# Chapter 15 Chemical Equilibrium

Chapter 15 Chemical Equilibrium 15.1 The Concept of Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. The Concept of Equilibrium

As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow.

N2O4 (g) 2 NO2 (g) p.630 GIST: a) Which quantities are equal in a dynamic equilibrium? b) If the rate constant for the forward reaction is larger than the rate constant for the reverse reaction, will the constant be greater than or smaller than 1? 15.2 The Equilibrium Constant

The Equilibrium Constant Forward reaction: N2O4 (g) 2 NO2 (g) Rate Law: Rate = kf [N2O4]

The Equilibrium Constant Reverse reaction: 2 NO2 (g) N2O4 (g) Rate Law: Rate = kr [NO2]2

The Equilibrium Constant Therefore, at equilibrium Ratef = Rater kf [N2O4] = kr [NO2]2 Rewriting this, it becomes

kf [NO2]2 = kr [N2O4] The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes Keq =

kf kr [NO2]2 = [N2O4] The Equilibrium Constant Consider the generalized reaction

aA + bB cC + dD The equilibrium expression for this reaction would be [C]c[D]d Kc =

[A]a[B]b Writing Equilibrium Constant Expressions: Exercise 15.1 Write the equilibrium-constant expression for the following reactions: A) 2 O3(g) 3 O2(g)

B) 2 NO(g) + Cl2(g) 2 NOCl(g) C) Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) D) H2(g) + I2(g) 2HI(g)

E) Cd2+(aq) + 4 Br-(aq) CdBr42-(aq) How do you know when equilibrium has been reached in a chemical reaction? Equilibrium Can Be Reached from Either Direction

As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are. Kc = [NO2]2 [N2O4] Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

Equilibrium Can Be Reached from Either Direction It doesnt matter whether we start with N2 and H2 or whether we start with NH3: we will have the same proportions of all three substances at equilibrium. The Equilibrium Constant Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be

written (PCc) (PDd) Kp = (PAa) (PBb) Relationship Between Kc and Kp From the Ideal Gas Law we know that PV = nRT

Rearranging it, we get n P= RT V Relationship Between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc

and Kp becomes where Kp = Kc (RT)n n = (moles of gaseous product) - (moles of gaseous reactant) Ex 15.2 Converting between Kc and Kp In the synthesis of ammonia from nitrogen and hydrogen,

N2(g) + 3 H2(g) 2 NH3(g) Kc = 9.60 at 300C. Calculate Kp for this reaction at this temperature. Practice For the equilibrium 2 SO3(g) 2 SO2(g) + O2(g), Kc is 4.08 x 10-3 at 1000 K. Calculate the value for Kp. 15.3 Interpreting and Working with

Equilibrium Constants What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. What Does the Value of K Mean?

If K>>1, the reaction is product-favored; product predominates at equilibrium. If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.

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

Practice Exercise Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. [NO2]2 = 0.212 at 100 C N2O4 (g) 2 NO2 (g) Kc = [N2O4]

2 NO2 (g) N2O4 (g) Kc = [N2O4] = 4.72 at 100 C 2 [NO2] Manipulating Equilibrium

Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that2 number. [NO2] = 0.212 at 100 C N2O4(g) 2 NO2(g) Kc = [N2O4] 2 N2O4(g) 4 NO2(g) [NO2]4

2 Kc = = (0.212) at 100 C 2 [N2O4] Sample Exercise 15.4 Evaluating an Equilibrium Constant When an Equation is Reversed The equilibrium constant for the reaction of N2 30

with O2 to form NO equals Kc = 1 10 at 25 C: Using this information, write the equilibrium constant expression and calculate the equilibrium constant for the following reaction: Practice For the formation of NH3from N2 and H2,

N2(g) + 3 H2(g) 2 NH3(g), Kp = 4.34 x 10-3 at 300C. What is the value of Kp for the reverse reaction? Manipulating Equilibrium Constants 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. p.638 GIST

How does the magnitude of the equilibrium constant, Kp, for the reaction 2HI(g) H2(g) + I2(g) change if the equilibrium is written 6HI(g) 3H2(g) + 3H2(g)? To summarize

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. Exercise 15.5 Given the following information: HF(aq) H+(aq) + F-(aq) Kc = 6.8 x 10-4 H2C2O4(aq) 2 H+(aq) + C2O42-(aq) Kc = 3.8 x 10-6 determine the value of Kc for the reaction 2 HF(aq) + C2O42-(aq) 2 F-(aq) + H2C2O4(aq)

Practice Given that, at 700 K, Kp = 54.0 for the reaction H2(g) + I2(g) 2 HI(g) and Kp = 1.04 x 10-4 for the reaction N2(g) + 3 H2(g) 2 NH3(g), determine the value of Kp for the reaction 2 NH3(g) + 3 I2(g) 6 HI(g) + N2(g) at 700 K. 15.4 Heterogeneous Equilibria

The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by multiplying the density of the substance by its molar mass and both of these are constants at constant temperature. The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium

expression. PbCl2 (s) Pb 2+ (aq) Kc = [Pb2+] [Cl-]2

- + 2 Cl (aq) CaCO3 (s) CO2 (g) + CaO(s) As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same. P.639 & 641 GIST Write

the equilibrium constant expression for the evaporation of water H2O(l) H2O(g), in terms of partial pressures, Kp. Write the equilibrium expression for the following: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Exercise 15.6 Write the equilibrium constant expression for

Kc for each of the following reactions: A) CO2(g) + H2(g) CO(g) + H2O(l) B) SnO2(s) + 2 CO(g) Sn(s) + 2 CO2(g) Practice Write

the following equilibrium-constant expressions: A) Kc for Cr(s) + 3 Ag+(aq) Cr3+(aq) + 3 Ag(s) B) Kp for 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) Exercise 15.7 Each of the following mixtures was placed in a closed container and allowed to stand.

Which is capable of attaining the equilibrium CaCO3(s) CaO(s) + CO2(g): A) pure CaCO3 B) CaO and a CO2 pressure greater than the value of Kp

C) some CaCO3 and a CO2 pressure greater than the value of Kp D) CaCO3 and CaO Practice When added to Fe3O4(s) in a closed container,

which one of the following substances H 2(g), H2O(g), O2(g) will allow equilibrium to be established in the reaction 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)? 15.5 Calculating Equilibrium Constants Exercise 15.8 A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 C. The equilibrium mixture of gases

was analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate the equilibrium constant Kp for the reaction. 3H2(g) + N2(g) 2NH3(g) Practice An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25C: [HC2H3O2] = 1.65 x 10-2 M; [H+] =

5.44 x 10-4 M; and [C2H3O2-] = 5.44 x 10-4 M. Calculate the equilibrium constant, Kc, for the ionization of acetic acid at 25C: HC2H3O2(aq) H+(aq) + C2H3O2-(aq) Equilibrium Constant if we know the value of K, we can predict:

tendency of a reaction to occur if a set of concentrations could be at equilibrium equilibrium position, given initial concentrations Equilibrium Constant If you start a reaction with only reactants: concentration of reactants will decrease by a certain amount

concentration of products will increase by a same amount Exercise 15.9 A closed system initially containing 1.000 x 10 -3 M H2 and 2.000 x 10-3 M I2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448 C for the reaction taking place, which is

H2 (g) + I2 (g) 2 HI (g) What Do We Know? Initially [H2], M

[I2], M [HI], M 1.000 x 10-3 2.000 x 10-3 0 Change At equilibrium

1.87 x 10-3 Practice 15.9 Sulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g) 2SO2(g) + O2(g) Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200

atm. Calculate the value of Kp at 1000 K. 15.6 Applications of Equilibrium Constants The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. To calculate Q, one substitutes the initial

concentrations on reactants and products into the equilibrium expression. If Q = K, the system is at equilibrium. If Q > K, there is too much product, and the equilibrium shifts to the left. If Q < K, there is too much reactant, and the equilibrium

shifts to the right. Example For the synthesis of ammonia at 500C, the equilibrium constant is 6.0 x 10-2. Predict the direction the system will shift to reach equilibrium in the following case: N2(g) + 3H2(g) 2NH3(g) 2

[ NH 3 ] 2 K 6.0 10 3 [ N 2 ][ H 2 ] Example [NH3]0 = 1.0x10-3 M,

[N2]0=1.0x10-5 M [H2]0=2.0x10-3 M 3 2 [1.0 10 ] 7 Q 1.3 10 5 3 3 [1.0 10 ][2.0 10 ] Q > K so forms reactants, shifts to left

Exercise 15.10 At 448C the equilibrium constant Kc for the reaction H2(g) + I2(g) 2 HI(g) is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448C if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol H2, and 3.0 x 10-2 mol of I2 in a 2.00-L container. Practice

At 1000 K the value of Kp for the reaction 2 SO3(g) 2 SO2(g) + O2(g) is 0.338. Calculate the value for Qp and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are PSO3 = 0.16 atm; PSO2 = 0.41 atm; PO2 = 2.5 atm. Example

In the gas phase, dinitrogen tetroxide decomposes to gaseous nitrogen dioxide: N2O4(g) 2NO2(g) Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a T where KP = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm.

Calculate the equilibrium pressure of NO2. Example KP PNO2 2

PN 2O4 0.133 2 PNO2 K P PN 2O4 (0.133)(2.71) 0.360 PNO2 0.360 0.600 Exercise 15.11 For

the Haber process, N2(g) + 3 H2(g) 2 NH3(g), Kp = 1.45 x 10-5 at 500C. In an equilibrium mixture of the three gases at 500C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH 3 in this equilibrium mixture? Practice At 500 K, the reaction PCl5(g) PCl3(g) + Cl2(g) has Kp = 0.497. In an equilibrium mixture at

500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture? Example At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl3(g) and 8.70x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00x10-3 mol Cl2(g) was found in

the flask. PCl5(g) PCl3(g) + Cl2(g) Calculate the equilibrium concentrations of all the species and the value of K. Example PCl5(g)

PCl3(g) + Cl2(g) I 8.70x10-3 0.298

0 C -x +x +x E

8.70x10-3-x = (8.70-2.00) x10-3 = 6.70x10-3 0.298+x = 0.298+2.00x10-3 = 0.300 x = 2.00x10-3 -3

(0.300)(2.00x10 ) 2 K 8.96 10 -3 6.70x10 Exercise 15.12 A 1.000-L flask is filled with 1.000 mol of H2

and 2.000 mol of I2 at 448 C. The value of the equilibrium constant Kc for the reaction at 448 C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter? H2(g) + I2(g) 2HI(g) Practice For the equilibrium PCl5(g) PCl3(g) + Cl2(g), the equilibrium constant Kp has a value of

0.497 at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure 1.66 atm. What are the equilibrium pressures of PCl5, PCl3 and Cl2 at this temperature? Approximation example At a particular temperature, K = 4.0 x 10-7 for the reaction N2O4(g) 2NO2(g)

In an experiment, 1.0 mol N2O4 is placed in a 10.0 L vessel. Calculate the concentrations of N2O4 and NO2 when this reaction reaches equilibrium. 15.7 Le Chteliers Principle Le Chteliers Principle 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 equilibrium position so as to counteract the effect of the disturbance. The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance. The Haber Process If H2 is added to the system, N2 will be consumed and the two

reagents will form more NH3. The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid. Changing Concentration

system will shift away from the added component or towards a removed component Ex: N2 + 3H2 2NH3 if more N2 is added, then equilibrium position shifts to right if some NH3 is removed, then equilibrium position

shifts to right Change in Pressure adding or removing gaseous reactant or product is same as changing conc. adding inert or uninvolved gas increase the total pressure

doesnt effect the equilibrium position Change in Pressure changing the volume decrease V (increase in pressure) decrease in # gas molecules

shifts towards the side of the reaction with less gas molecules increase V (decrease in pressure) increase in # of gas molecules shifts towards the side of the reaction with more gas molecules Change in Temperature all other changes alter the concentration at

equilibrium position but dont actually change value of K value of K does change with temperature Change in Temperature if energy is added, the reaction will shift in direction that consumes energy treat energy as a

reactant: for endothermic reactions product: for exothermic reactions As4O6(s) + 6C(s) As4(g) + 6CO(g)

to right remove As4O6 to left remove As4

to right decrease volume to left add Ne gas no shift

P4(s) + 6Cl2(g) 4PCl3(l) decrease volume to right increase volume to left add P4

to right remove Cl2 to left add Kr gas no shift add PCl3

to left energy + N2(g) + O2(g) 2NO(g) endo or exo? H=181 kJ endothermic increase temp

to right increase volume no shift decrease temp to left The Effect of Changes in Temperature Co(H2O)62+(aq) + 4 Cl(aq)

CoCl4 (aq) + 6 H2O (l) N2(g) + 3H2(g) 2NH3(g) Section 15.7 GIST What happens to the equilibrium 2NO(g) + O2(g) 2NO2(g) if a)

O2 is added b) NO is removed What happens to the equilibrium 2SO2(g) + O2(g) 2SO3(g) if the volume of the system is increased? Use Le Chteliers principle to explain why the vapor pressure of a liquid increases with increasing temperature. Sample Exercise 15.13 Using Le Chteliers Principal to Predict shifts in Equilibrium

Consider the equilibrium In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased? Practice Exercise 15.13 For

the reaction PCl5(g) PCl3(g) + Cl2(g) H = 87.9 kJ, in which direction will the equilibrium shift when: A) Cl2 is removed B) the temperature is decreased C) the volume of the reaction system is increased

D) PCl3 is added Sample Exercise15.14 Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction N2(g) + 3 H2(g) 2 NH3(g)

Determine how the equilibrium constant for this reaction should change with temperature. Practice Using the thermodynamic data in Appendix C, determine the enthalpy change for the

reaction 2 POCl3(g) 2 PCl3(g) + O2(g) Use this result to determine how the equilibrium constant for the reaction should change with temperature. Catalysts Catalysts increase the rate of both the forward and reverse reactions.

Catalysts When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered. p.655 GIST: Does the addition of a catalyst have any effect on the position of an equilibrium?

Integrative Exercise At temperatures near 800C, steam pased over hot coke (a form of carbon from coal) reacts to form CO and H 2 C(s) + H2O(g) CO(g) + H2(g) producing an important industrial fuel called water gas.

A) At 800C the equilibrium constant is Kp = 14.1. What are the equilibrium partial pressures of H2O, CO and H2 in the equilibrium mixture if we start with solid carbon and 0.100 mol of H2O in a 1.00-L vessel? B) What is the minimum amount of carbon required to achieve equilibrium under these conditions? C) What is the total pressure in the vessel at equilibrium? D) At 25C the value of Kp for this reaction is 1.7 x 10-21. Is the reaction exothermic or endothermic?

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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