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Chapter 15: Chemical Equilibrium

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Chapter 15: Chemical Equilibrium

15.1 The Concept of Equilibrium

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15.2 The Equilibrium Constant

LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an

explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

N​2​ (g) + 3H​2​ (g) ⇌ 2NH​3​ (g)

General equilibrium equation

aA + bB ⇌ dD + eE

The Equilibrium Constant in term of concentration

K​c​ =

Sample Exercise 15.1

Sample Exercise 15.1​​ ​​Writing Equilibrium-Constant ExpressionsWriting Equilibrium-Constant Expressions (a)(a)​​ 2 O​3​(​g​) ⇌ 3 O​2​(​g​)

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(b)(b)​​ 2 NO(​g​) + Cl​2​(​g​) ⇌ 2 NOCl(​g​)

(c)(c)​​ Ag​+​(​aq​) + 2 NH​3​(​aq​) ⇌ Ag(NH​3​)​2​+​(​aq​)

Evaluating K​c

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The Equilibrium constant in terms of pressure

aA + bB ⇌ dD + eE

K​p​ =

K​

p​

= K​

c​

(RT)​

n

R is .0821

T must be in Kelvin When would K​p​ = K​c​?

Sample Exercise 15.2

Sample Exercise 15.2 ​​​​Converting between Converting between ​​KK​​cc​​ and and ​​KK​​pp

In the synthesis of ammonia from hydrogen and nitrogen N​2​ (g)​ + 3H​2 ​(g) ​ 2NH​3​ ​(g)

K​c​ = 9.60 at 300​o​C, Calculate K​p​ for this reaction at this temperature.

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15.3 Understanding and Working with Equilibrium Constants

LO 6.7 The student is able, for a reversible reaction that has a large or small K, to determine which chemical species will have very large versus very small concentrations at equilibrium.

The magnitude of equilibrium constants.

Sample Exercise 15.3

Sample Exercise 15.3​​ ​​Interpreting the Magnitude of an Equilibrium ConstantInterpreting the Magnitude of an Equilibrium Constant The following diagrams represent three systems at equilibrium, all in the same-size containers. ​(a) (a) ​​Without doing any calculations, rank the systems in order of increasing ​K​c​. ​(b)(b)​​ If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate ​K​c​ for each system.

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Practice Exercise 2 Practice Exercise 2

For the reaction H​2​ (g) + I​2​(g) ⇌ 2HI (g) , K​p​ = 794 at 289K and K​p​ = 54 at 700K. Is the formation of HI favored more at the higher or lower temperature?

LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition of two reactions), determine the effects of that manipulation on Q or K.

The direction of the chemical equation and K

N​2​O​4 ​(g) ⇌ 2NO​2​ (g)

K​c​ = = .212

What about

2NO​2​ (g) ⇌ N​2​O​4 ​(g)

Relating Chemical Equation Stoichiometry and Equilibrium Constants What if N​2​O​4 ​(g) ⇌ 2NO​2​ (g) were multiplied by 2??

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Equations can also be added together like they were in the kinetics chapters...The K​eq ​of the ​SUMSUM​​ of the equations is the ​PRODUCTPRODUCT​​ of the K​eqs

Given that, at 700K, K​p​= 54.0 for the reaction H​2​(g) + I​2​(g) ⇌2HI (g) , and K​p​ = 1.04x10​-4​ for the reaction N​2​ (g) + 2H​2 ​(g) ⇌ 2NH​3​ (g), determine the value of K​p

for the reaction 2 NH​3 ​(g) + 3I​2​ (g) ⇌6HI + N​2​ (g) at 700K

15.4 Heterogeneous Equilibria

Decomposition of calcium carbonate into calcium oxide and carbon dioxide gas.

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Sample Exercise 15.5

Sample Exercise 15.5​​ ​​Writing Equilibrium-Constant Expressions forWriting Equilibrium-Constant Expressions for Heterogeneous Reactions

Heterogeneous Reactions

Write the equilibrium-constant expression ​K​c​ for (a)(a)​​ CO​2​(g) ​+ H​2​(g)​CO​(g) ​+ H​2​O​(l)

(b)(b)​​ SnO​2​(s) ​+ 2 CO​(g)​Sn​(s) ​+ 2 CO​2​(g)

15.5 Calculating equilibrium Constants

LO 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained, calculate the equilibrium constant, K.

A closed system initially containing 1.000x10​-3​ M H​2​ and 2.000 x10​-3​ M I​2​ at 448​o​C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87x10​-3​ M. Calculate K​c​ at 448​o​C for the reaction taking place

H​2​ ​(g)​ + I​2​ ​(g)​ ⇌ 2HI ​(g)

[H​2​] [I​2​] [HI]

II​​nitial CC​​hange

EE​​quilibrium

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Sulfur trioxide decomposes at high temperature in a sealed container:

2SO​3​ ​(g) ​ 2SO​2​ ​(g) ​+ O​2​ (g)

Initially, the vessel is charged at 1000K with SO​3​ (g) at a partial pressure of 0.500 atm. At equilibrium, the SO​3​ partial pressure is .200 atm. Calculate the value of K​p​ at 1000K

[SO​3​ ] [SO​2​] [O​2​] II​​nitial

CC​​hange

EE​​quilibrium

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15.6 Applications of Equilibrium Constants

LO 6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K, use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants as equilibrium is approached.

Predicting the direction of a reaction

Q = reaction quotient a number obtained from

substituting starting concentrations into the equilibrium constant

expression...

if Q>K

if Q<K

if Q=K

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At 1000K, the value of K​p​ for the reaction 2SO​3​ (g) ⇌ 2SO​2​ (g) + O​2​ (g) is .338.

Calculate the value of Q​p​, and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are P​SO3​ = .16 atm, P​SO2​ =.41 atm, and P​O2​ = 2.5 atm.

Calculating equilibrium concentrations

LO 6.6 The student can, given a set of initial conditions (concentrations or partial

pressures) and the equilibrium constant, K, use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the conditions at equilibrium for a system involving a single reversible reaction.

AT 500K the reaction PCl​5​ (g) ⇌ PCl​3​ (g) + Cl​2​ (g) has K​p​ = .497. In an equilibrium mixture at 500K, the partial pressure of PCl​5​ is .860 atms and that of PCl​3​(g) is .350 atm. What is the partial pressure of Cl​2​ in the equilibrium mixture?

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A 1.000 L flask is filled with 1.000 mol of H​2​ and 2.000 mol of I​2 ​at 448​o​C. The value of the equilibrium constant K​c​ is 50.5. What are the concentration of H​2​, I​2​, and HI in the flask at equilibrium?

H​2​ (g) + I​2​ (g) ⇌ 2HI (g)

1. Set up ICE chart with initial and equilibrium concentration of all species.

H​2​ (g) + I​2​ (g) ↔ 2HI (g) Initial

Change Equilibrium

2. Write the change in equilibrium as “x”, account for stoichiometry.

3. Calculate the equilibrium concentrations

4. Plug equilibrium concentrations into equilibrium expression setting it equal to the equilibrium constant.

5. Make equal to 0 and use the quadratic formula to solve for “x”. One of these will not make sense.

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6. Go back to the equilibrium line of the ICE chart to solve for the equilibrium concentrations.

For the equilibrium PCl​5​ (g) ⇌ PCl​3​ (g) + Cl​2​ (g) the equilibrium constant K​p​ has the value of .497 at 500 K. A gas cylinder at 500 K is charged with PCl​5​ (g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl​5​, PCl​3​, and Cl​2 ​at this temperature?

Initial Change Equilibrium

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Practice Free Response. (based on 2003B) After a 1.0 mole sample of HI​(g) ​is placed into an evacuated 1.0 L container at 700. K, the reaction represented below occurs. 2HI​(g)​ ⇌ H​2​(g) ​+ I​2​(g) ​The concentration of HI​(g)​ as a function of time is shown on the graph to the right.

(a) Write the expression for the equilibrium constant K​c​, for the reaction. (1)

(b) What is [HI] at equilibrium? (1)

(c) Determine the equilibrium concentrations of H​2​(g) ​and I​2​(g)​. (1)

(d) On the graph above, make a sketch that shows how the concentration of H​2​(g) changes over time. (2)

(e) Calculate the value of K​c​ for the reaction at 700. K. (1)

(f) At 1000 K, the value of Kc for the reaction is 2.6 x 10​-2 ​. In an experiment, 0.75 mole of HI​(g)​, 0.10 mole of H​2​(g)​ and 0.50 mole of I​2​(g)​ are placed in a 1.0 L container and allowed to reach equilibrium at 1000 K. Determine whether the equilibrium concentration of HI​(g) ​will be greater than, equal to, or less than the initial concentration of HI​(g)​. Justify your answer. (2)

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15.7 Le Chatelier's Principle

LO 6.3 The student can connect kinetics to equilibrium by using reasoning about

equilibrium, such as LeChatelier’s principle, to infer the relative rates of the forward and reverse reactions.

LO 6.8 The student is able to use Le Chatelier's principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium.

LO 6.9 The student is able to use Le Chatelier's principle to design a set of conditions that will optimize a desired outcome, such as product yield.

LO 6.10 The student is able to connect LeChatelier’s principle to the comparison of Q to K by explaining the effects of the stress on Q and K.

If a system at equilibrium is disturbed by a change in temperature, pressure If a system at equilibrium is disturbed by a change in temperature, pressure or the concentration of one of the components, the system will shift its

or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.

equilibrium position so as to counteract the effect of the disturbance.

Change in reactant or Product Concentration

SUMMARY -

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Effects of Volume and Pressure Changes

SUMMARY -

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Effect of Temperature Changes

SUMMARY -

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Effect of a catalyst

Sample Exercise 15.12

Sample Exercise 15.12​​ ​​Using Le Châtelier’s Principle to Predict Shifts inUsing Le Châtelier’s Principle to Predict Shifts in Equilibrium

Equilibrium

Consider the equilibrium

N​2​O​4​(​g​) ⇌ 2 NO​2​(​g​) ∆​H​° = 58.0 kJ

In which direction will the equilibrium shift when (a)(a)​​ N​2​O​4​ is added,

(b)(b)​​ NO​2​ is removed,

(c)(c)​​ the pressure is increased by addition of N​2​(​g​),

(d)(d)​​ the volume is increased,

(e)(e)​​ the temperature is decreased?

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Sample Exercise 15.13

Sample Exercise 15.13 ​​​​Predicting the Effect of Temperature on Predicting the Effect of Temperature on ​​KK (a)

(a) ​​Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction

N​2​(​g​) + 3 H​2​(​g​) ⇌ 2 NH​3​(​g​) (b)

(b) ​​Determine how the equilibrium constant for this reaction should change with temperature.

a) delta H = -92.38 kJ

b)

Sample Integrative Exercise

Sample Integrative Exercise ​​Putting Concepts TogetherPutting Concepts Together

At temperatures near 800 °C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H​2​:

C(​s​) + H​2​O(​g​) ⇌ CO(​g​) + H​2​(​g​)

The mixture of gases that results is an important industrial fuel called ​water gas​.

(a)(a)​​ At 800 °C the equilibrium constant for this reaction is ​K​p​ = 14.1. What are the equilibrium partial pressures of H​2​O, CO, and H​2​ in the equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H​2​O in a 1.00-L vessel?

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(b)(b)​​ What is the minimum amount of carbon required to achieve equilibrium under these conditions?

(c)(c)​​ What is the total pressure in the vessel at equilibrium?

(d)(d)​​ At 25 °C the value of ​K​p​ for this reaction is 1.7 × 10​–21​. Is the reaction exothermic or endothermic?

(e)(e)​​ To produce the maximum amount of CO and H​2​ at equilibrium, should the pressure of the system be increased or decreased?

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Practice Free Response.

Consider the reaction below that is allowed to reach equilibrium at a certain temperature.

CaCO​3​(s)​ ⇌ CaO​(s) ​+ CO​2​(g) ​ΔH = + 179 kJ/mol​rxn (a) Write the K​p​ expression for the reaction. (1)

(b) What are the optimum conditions of temperature and pressure, that would lead to a large % of CO​2​(g)​ in the equilibrium mixture? (2)

(b) In each case below, predict the shift in the equilibrium position that the applied stress would cause, and give an explanation in terms of Q and K to justify your choice. (9)

CaCO​3​(s)​ ⇌ CaO​(s)​ + CO​2​(g)​ ΔH = + 179 kJ/mol​rxn Stress applied to equilibrium

system (assume all other variables remain

unchanged)

Check only ONE box in

each row Explanation in terms of Q and K

Shift to

LHS No

change Shift to RHS

A. CO​2​(g) is added to the system

B. The volume of the container is

decreased

C. CaO(s) is removed from the system

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References

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