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. (Sec 13.1-13.7) 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. (Sec 13.4-13.6) LO 6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Châtelier’s Principle, to infer the relative rates of the forward and reverse reactions. (Sec 13.7) 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. (Sec 13.5) 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. (Sec 13.3, 13.5-13.6)

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. (Sec 13.5-13.6) 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. (Sec 13.2) LO 6.8 The student is able to use Le Châtelier’s Principle to predict the direction of the shift resulting from various possible stresses on a system at chemical equilibrium. (Sec 13.7) LO 6.9 The student is able to use Le Châtelier’s Principle to design a set of conditions that will optimize a desired outcome, such as product yield. (Sec 13.7) LO 6.10 The student is able to connect Le Châtelier’s Principle to the comparison of Q to K by explaining the effects of the stress on Q and K. (Sec 13.7)

AP Learning Objectives, Margin Notes and References 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.

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. In the figure above, equilibrium is finally reached in the third picture.

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. Once equilibrium is achieved, the amount of each reactant and product remains constant.

Equilibrium Is: On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. Macroscopically static  Microscopically dynamic Copyright © Cengage Learning. All rights reserved

Writing the Equation for an Equilibrium Reaction 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)

For the forward reaction N2O4(g) → 2 NO2(g) Comparing Rates For the forward reaction N2O4(g) → 2 NO2(g) The rate law is Rate = kf [N2O4] For the reverse reaction 2 NO2(g) → N2O4(g) The rate law is Rate = kr [NO2]2

The Meaning of Equilibrium Therefore, at equilibrium Ratef = Rater kf[N2O4] = kr[NO2]2 Rewriting this, it becomes the expression for the equilibrium constant, Keq.

Another Equilibrium— The Haber Process Consider the Haber Process, which is the industrial preparation of ammonia: N2(g) + 3 H2(g) ⇌ 2 NH3(g) The equilibrium constant depends on stoichiometry:

H2O(g) + CO(g) H2(g) + CO2(g) CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H2O(g) + CO(g) H2(g) + CO2(g) You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. The concentrations of each product will increase, the concentration of CO will decrease, and the concentration of water will be higher than the original equilibrium concentration, but lower than the initial total amount. Students may have many different answers (hydrogen goes up, but carbon dioxide in unchanged, etc.) Let them talk about this for a while – do not go over the answer until each group of students has come up with an explanation. This question also sets up LeChâtelier’s principle for later. Copyright © Cengage Learning. All rights reserved

H2O(g) + CO(g) H2(g) + CO2(g) CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H2O(g) + CO(g) H2(g) + CO2(g) You add more H2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. This is the opposite scenario of the previous slide. The concentrations of water and CO will increase. The concentration of carbon dioxide decreases and the concentration of hydrogen will be higher than the original equilibrium concentration, but lower than the initial total amount. Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References 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. 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.

Consider the following reaction at equilibrium: jA + kB lC + mD A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). l m [C] [D] K = [A] j [B] k Copyright © Cengage Learning. All rights reserved

Forward and reverse reaction rates, not reactant and product concentrations, that are equal. Relationship between magnitude of K and composition of an equilibrium mixture. Copyright © Cengage Learning. All rights reserved

a) ii<i<iii b)1.5, .11, 4

The Direction of the Chemical Equation and K The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction: Kc = = 0.212 at 100 C [NO2]2 [N2O4] N2O4(g) 2 NO2(g) ⇌ Kc = = 4.72 at 100 C [N2O4] [NO2]2 N2O4(g) 2 NO2(g) ⇌

Stoichiometry and Equilibrium Constants To find the new equilibrium constant of a reaction when the equation has been multiplied by a number, simply raise the original equilibrium constant to that power. Here, the stoichiometry is doubled; the constant is the squared! Kc = = 0.212 at 100 C [NO2]2 [N2O4] N2O4(g) 2 NO2(g) ⇌ Kc = = (0.212)2 at 100 C [NO2]4 [N2O4]2 4 NO2(g) 2 N2O4(g) ⇌

Consecutive Equilibria When two consecutive equilibria occur, the equations can be added to give a single equilibrium. The equilibrium constant of the new reaction is the product of the two constants: K3 = K1 × K2 Example 2 NOBr ⇌ 2 NO + Br2 K1 = 0.014 Br2 + Cl2 ⇌ 2 BrCl K2 = 7.2 2 NOBr + Cl2 ⇌ 2 NO + 2 BrCl K3 = K1 × K2 = 0.014 × 7.2 = 0.10

Conclusions About the Equilibrium Expression Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus Knew = (Koriginal)n. K values are usually written without units. Copyright © Cengage Learning. All rights reserved

Kc= (6.8x10-4)2 (1/3.8x10-6)= 0.12

Equilibrium position is a set of equilibrium concentrations. K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium concentrations. Copyright © Cengage Learning. All rights reserved

Equilibrium Can Be Reached from Either Direction N2O4(g) ⇌ 2 NO2(g) 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.

AP Learning Objectives, Margin Notes and References 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. 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.

Kc involves concentrations. Kp involves pressures. For a given reaction Kc is different from Kp . Copyright © Cengage Learning. All rights reserved

Consider the generalized reaction a A + b B ⇌ d D + e E The equilibrium expression for this reaction would be Also, since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

Example N2(g) + 3H2(g) 2NH3(g) Copyright © Cengage Learning. All rights reserved

Example N2(g) + 3H2(g) 2NH3(g) Equilibrium pressures at a certain temperature: Copyright © Cengage Learning. All rights reserved

Example N2(g) + 3H2(g) 2NH3(g) Copyright © Cengage Learning. All rights reserved

More with Gases and Equilibrium We can compare the equilibrium constant based on concentration to the one based on pressure. For gases, PV = nRT (the Ideal Gas Law). Rearranging, P = (n/V)RT; (n/V) is [ ].

The Relationship Between K and Kp The result is where R = 0.08206 L·atm/mol·K T = temperature (in Kelvin) If total number of moles of gases are the same on the left and right, then Kc=Kp Copyright © Cengage Learning. All rights reserved

Example N2(g) + 3H2(g) 2NH3(g) Using the value of Kp = 3.9 × 104, calculate the value of K at 35°C. Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References 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. 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.

Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N2(g) + 3H2(g) 2NH3(g) HCN(aq) H+(aq) + CN-(aq) Copyright © Cengage Learning. All rights reserved

Heterogeneous Equilibria Heterogeneous equilibria – involve more than one phase: 2KClO3(s) 2KCl(s) + 3O2(g) 2H2O(l) 2H2(g) + O2(g) Copyright © Cengage Learning. All rights reserved

Solutions and gases are always included in the expression. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids are constant. (Therefore solids and liquids are not included in the equilibrium expression!) Solutions and gases are always included in the expression. 2KClO3(s) 2KCl(s) + 3O2(g) Copyright © Cengage Learning. All rights reserved

The Decomposition of CaCO3— The equation for the reaction is CaCO3(s) ⇌ CaO(s) + CO2(g) This results in Kc = [CO2] and Kp = PCO2

CaCO3(s) ⇌ CaO(s) + CO2(g) Equilibirum will be established in all but c.

AP Learning Objectives, Margin Notes and References 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. 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. 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. 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. 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.

If K>>1, the reaction favors products; products predominate at equilibrium. If K<<1, the reaction favors reactants; reactants predominate at equilibrium. K indicates the tendency for a reaction to occur, but NOT the speed of the reaction!!!

The size of K is determined by thermodynamic factors such as E. It is important to note that the size of K and the time required to reach equilibrium are NOT directly related. The time to reach equilibrium depends on the reaction rate, which is determined by Ea. The size of K is determined by thermodynamic factors such as E. Figure 13.7 | (b) The reactants H2 and O2 have a strong tendency to form H2O.That is, H2O has lower energy than H2 and O2. However, the large activation energy Ea prevents the reaction at 25°C. In other words, the magnitude of K for the reaction depends on ΔE, but the reaction rate depends on Ea.

Deducing Equilibrium Concentrations Tabulate all known initial and equilibrium concentrations. For anything for which initial and equilibrium concentrations are known, calculate the change. Use the balanced equation to find change for all other reactants and products. Use initial concentrations and changes to find equilibrium concentration of all species. Calculate the equilibrium constant using the equilibrium concentrations.

An Example A closed system initially containing 1.000 × 10–3 M H2 and 2.000×10–3 M I2 at 448 °C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 × 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? [H2], M [I2], M [HI], M Initially 1.000 × 10–3 2.000 × 10–3 Change At equilibrium 1.87 × 10–3

[HI] Increases by 1.87 × 10−3 M [H2], M [I2], M [HI], M Initially 1.000 × 10–3 2.000 × 10–3 Change +1.87 × 10−3 At equilibrium 1.87 × 10–3

Stoichiometry tells us [H2] and [I2] decrease by half as much. [H2], M [I2], M [HI], M Initially 1.000 × 10–3 2.000 × 10–3 Change −9.35 × 10−4 +1.87 × 10–3 At equilibrium 1.87 × 10–3

We can now calculate the equilibrium concentrations of all three compounds. [H2], M [I2], M [HI], M Initially 1.000 × 10–3 2.000 × 10–3 Change –9.35 × 10–4 +1.87 × 10–3 At equilibrium 6.5 × 10−5 1.065 × 10−3 1.87 × 10–3

And, therefore, the equilibrium constant… Kc = [HI]2 [H2] [I2] = 51 = (1.87 × 10–3)2 (6.5 × 10–5)(1.065 × 10–3)

Is a Mixture in Equilibrium? Which Way Does the Reaction Go? To answer these questions, we calculate the reaction quotient, Q. Q looks like the equilibrium constant, K, but the values used to calculate it are the current conditions, not necessarily those for equilibrium. To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression.

If Q = K, the reaction is in equilibrium. Comparing Q and K Nature wants Q = K. If Q = K, the reaction is in equilibrium.

If Q < K, nature will make the reaction proceed to products If Q < K, nature will make the reaction proceed to products. Consuming reactants and forming products, to attain equilibrium If Q > K, nature will make the reaction proceed to reactants. Consuming products and forming reactants, until equilibrium is achieved

a) Left b) no shift c) right

AP Learning Objectives, Margin Notes and References 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. 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. 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. 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.

Solving Equilibrium Problems Write the balanced equation for the reaction. Write the equilibrium expression using the law of mass action. List the initial concentrations. Calculate Q, and determine the direction of the shift to equilibrium. Copyright © Cengage Learning. All rights reserved

Solving Equilibrium Problems 5) Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. Check your calculated equilibrium concentrations by making sure they give the correct value of K. Copyright © Cengage Learning. All rights reserved

Calculating Equilibrium Concentrations If you know the equilibrium constant, you can find equilibrium concentrations from initial concentrations and changes (based on stoichiometry). You will set up a table similar to the ones used to find the equilibrium concentration, but the “change in concentration” row will simple be a factor of “x” based on the stoichiometry.

Find Q to determine shift. Square root both sides Reactants = 0 Find Q to determine shift. Square root both sides Reactants = 0.472M products = 5.056M

An Example A 1.000 L flask is filled with 1.000 mol of H2(g) and 2.000 mol of I2(g) at 448 °C. Given a Kc of 50.5 at 448 °C, what are the equilibrium concentrations of H2, I2, and HI? H2(g) + I2(g) ⇌ 2 HI(g) initial concentration (M) 1.000 2.000 change in concentration (M) –x +2x equilibrium concentration (M) 1.000 – x 2.000 – x 2x

Example (continued) Set up the equilibrium constant expression, filling in equilibrium concentrations from the table. Solving for x is done using the quadratic formula, resulting in x = 2.323 or 0.935. See appendix A1.4

Example (completed) Since x must be subtracted from 1.000 M, 2.323 makes no physical sense. (It results in a negative concentration!) The value must be 0.935. So [H2]eq = 1.000 – 0.935 = 0.065 M [I2]eq = 2.000 – 0.935 = 1.065 M [HI]eq = 2(0.935) = 1.87 M This problem could have been given with pressures instead of concentrations.

If K <<1, then can simplify the problem by assuming the reaction does not proceed significantly to the left. You may delete the –X term in the denominator (from the reactant). Once you solve for X, you should confirm that the value of X is <5% of the initial amount of reactant to confirm the Mount of reactant lost compared to what you started with is insignificant. See example page 631

CONCEPT CHECK! A 1.00 mol sample of N2O4(g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N2O4(g) 2NO2(g) K = 4.00 × 10-4 Calculate the equilibrium concentrations of: N2O4(g) and NO2(g). Concentration of N2O4 = 0.097 M Concentration of NO2 = 6.32 × 10-3 M Concentration of N2O4 = 0.097 M Concentration of NO2 = 6.32 x 10-3 M (without quadratic) or Concentration of N2O4 = 0.094 M Concentration of NO2 = 6.22 x 10-3 M (with quadratic) Use this problem to discuss the 5% allowable error (so we can assume x is negligible). Make sure the students understand we are NOT saying x is equal to zero but that x is negligible. This is a subtle but very important point. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References 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. LO 6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Châtelier’s Principle, to infer the relative rates of the forward and reverse reactions. LO 6.8 The student is able to use Le Châtelier’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 Châtelier’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 Le Châtelier’s Principle to the comparison of Q to K by explaining the effects of the stress on Q and K. Additional AP References LO 6.9 (see APEC #13, “Le Châtelier’s Principle”)

Studying the factors that control the position of equilibrium. If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Copyright © Cengage Learning. All rights reserved

Although max NH3 production is favored at low temperature and high pressure, this is not ideal. The reaction rate to obtain equilibrium is too slow. This stresses the need to study both the thermodynamics and kinetics of a chemical reaction.

Effects of Changes on the System Concentration (reactant or product): The system will shift away from the added component. If a component is removed, the will shift to replace the removed component. (K not affected by change in concentration!!) Effects of Changes on the System Figure 13.9 | (a) The initial equilibrium mixture of N2, H2, and NH3. (b) Addition of N2. (c) The new equilibrium position for the system containing more N2 (due to addition of N2), less H2, and more NH3 than the mixture in (a). Copyright © Cengage Learning. All rights reserved

To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Effects of Changes on the System 2. Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product). Copyright © Cengage Learning. All rights reserved

Is the reaction endothermic or exothermic as written? That matters! Change in Temperature Is the reaction endothermic or exothermic as written? That matters! Endothermic: Heats acts like a reactant; adding heat drives a reaction toward products. Exothermic: Heat acts like a product; adding heat drives a reaction toward reactants.

An Endothermic Equilibrium

An Exothermic Equilibrium The Haber Process for producing ammonia from the elements is exothermic. One would think that cooling down the reactants would result in more product. However, the activation energy for this reaction is high! This is the one instance where a system in equilibrium can be affected by a catalyst!

Catalysts increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. Activation energy is lowered, allowing equilibrium to be established at lower temperatures.

Effects of Changes on the System Pressure: (K not affected by a change in pressure) The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. Addition of inert gas does not affect the equilibrium position. Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas. Copyright © Cengage Learning. All rights reserved