COLLEGE PHYSICS CHEMISTRY 2e Chapter # Chapter Title Chapter 13

COLLEGE PHYSICS CHEMISTRY 2e Chapter # Chapter Title Chapter 13 FUNDAMENTAL EQUILIBRIUM CONCEPTS PowerPoint Image Slideshow

Chapter 13 Outline 13.1 Chemical Equilibrium 13.2 Equilibrium Constants 13.3 Shifting Equilibria: Le Châtelier’s Principle 13.4 Equilibrium Calculations

Figure 13.1 Transport of carbon dioxide in the body involves several reversible chemical reactions, including hydrolysis and acid ionization (among others).

Learning Objectives 13.1 Chemical Equilibria Describe the nature of equilibrium systems Explain the dynamic nature of a chemical equilibrium

Chemical Equilibria Reactions with an appreciable reverse reaction are often best represented as an equilibrium.

Chemical Equilibria N2O4 is colorless. NO2 is brown. When N2O4 is placed in a closed container at 100 C, a reddish-brown color develops due to the formation of NO2. The forward reaction occurs. As NO2 builds up, it can react to form N2O4 The reverse reaction occurs. At equilibrium, the amounts of reactants and products stop changing.

Figure 13.2 (a) A sealed tube containing colorless N2O4 darkens as it decomposes to yield brown NO2. (b) Changes in concentration over time as the decomposition reaction achieves equilibrium. (c) At equilibrium, the forward and reverse reaction rates are equal.

Establishment of Equilibrium N 2 O4 (𝑔) 2 N O2 (𝑔) The rate of the forward reaction starts out fast, but slows down as the concentration of N2O4 decreases. The rate of the reverse reaction starts out slow, but speeds up as the concentration of NO2 increases. Once the rates of both reactions are equal, equilibrium is established. No further changes in concentration of either gas occurs unless the temperature or the volume of the container is changed.

Common Equilibrium Misconceptions 1) The amount of reactants and products are ordinarily not equal to each other at equilibrium. 2) Although the amount of reactants and products remains constant at equilibrium, the system is not static at equilibrium. Chemical equilibrium is a dynamic process.

Figure 13.4 A sealed tube containing an equilibrium mixture of liquid and gaseous bromine. (credit: http://images-ofelements.com/bromine.php)

Learning Objectives 13.2 Equilibrium Constants Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures Relate the magnitude of an equilibrium constant to properties of the chemical system

Equilibrium Constants The symbol , placed between reactants and products, is used to designate reversible reactions. The reaction quotient, Q, allows us to mathematically express the concentrations (or pressures) of reactants and products present at any point in a reversible reaction.

The Reaction Quotient, Q Consider the general reaction: 𝑚A 𝑛B 𝑥C 𝑦 D A, B, C, D are either gases or aqueous species. m, n, x, y are the coefficients in the balanced equation. When using amounts expressed in concentration, the reaction quotient is called Qc. 𝑄𝐶 ¿¿

The Concentration Reaction Quotient, Qc All concentrations must be expressed in Molarity. Product concentrations are in the numerator (multiplied together). Reactant concentrations are in the denominator (multiplied together). Each concentration is raised to the power of its coefficient from the balanced equation. Solids are always omitted from the expressions for Q and K!

Write the expression for the Qc for each equation. (b) 4NH3 (g) 5O2 (g) 4NO(g) 6H2 O(g) (c) N2 O4 (g) 2NO2 (g) (d) CO2 (g) H2 (g) CO(g) H2O(g) (e) NH4Cl(s) NH3 (g) HCl(g) (f) 2Pb(NO3) 2 (s) 2PbO(s) 4NO2 (g) O2 (g)

The Value of the Reaction Quotient, Q The numeric value of Qc for a given reaction can vary prior to equilibrium. The value of Qc depends on the concentration of products and reactants present at that particular moment. We can calculate Qc at any point in a reaction. We will often calculate Qc at the start of the reaction using initial concentrations.

Figure 13.5 Changes in concentrations and Qc for a chemical equilibrium achieved beginning with a (a) mixture of reactants only and (b) products only.

The Equilibrium Constant, K The value of Q when the reaction is at equilibrium is called the equilibrium constant (K). aA bB cC dD 𝑄𝐶 at equilibrium 𝐾 𝑐 ¿¿ Be careful not confuse with the kinetic rate constant (k).

Q and K aA bB cC dD 𝑄𝐶 ¿ ¿ The equilibrium constant (K) has the same form as the reaction quotient (Q). For K, the concentrations must be those at equilibrium. For Q, the concentrations can be those at any point in the reaction, not necessarily when at equilibrium.

The Equilibrium Constant, K The value of the equilibrium constant is independent of the starting amounts of reactants and products. The value of the equilibrium constant is dependent on the temperature of the system. ONLY a change in temperature can change the value of the equilibrium constant! K and Q are unitless values. The magnitude of an equilibrium constant indicates the extent of a reaction.

The Magnitude of the Equilibrium Constant If K is very small, the mixture contains mostly reactants at equilibrium. If K is very large, the mixture contains mostly products at equilibrium. The value of K gives no indication as to whether the reaction is fast or slow.

Q, K, and the Direction of Reaction aA bB cC dD 𝑄𝐶 ¿¿ A system that is not at equilibrium will proceed in the direction that establishes equilibrium. By comparing Q to K, it is possible to determine which direction the system will proceed to achieve equilibrium.

Q, K, and the Direction of Reaction aA bB cC dD 𝑄𝐶 ¿¿ When Q K Reaction must shift FORWARD When Q K Reaction must shift BACKWARD When Q K Reaction is at equilibrium, and will maintain Constant concentration

Changes in Reactant and Product Concentrations CO (𝑔) H 2 O(𝑔) CO2 (𝑔) H2 (𝑔) There are many ways to approach the equilibrium state. Mixture 1: Start with just CO and H2O. Mixture 2: Start with just CO2 and H2. Mixture 3: Start with CO, H2O, CO2, and H2. The value of the equilibrium constant is independent of the starting amounts of the reactants and products. If the reaction initially contained 4.0 M CO, and 2.0 M H2 only, what would happen?

Homogenous Equilibrium A homogenous equilibrium is one in which all of the reactants and products are present in the same phase. Most commonly are either liquid or gaseous phases. Reaction quotients include concentration or pressure terms only for gaseous and solute species. For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (Kc) or partial pressures (Kp) of the reactants and products.

Homogenous Equilibria HF(𝑎𝑞) H2 O(𝑙) H 3 O – ¿(𝑎𝑞 ) F (𝑎𝑞)¿ C 2 H 6 (𝑔) C2 H 4 (𝑔) H2 (𝑔) 𝐾𝑝 ( 𝑃𝐶 2 𝐻4 )( 𝑃 𝐻 ) 𝑃𝐶 2 2 𝐻6 NOTE: In the Kc, we MUST use brackets. [ ] not ( ) In the Kp we must NOT use brackets!!!

Kc and Kp For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (Kc) or partial pressures (Kp) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity: 𝑃𝑉 𝑛𝑅𝑇

Kc and Kp The relationship between Kc and Kp: 𝐾 𝑝 𝐾 𝑐 ¿ Δn is the change in the number of moles of gas. R 0.08206 L atm/mol K T is temperature in Kelvin. Solve the following problem: H2(g) Cl2(g) 2 HCl(g) At 1000 K, the Kc for this reaction is found to be 6.00 x 106 . What is the value of the Kp at that T?

Did you get the answer, 6.00 x 106 ? Did you waste a lot of time on it? If the sum of the exponents in the numerator the sum of the exponents in the denominator, then n is 0. In that case Kc Kp.

Heterogeneous Equilibrium Some reaction mixtures contain reactants and products that are in two or more different phases. These systems are called heterogeneous equilibria. N H 4 Cl (𝑠) N H3 (𝑔) HCl(𝑔) Pure solids and pure liquids do not appear in the K expression. The position of equilibrium is independent of the amount of solid or liquid present, as long as at least some is present in the reaction mixture.

Learning Objectives 13.3 Shifting Equilibria: Le Châtelier’s Principle Describe the ways in which an equilibrium system can be stressed Predict the response of a stressed equilibrium using Le Chatelier’s principle

Shifting Equilibria: Le Châtelier’s Principle Le Châtelier’s Principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. At equilibrium Q K. The disturbance causes a change in Q. The reaction will shift to re-establish Q K. In the case of a temperature change, the disturbance changes the value of K. The direction of that change depends on whether the reaction is exothermic or endothermic.

Adding or Removing a Reactant or Product If a chemical equilibrium is disturbed by adding a reactant or product, the system will proceed in the direction that consumes part of the added species. If a chemical equilibrium is disturbed by removing a reactant or product, the system will proceed in the direction that restores part of the removed species. The system responds in the way that restores equilibrium and therefore allows Q K again. If what is added or removed is a SOLID, the reaction does not shift at all!!! However, while the amount of solid does not affect the equilibrium, any shift in equilibrium DOES change the amount of solid.

Adding or Removing a Pure Liquid or Solid Adding or removing a pure liquid or solid has no effect on the system unless all of the liquid or solid is removed. This is because pure liquids and solids do not appear in the equilibrium expression.

Temperature and the Equilibrium Constant Changing the temperature of a system results in a change in the value of the equilibrium constant. If the forward reaction is exothermic, then K decreases as T increases. If the forward reaction is endothermic, then K increases as T increases. X K T othermic Silly mnemonic, on the “X” in “exo.” As T goes up, K goes down. JUST KNOW IT!!

Summarizing factors that shift equilibria. 1. Adding more reactant (but not solids!) will cause the reaction to shift so as to use that reactant up. The product concentration will INCREASE, as reactant decreases. Reaction shifts FORWARD. K does NOT change. Adding more product will have the OPPOSITE impact. Adding more reactant will Decrease the value of Q. When Q, K, reactions shift so as to produce product. 2. Removing reactant has the OPPOSITE effect. Reaction shifts left. 3. Adding product INCREASES Q. Reaction shifts to form more reactant. 4. Removing product has the same effect as adding reactant. (why?)

Effect of Temperature 1. An increase in temperature will change K. It will INCREASE K for an endo, and DECREASE K for an exo. 2. It will shift the reaction so as to favor whichever direction is ENDOthermic. 3. A decrease in temperature will also change K. It will Increase K for an exo, and decrease K for an endo. 4. It will shift the reaction so as to favor the EXOthermic direction. Effect of a change in volume. A decrease in volume will shift the equilibrium so as to produce FEWER moles of gas. It will NOT change K.

N2(g) 3 H2(g) 2 NH3(g) DH -92kJ/mol rxn 1. If additional hydrogen is added to the mixture at equilibrium, A) the conc. Of N2(g) B) the conc. of NH3(g) C) the value of Keq 2. The volume of the system is decreased. A) the amount of N2(g) B) the amount of NH3(g) C) the value of Keq 3. How are those 3 variables affected if T

Catalysts do not Affect Equilibrium A catalyst speeds up the rate of a reaction. For reversible reactions, catalysts increase the rates of the forward and reverse reactions. Result: A catalyst causes the system to reach equilibrium more quickly. But a catalyst does not affect the equilibrium concentrations or value of the equilibrium constant. And what is the ONLY thing that will change that value?

Learning Objectives 13.4 Equilibrium Calculations Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches

Equilibrium Calculations Types of Equilibrium Calculations: Calculation of an equilibrium constant Calculation of missing equilibrium concentration or partial pressure Calculation of equilibrium concentrations (or partial pressures) from initial concentrations (or partial pressures)

Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction: N2O4(g) 2NO2(g) 2NO2(g) N2O4(g) [NO2]2 Kc 0.212 at 100 C [N2O4] Kc [N2O4] 4.72 at 100 C 2 [NO2] 2012 Pearson Education, Inc.

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 that number: N2O4(g) 2N2O4(g) 2NO2(g) [NO2]2 Kc 0.212 at 100 C [N2O4] 4NO2(g) [NO2]4 2 Kc (0.212) at 100 C 2 [N2O4] 2012 Pearson Education, Inc.

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. Practice Exercise 1. For the reaction For , Kp 4.34 10 3 at 300 C .What is the value of Kp for the reverse reaction? 2012 Pearson Education, Inc.

Sample Exercise 15.5 Combining Equilibrium Expressions Given the reactions determine the value of Kc for the reaction Solution Analyze We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two. Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equations to come up with the steps that will add to give us the desired equation.

Sample Exercise 15.5 Combining Equilibrium Expressions Continued Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives Now we have two equations that sum to give the net equation, and we can multiply the individual Kc values to get the desired equilibrium constant.

Sample Exercise 15.5 Combining Equilibrium Expressions Continued Practice Exercise Given that, at 700 K, Kp 54.0 for the reaction , determine the value of Kp for the reaction at 700 K. Answer: and Kp 1.04 10 4 for the reaction 6 HI(g) N2(g)

1, Answer: 2.30 102 2.

Solving Equilibrium Problems. All equilibrium problems require the correct expression for K (or Q). They can be divided into 4 types. Two are very easy, the third more complicated, the fourth more complicated, but unfortunately, more common. 1. All concentrations (or pressures ) are given, find K. Simply substitute the given values into the equilibrium expression.

Sample Exercise 15.8 Calculating K When All Equilibrium Concentrations Are Known After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium at 472 C, it is 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 Solution Analyze We are given a balanced equation and equilibrium partial pressures and are asked to calculate the value of the equilibrium constant. Plan Using the balanced equation, we write the equilibrium-constant expression. We then substitute the equilibrium partial pressures into the expression and solve for Kp. Solve Practice Exercise An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 C: 2 4 4 [CH3COOH] 1.65 10 M; [H ] 5.44 10 M; and [CH3COO ] 5.44 10 M. Calculate the equilibrium constant Kc for the ionization of acetic acid at 25 C. The reaction is

K is given. All but one of the equilibrium concentrations are given, find the other one. Type 2.

Type 2. Calculating Equilibrium Concentrations when K and all but one of them is known For the Haber process, , Kp 1.45 10 5 at 500 C. In an equilibrium mixture of the three gases at 500 C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture? Solution Analyze We are given an equilibrium constant, Kp, and the equilibrium partial pressures of two of the three substances in the equation (N2 and H2), and we are asked to calculate the equilibrium partial pressure for the third substance (NH3). Solve We tabulate the equilibrium pressures: Because we do not know the equilibrium pressure of NH3, we represent it with x. At equilibrium the pressures must satisfy the equilibrium-constant expression: We now rearrange the equation to solve for x: Plan We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that we know. Then we can solve for the only unknown in the equation.

Practice Exercise 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? Answer: 1.22 atm

TYPE 3 Finding K, when all the initial concentrations are known, and ONE equilibrium concentration is given. It is necessary to set up a reaction table, sometimes called an “ICE chart” to find ALL the equilibrium concentrations. THEN, you can put those values into the equilibrim equation to find K. (See also, problem 13.7 in text)

An Equilibrium Problem 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) 2012 Pearson Education, Inc. 2HI(g)

Below is an ICE chart for this reaction. I normally include the coefficients. Initially [H2], M [I2], M 2 [HI], M 1.000 x 10 3 2.000 x 10 3 0 Change At equilibrium 1.87 x 10 3

[HI] Increases by 1.87 x 10 3 M Initially [H2], M [I2], M 2 [HI], M 1.000 x 10 3 2.000 x 10 3 0 Change 1.87 x 10 3 At equilibrium 1.87 x 10 3 2012 Pearson Education, Inc.

We can now calculate the equilibrium concentrations of all three compounds [H2], M [I2], M 2[HI], M Initially 1.000 x 10 3 2.000 x 10 3 0 Change 9.35 x 10 4 9.35 x 10 4 1.87 x 10 3 At equilibrium 6.5 x 10 5 1.065 x 10 3 1.87 x 10 3 2012 Pearson Education, Inc.

and, therefore, the equilibrium constant: [HI]2 Kc [H2] [I2] (1.87 x 10 3)2 (6.5 x 10 5)(1.065 x 10 3) 51 2012 Pearson Education, Inc.

TYPE FOUR. K is known, initial conc. are known, find one or more FINAL concentration. Example 13.9 in text, with a different given value of K. PCl5(g) PCl3(g) Cl2(g) At a certain temperature, Kc is 2.1 x 10-4 for this reaction. If the initial [PCl5(g) is 1.00 molar, with no products present, what are the equilibrium concentrations of the three components?

x Page 746 in text, example 13.9 PCl5(g) PCl3(g) Cl2(g) Let x change in PCl5 Initial 1.00 Change -x Equilibrium 1.00-x 0 x x 0 x x Plugging into the expression for K gives us 2 /(1-x) 2.1 x 10-4 Quadratic. BUT . x Since K is small, very little product is formed. Let’s assume that x is very small compared to 1.00. We are saying 1-x 1. X 0.014 So at equilibrium, the concentrations are 0.99, 0.014, and 0.014. Were we right to say that 1-x 1? The usual guideline is that “x” should not change the . of reactant by more than 5 %. If you look at the problem in the text, the K was much higher, so this “simplifying assumption” produces a small error