the ratio of products to reactants. • The larger the K = the more products are present at equilibrium. • If K >> 1, then products dominate at equilibrium and equilibrium

lies to the right. • The smaller the K= the more reactants are present at equilibrium. • If K << 1, then reactants dominate at equilibrium and the

equilibrium lies to the left. ***The equilibrium constant for a reaction in one direction is the

reciprocal of the equilibrium constant of the reaction in the opposite direction.

Practice Problem #4

(a) 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?

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

Write the equilibrium constant expression (Kc) for this reaction

Does the reaction favor the products or reactants?

Does it lie to the left or to the right?

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

• The equilibrium constant of a reaction in the reverse direction is the inverse of the

equilibrium constant of the reaction in the forward direction.

• The equilibrium constant of a reaction that has been multiplied by a number is the

equilibrium constant raised to a power equal to that number.

• The equilibrium constant for a net reaction made up of two or more steps is the product of the

equilibrium constants for the individual steps.

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NOK c

9

Practice Problem #5

(a) 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)

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 10

5

3.8 x 10-6

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 10

5)= 0.12

Does the reaction favor the products or reactants? Does it lie to the left or to the right?

Practice Problem #6

(a) 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)

15.4 Heterogeneous Equilibria SG Questions – Read p. 641-644. Answer the following questions

1. What is homogenous equilibria

2. Explain the meaning of heterogenous equilibria.

3. Is there a certain way to write the K expression for a heterogenous reaction?

4. What is not included the equilibrium expressions and why ?

• Equilibria in which all reactants and products are present in the same phase are called

homogeneous equilibria. • Equilibria in which one or more reactants or products are present in a different phase are

called heterogeneous equilibria. • 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) Pb

2+(aq) + 2Cl

–(aq) Kc = [Pb

2+][Cl

–]

2

H2O(l) + CO32–

(aq) OH–(aq) + HCO3

–(aq)

][ 2COK c

][

]][[2

3

3

CO

HCOOHKc

2COp PK

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Practice Problem #7:

Write the Kc and Kp equilibrium expressions for the following.

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

2Ag +2

(aq) + Zn (s)

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

15.5 Calculating Equilibrium Constants

SG Questions – Read p. 644-646. Answer the following questions

1. What must be known in order to calculate K?

Practice Problem #8: A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472

oC. 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.

***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: • Tabulate initial and equilibrium concentrations (or partial pressures) for all species in the

equilibrium. • If an initial and an equilibrium concentration 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. • Deduce the equilibrium concentrations of all species. • Use these to calculate the value of the equilibrium constant.

Practice Problem #9

A closed systeminitially 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 10-3

M. Calculate Kc for the reaction at 448

oC.

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 10-3

M.

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 10-3

M – 0M = +1.87 x 10-3

M ( + = 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 10-3

M.

Equilibrium 1.87 x 10-3

M.

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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 10-3

M -0.935 x 10-3

M. + 1.87 x 10-3

M.

Equilibrium 1.87 x 10-3

M.

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 10-3

M -0.935 x 10-3

M. + 1.87 x 10-3

M.

Equilibrium 0.065 x 10-3

M 1.065 x 10-3

M 1.87 x 10-3

M.

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

15.6 Applications of Equilibrium Constants SG Questions – Read p. 646-650. Answer the following questions

1. What is the reaction quotient?

2. How do you predict the direction of a reaction?

3. How is the reaction quotient found?

4. Explain the relationship between Q and K

5. There are 3 possible situations for Q and K, what are they?

6. Understand figure 15.8, no need to copy it.

7. How can you predict the direction of a reaction given the equilibrium constant and the

concentrations of reactants and products?

8. What affects Q ?

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 Q < K, concentration of products are too small and the reactants too large. It will

achieve equilibrium when more products are formed ; it moves from left to right.

• If Q > K, concentration of products are too large and the reactants too small. It will achieve equilibrium when more reactants are formed ; 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

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Practice Problem #11 Predicting the Directionto 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.

Practice Problem #12 Calculating Equilibrium Concentrations (p. 647)

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?

Practice Problem #13 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)

15.7 Le Châtelier’s Principle (VERY IMPORTANT) SG Questions – Read p. 650-659. Answer the following questions

1. Explain Le Chatelier principle and its purpose.

2. What are the effects of adding a reactant or product to the system?

3. How is a system affected by a change in volume of a reactant or product according to Le

Chatelier?

4. Do changes in concentration or pressure affect the value of K

5. How does temperature affect equilibrium concentrations and constants?

6. How does temperature affect an endothermic systems equilibrium?

7. How does temperature affect an exothermic systems equilibrium?

8. Copy table on pg 651.

9. What is the effect of a catalyst in a system at equilibrium?

10. Copy figure 15.4

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• Le Châtelier’s principle: If a system at equilibrium is disturbed by a

change in temperature, a change in pressure, or a change in the concentration of one or more components, the system will shift its equilibrium position in such a way as to counteract the effects of the disturbance.

1. Change in Reactant or Product Concentration

• If a chemical system is at equilibrium and we add/ remove a product or reactant, the reaction will shift so as to reestablish equilibrium.

• For example, consider the Haber process again:

N2(g) + 3H2(g) 2NH3(g) • If H2 is added while the system is at equilibrium, Q < K. • The system must respond to counteract the added H2 (Le Châtelier’s principle).

• That is, the system must consume the H2 and produce products until a new equilibrium is established.

• Therefore, [H2] and [N2] will decrease and [NH3] will increase until Q = K.

2. Effects of Volume and Pressure Changes – Gas laws play a role!!

• If the equilibrium involves gaseous products or reactants, the concentration of these species

will be changed if we change the volume of the container. • For example, if we decrease the volume of the container, the partial pressures of each

gaseous species will increase. (Boyles Law) • Le Châtelier’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.

Consider the following system:

N2O4(g) 2NO2(g) The instant the pressure increases, the concentration of both gases increases and the system is not at equilibrium. • The system changes and favors the side with the fewer moles of gas: the left side • In a reaction with the same number of moles of gas in the products and reactants, changing the pressure has no effect on the equilibrium. • In addition, no change will occur if we increase the total gas pressure by the addition of

a gas that is not involved in the reaction. Changing P or V DOES NOT change Kc or Kp = no effect

3. Effect of Temperature Changes

Remember: (∆H = change in heat)

Endothermic Reactants + heat products ∆H > 0

Exothermic Reactantc products + heat ∆H < 0

The equilibrium constant (Kc or Kp) is temperature dependent!!!!! • How will a change in temperature alter a system at equilibrium? • It depends on the particular reaction. • For example, consider the endothermic reaction:

Co(H2O)62+

(aq) + 4Cl–(aq) CoCl4

2–(aq) + 6H2O(l) ∆H > 0

• This reaction is endothermic. • For an endothermic reaction (∆H > 0), heat can be considered as a reactant. • Thus, adding heat causes a shift in the forward direction. • The room-temperature equilibrium mixture is placed in a beaker of ice water. • This indicates a shift toward reactants • In this case, by cooling the system we are removing a reactant (heat). • Thus, the reaction is shifted in the reverse reaction.

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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 ↑

4. The 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 effect the composition of the equilibrium mixture.

Practice Problem #13 Using Le Chatelier’s Principle to predict shifts in equilibrium

N2O4(g) 2NO2(g) ∆H = 58.0 kJ In which direction will the equilibrium shift when the following occur: Change Shifts Reason

(a) N2O4 is added

(b) NO2 is removed

(c) The total pressure is increased by adding N2 (g)

(d) The volume is increased

(e) The temperature is decreased

Practice Problem #14 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

Practice Problem #15 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

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Practice Problem #16 Using Le Chatelier’s Principle to predict shifts in equilibrium

H2(g) + I2(g) 2HI (g) ∆H = 51.0 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