# Chapter 17 Equilibrium: The Extent of Chemical Reactions

Chapter 17 Equilibrium: The Extent of Chemical Reactions 17-1 Equilibrium: The Extent of Chemical Reactions 17.1 The Equilibrium State and the Equilibrium Constant 17.2 The Reaction Quotient and the Equilibrium Constant 17.3 Expressing Equilibria with Pressure Terms: Relation between Kc and Kp 17.4 Comparing Q and K to Determine Reaction Direction 17.5 How to Solve Equilibrium Problems 17.6 Reaction Conditions and Equilibrium: Le Chteliers Principle 17-2 The Equilibrium State All reactions are reversible and under suitable conditions will reach a state of equilibrium. At equilibrium, the concentrations of products and reactants no longer change because the rates of the forward and reverse reactions are equal.

At equilibrium: rateforward = ratereverse Chemical equilibrium is a dynamic state because reactions continue to occur, but because they occur at the same rate, no net change is observed on the macroscopic level. 17-3 Figure 17.1 Reaching equilibrium on the macroscopic and molecular levels. N2O4(g) 17-4 2NO2(g) The Equilibrium Constant Consider the reaction N2O4(g) 2NO2(g) At equilibrium ratefwd = raterev

2 so k[N2O4]eq = k[NO2] eq 2 kfwd [NO ] 2 eq = then krev [N2O4]eq The ratio of constants gives a new constant, the equilibrium constant: 2 kfwd = [NO2] eq K= krev [N2O4]eq

17-5 K and the extent of reaction K reflects a particular ratio of product concentrations to reactant concentrations for a reaction. K therefore indicates the extent of a reaction, i.e., how far a reaction proceeds towards the products at a given temperature. A small value for K indicates that the reaction yields little product before reaching equilibrium. The reaction favors the reactants. A large value for K indicates that the reaction reaches equilibrium with very little reactant remaining. The reaction favors the products. 17-6 Figure 17.2 The range of equilibrium constants. small K The reaction mixture contains mostly

reactants. large K The reaction mixture contains mostly products. 17-7 intermediate K The Reaction Quotient Q For the general reaction aA + bB [C]c[D]d the reaction quotient Q = [A]a[B]b cC + dD Q gives the ratio of product concentrations to reactant concentrations at any point in a reaction. At equilibrium: Q = K For a particular system and temperature, the same equilibrium state is attained regardless of starting

concentrations. The value of Q indicates how close the reaction is to equilibrium, and in which direction it must proceed to reach equilibrium. 17-8 Figure 17.3 17-9 The change in Q during the N2O4-NO2 reaction. Table 17.1 Initial and Equilibrium Concentration Ratios for the N2O4-NO2 System at 200C (473 K) Initial Expt [N2O4] [NO2]

Equilibrium Q, [NO2]2 [N2O4]eq [NO2]eq [N2O4] K, [NO2] 2eq [N2O4]eq 1 0.1000 0.0000 0.0000 0.0357 0.193

10.4 2 0.0000 0.1000 0.000924 0.0982 10.4 3 0.0500 0.0500

0.0500 0.00204 0.146 10.4 4 0.0750 0.0250 0.0833 0.00275 0.170 10.5 17-10

Sample Problem 17.1 Writing the Reaction Quotient from the Balanced Equation PROBLEM: Write the reaction quotient, Qc, for each of the following reactions: NO2(g) + O2(g) (a) The decomposition of dinitrogen pentaoxide, N2O5(g) CO2(g) + H2O(g) (b) The combustion of propane, C3H8(g) + O2(g) PLAN: We balance the equations and then construct the reaction quotient. SOLUTION: (a) 2N2O5(g) (b) C3H8(g) + 5 O2(g) 17-11 4NO2(g) + O2(g)

Qc = [NO2]4[O2] [N2O5]2 3 CO2(g) + 4 H2O(g) Qc = [CO2]3[H2O]4 [C3H8][O2]5 Forms of K and Q For an overall reaction that is the sum of two more individual reactions: Qoverall = Q1 x Q2 x Q3 x .. and Koverall = K1 x K2 x K3 x The form of Q and K depend on the direction in which the balanced equation is written: Qc(rev) = 17-12 1 Qc(fwd)

Kc(rev) = 1 Kc(fwd) Forms of K and Q If the coefficients of a balanced equation are multiplied by a common factor, c d [C] [D] Q' = Qn = [A]a[B]b 17-13 n and K' = Kn Sample Problem 17.2

Writing the Reaction Quotient and Finding K for an Overall Reaction PROBLEM: Nitrogen dioxide is a toxic pollutant that contributes to photochemical smog. One way it forms is through the following sequence: (1) N2(g) + O2(g) 2NO(g) K c1 = 4.3x10-25 (2) 2NO(g) + O2(g) 2NO2(g) K c2 = 6.4x109 (a) Show that the overall Qc for this reaction sequence is the same as the product of the Qc's of the individual reactions. (b) Given that both reactions occur at the same temperature, find Kc for the overall reaction. PLAN: We first write the overall reaction by adding the individual reactions and then write the overall Qc. Then we write the Qc

for each step. We add the steps and multiply their Qc's, canceling common terms. We can then calculate the overall Kc. 17-14 Sample Problem 17.2 SOLUTION: (a) (1) N2(g) + O2(g) (2) 2NO(g) + O2(g) N2(g) + 2O2(g) Qc(overall) = 2NO(g) K c1 = 4.3x10-25 2NO2(g) K c2 = 6.4x109 2NO2(g)

[NO2]2 [N2][O2]2 [NO]2 Qc1 = [N2][O2] [NO2]2 Qc2 = [NO2]2 [NO]2[O2] [NO2]2 [NO] = Qc1 x Qc2 = x [N2][O2] [N2] [O2] 2 [NO]2[O2] 2

(b) 17-15 Kc = Kc1 x Kc2 = (4.3x10-25) x (6.4x109) = 2.8x10-15 Sample Problem 17.3 Finding the Equilibrium Constant for an Equation Multiplied by a Common Factor PROBLEM: For the ammonia formation reaction, the reference equation is N2(g) + 3H2(g) 2NH3(g) Kc is 2.4x10-3 at 1000 K. What are the values of Kc for the following balanced equations: (a) 13 N2(g) +H2(g)

2 NH (g) 3 3 (b) NH3(g) 1 3 H (g) N (g) + 2 2 2 2 PLAN: We compare each equation with the reference equation to see how the direction and/or coefficients have changed. SOLUTION: (a) The reference equation is multiplied by , so Kc(ref) will be to the power. Kc = [Kc(ref)]1/3 = (2.4x10-3)1/3 = 0.13

17-16 Sample Problem 17.3 (b) This equation is one-half the reverse of the reference equation, so Kc is the reciprocal of Kc(ref) raised to the power. Kc = [Kc(ref)]-1/2 = (2.4x10-3)-1/2 17-17 = 20 K and Q for hetereogeneous equilibrium A hetereogeneous equilibrium involves reactants and/ or products in different phases. CaCO3(s) CaO(s) + CO2(g) A pure solid or liquid always has the same concentration, i.e., the same number of moles per liter of solid or liquid. The expressions for Q and K include only species whose concentrations change as the reaction approaches equilbrium.

Pure solids and liquids are omitted from the expression for Q or K.. For the above reaction, Q = [CO2] 17-18 Figure 17.4 The reaction quotient for a heterogeneous system depends only on concentrations that change. solids do not change their concentrations 17-19 Table 17.2 17-20 Ways of Expressing Q and Calculating K Expressing Equilibria with Pressure Terms Kc and Kp

K for a reaction may be expressed using partial pressures of gaseous reactants instead of molarity. The partial pressure of each gas is directly proportional to its molarity. 2NO(g) + O2(g) 2NO2(g) 2 Qp = P NO2 P 2NO x PO Qc = [NO2]2 [NO]2 [O2] 2 Kp = Kc (RT)n(gas) If the amount (mol) of gas does not change in the reaction,

ngas = 0 and Kp = Kc. 17-21 Sample Problem 17.4 Converting Between Kc and Kp PROBLEM: A chemical engineer injects limestone (CaCO3) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO2 from the gas and forms gypsum (CaSO42H2O). Find Kc for the following reaction, if CO2 pressure is in atmospheres. CaCO3(s) CaO(s) + CO2(g) K p = 2.1x10-4 (at 1000 K) PLAN: We know Kp, so to convert between Kp, we must first determine ngas from the balanced equation before we calculate Kc. Since pressure is in atmospheres, R = 0.0821 Latm/molK. SOLUTION: ngas = 1 - 0 since there is one gaseous product and no gaseous reactants. Kp = Kc(RT)n Kc = Kp/(RT)n = (2.1x10-4)(0.0821x1000)-1 = 2.6x10-6

17-22 Determining the Direction of Reaction The value of Q indicates the direction in which a reaction must proceed to reach equilibrium. If Q < K, the reactants must increase and the products decrease; reactants products until equilibrium is reached. If Q > K, the reactants must decrease and the products increase; products reactants until equilibrium is reached. If Q = K, the system is at equilibrium and no further net change takes place. 17-23 Figure 17.5 Reaction direction and the relative sizes of Q and K. Q>K Q

17-24 Q=K Sample Problem 17.5 Using Molecular Scenes to Determine Reaction Direction PROBLEM: For the reaction A(g) B(g), the equilibrium mixture at 175C is [A] = 2.8x10-4 M and [B] = 1.2x10-4 M. The molecular scenes below represent mixtures at various times during runs 1-4 of this reaction (A is red; B is blue). Does the reaction progress to the right or left or not at all for each mixture to reach equilibrium? PLAN: We must compare Qc with Kc to determine the reaction direction, so we use the given equilibrium concentrations to find Kc. Then we count spheres and calculate Q for each mixture. 17-25

Sample Problem 17.5 SOLUTION: [B] 1.2x10-4 Kc = = 0.43 = [A] 2.8x10-4 Counting the red and blue spheres to calculate Qc for each mixture: 1. Qc = 8/2 = 4.0 2. Qc = 3/7 = 0.43 3. Qc = 4.6 = 0.67 4. Qc = 2/8 = 0.25 Comparing Qc with Kc to determine reaction direction: 1. Qc > Kc; reaction proceeds to the left. 2. Qc = Kc; no net change. 3. Qc > Kc; reaction proceeds to the left. 4. Qc < Kc; reaction proceeds to the right. 17-26 Sample Problem 17.6

Using Concentrations to Determine Reaction Direction PROBLEM: For the reaction N2O4(g) 2NO2(g), Kc = 0.21 at 100C. At a point during the reaction, [N2O4] = 0.12 M and [NO2] = 0.55 M. Is the reaction at equilibrium? If not, in which direction is it progressing? PLAN: We write an expression for Qc, find its value by substituting the giving concentrations, and compare the value with the given Kc. SOLUTION: [NO2]2 Qc = [N2O4] = (0.55)2 (0.12) = 2.5 Qc > Kc, therefore the reaction is not at equilibrium and will proceed from right to left, from products to reactants, until Qc = Kc.

17-27 Solving Equilibrium Problems If equilibrium quantities are given, we simply substitute these into the expression for Kc to calculate its value. If only some equilibrium quantities are given, we use a reaction table to calculate them and find Kc. A reaction table shows the balanced equation, the initial quantities of reactants and products, the changes in these quantities during the reaction, and the equilibrium quantities. 17-28 Example: In a study of carbon oxidation, an evacuated vessel containing a small amount of powdered graphite is heated to 1080 K. Gaseous CO2 is added to a pressure of 0.458 atm and CO forms. At equilibrium, the total pressure is 0.757 atm. Calculate Kp. CO2(g) + C(graphite) Pressure (atm)

Initial Change Equilibrium 2CO(g) CO2 (g) + C(graphite) 0.458 - 0 -x - +2x 0.458 -x

- 2x 2CO(g) The amount of CO2 will decrease. If we let the decrease in CO2 be x, then the increase in CO will be +2x. Equilibrium amounts are calculated by adding the change to the initial amount. 17-29 Once we have the equilibrium amounts expressed in terms of x, we use the other information given in the problem to solve for x. The total pressure at equilibrium is 0.757 atm = PCO2(eq) + PCO(eq) 0.757 atm = 0.458 x + 2x (from reaction table) 0.757 atm = 0.458 + x x = 0.757 0.458 = 0.299 atm

At equilibrium PCO2(eq) = 0.458 x = 0.458 0.299 = 0.159 atm PCO = 2x = 2(0.299) = 0.598 atm Kp = 17-30 2 PCO 2(eq) PCO(eq) 2 = 0.598 0.159 = 2.25 Sample Problem 17.7 Calculating Kc from Concentration Data PROBLEM: In order to study hydrogen halide decomposition, a

researcher fills an evacuated 2.00-L flask with 0.200 mol of HI gas and allows the reaction to proceed at 453C. 2HI(g) H2(g) + I2(g) At equilibrium, [HI] = 0.078 M. Calculate Kc. PLAN: To calculate Kc we need equilibrium concentrations. We can find the initial [HI] from the amount and the flask volume, and we are given [HI] at equilibrium. From the balanced equation, when 2x mol of HI reacts, x mol of H2 and x mol of I2 form. We use this information to set up a reaction table. SOLUTION: [HI] = 17-31 Calculating [HI]: 0.200 mol 2.00 L = 0.100 M Sample Problem 17.7 Concentration (M)

2HI(g) H2(g) + I2(g) Initial 0.100 0 0 - 2x +x +x 0.100 - x

x x Change Equilibrium [HI] = 0.078 = 0.100 - 2x; x = 0.011 M Qc = 17-32 [H2] [I2] [HI] 2 = (0.011)(0.011) (0.078) 2

= 0.020 = Kc Sample Problem 17.8 PROBLEM: Determining Equilibrium Concentrations from Kc In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous CH4 and H2O in a 0.32-L flask at 1200 K. At equilibrium the flask contains 0.26 mol of CO, 0.091 mol of H2, and 0.041 mol of CH4. What is the [H2O] at equilibrium? Kc = 0.26 for this process at 1200 K. PLAN: First we write the balanced equation and the expression for Qc. We calculate the equilibrium concentrations from the given numbers of moles and the flask volume. We then use the value of Kc to solve for [H2O]. SOLUTION: CH4(g) + H2O(g)

CO(g) + 3H2(g) 0.041 mol 0.26 mol [CH4]eq = = 0.13 M [CO]eq = = 0.81 M 0.32 L 0.32 L 0.091 mol [H2]eq = = 0.28 M 0.32 L 17-33 Sample Problem 17.8 Qc = [H2O] = 17-34

[CO][H2]3 [CH4][H2O] [CO][H2]3 [CH4] Kc = (0.81)(0.28)3 (0.13)(0.26) = 0.53 M Sample Problem 17.9 Determining Equilibrium Concentrations from Initial Concentrations and Kc PROBLEM: Fuel engineers use the extent of the change from CO and H2O to CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.250 mol of H2O are placed in a 125-mL flask at 900 K, what is the composition of the equilibrium mixture? At this temperature, Kc is 1.56.

PLAN: We have to find the concentrations of all species at equilibrium and then substitute into a Kc expression. First we write a balanced equation, calculate initial concentrations and set up a reaction table. SOLUTION: CO(g) + H2O(g) CO2(g) + H2(g) 0.250 mol Calculating initial concentrations,[CO] = [H2O] = 0.125 L 17-35 = 2.00 M Sample Problem 17.9 Concentration (M) Initial CO(g) + H2O(g)

2.00 2.00 0 0 -x -x +x +x 2.00 - x 2.00 - x x x

Change Equilibrium Qc = K c = 1.56 = (x)(x) [CO2][H2] (x)2 = = [CO][H2O] (2.00 - x) (2.00 - x) (2.00 - x)2 x 2.00- x = 1.25 [CO] = [H2O] = 0.89 M 17-36

CO2(g) + H2(g) x = 1.11 M 2.00 - x = 0.89 M [CO2] = [H2] = 1.11 M Sample Problem 17.10 Making a Simplifying Assumption to Calculate Equilibrium Concentrations PROBLEM: Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction COCl2(g) CO(g) + Cl2(g); Kc = 8.3x10-4 at 360C. Calculate [CO], [Cl2], and [COCl2] when the following amounts of phosgene decompose and reach equilibrium in a 10.0-L flask. (a) 5.00 mol COCl2 (b) 0.100 mol COCl2

PLAN: We know from the balanced equation that when x mol of COCl2 decomposes, x mol of CO and x mol of Cl2 form. We calculate initial concentrations, define x, set up a reaction table, and substituted the values into the expression for Qc. Since Kc is very small, we can assume that x is negligible, which simplifies the expression. We must check this assumption when we have solved for x. 17-37 Sample Problem 17.10 SOLUTION: 0.500 mol (a) Calculating initial concentrations,[COCl2] = = 0.500 M 10.0 L Let x = amount of COCl2 that reacts: Concentration (M) COCl2 (g) Initial

0.500 Change Equilibrium Kc = 17-38 [CO][Cl2] [COCl2] -x 0.500 - x x2 = = 8.3x10-4 0.500 - x CO(g) + Cl2(g) 0

0 +x +x x x Sample Problem 17.10 Since Kc is very small, the reaction does not proceed very far to the right, so let's assume that x can be neglected when we calculate [COCl2]eqm: 2 x Kc = 8.3x10-4 0.500 x2 (8.3x10-4)(0.500) so x 2.0x10-2 At equilibrium, [COCl2] = 0.500 2.0x10-2 = 7.5x10-3 M

and [Cl2] = [CO] = 2.0x10-2 M Check that the assumption is justified: x 2.0x10-2 x 100 = x 100 = 4%, which is < 5%, so the [COCl2]init 0.500 assumption is justified. 17-39 Sample Problem 17.10 0.100 mol (b) Calculating initial concentrations,[COCl2] = = 0.0100 M 10.0 L [CO][Cl2] x2 Kc = = = 8.3x10-4 0.0100 - x

[COCl2] If we assume that 0.0100 x 0.100 2 x then Kc = 8.3x10-4 and x 2.9x10-3 0.0100 Checking the assumption: 2.9x10-3 x 100 = 29%, which is > 5%, so the assumption is not 0.0100 justified. 17-40 Sample Problem 17.10 Using the quadratic formula we get x2 + (8.3x10-4)x (8.3x10-6) = 0 x = 2.5x10-3 and 0.0100 - x = 7.5x10-3 M [CO] = [Cl2] = 2.5x10-3 M [COCl2] = 7.5x10-3 M 17-41

The Simplifying Assumption We assume that x([A]reacting can be neglected if Kc is relatively small and/or [A]init is relatively large. If [A]init > 400, the assumption is justified; neglecting x Kc introduces an error < 5%. If [A]init < 400, the assumption is not justified; neglecting x Kc introduces an error > 5%. 17-42 Sample Problem 17.11 Predicting Reaction Direction and Calculating Equilibrium Concentrations PROBLEM: The research and development unit of a chemical company is studying the reaction of CH4 and H2S, two components of natural gas: CH4(g) + 2H2S(g) CS2(g) + 4H2(g)

In one experiment, 1.00 mol of CH4, 1.00 mol of CS2, 2.00 mol of H2S, and 2.00 mol of H2 are mixed in a 250-mL vessel at 960C. At this temperature, Kc = 0.036. (a) In which direction will the reaction proceed to reach equilibrium? (b) If [CH4] = 5.56 M at equilibrium, what are the equilibrium concentrations of the other substances? PLAN: (a) To find the direction of reaction we determine the initial concentrations from the given amounts and volume, calculate Qc and compare it with Kc. (b) Based on this information, we determine the sign of each concentration change for the reaction table and hence calculate equilibrium concentrations. 17-43 Sample Problem 17.11 SOLUTION: 1.00 mol (a) Calculating initial concentrations,[CH4] = 0.250 L = 4.00 M

[H2S] = 8.00 M, [CS2] = 4.00 M and [H2] = 8.00 M Qc = [CS2][H2]4 [CH4][H2S] 2 = (4.0)(8.0)4 (4.0)(8.0) 2 = 64 Since Qc > Kc, the reaction will proceed to the left. The reactant concentrations will decrease and the product concentrations will increase. 17-44

Sample Problem 17.11 (b) We use this information to construct our reaction table. Concentration (M) CH4(g) + 2H2S(g) CS2(g) + 4H2(g) Initial 4.00 8.00 4.00 8.00 Change +x +2x

-x -4x Equilibrium 4.00 + x 8.00 + 2x 4.00 - x 8.00 - 4x At equilibrium [CH4] = 5.56 M, so 4.00 + x = 5.56 and x = 1.56 [H2S] = 8.00 + 2x = 11.12 M [CS2] = 4.00 - x = 2.44 M [H2] = 8.00 - 4x = 1.76 M 17-45

17.6 Steps in solving equilibrium problems. PRELIMINARY SETTING UP 1. Write the balanced equation. 2. Write the reaction quotient, Q. 3. Convert all amounts into the correct units (M or atm). WORKING ON THE REACTION TABLE 4. When reaction direction is not known, compare Q with K. 5. Construct a reaction table. Check the sign of x, the change in the concentration (or pressure). 17-46 17.6 continued

SOLVING FOR x AND EQUILIBRIUM QUANTITIES 6. Substitute the quantities into Q. 7. To simplify the math, assume that x is negligible: ([A]init x = [A]eq [A]init) 6. Solve for x. Check that assumption is justified (<5% error). If not, solve quadratic equation for x. 7. Find the equilibrium quantities. Check to see that calculated values give the known K. 17-47 Le Chteliers Principle When a chemical system at equilibrium is disturbed, it reattains equilibrium by undergoing a net reaction that reduces the effect of the disturbance. A system is disturbed when a change in conditions forces it temporarily out of equilibrium.

The system responds to a disturbance by a shift in the equilibrium position. A shift to the left is a net reaction from product to reactant. A shift to the right is a net reaction from reactant to product. 17-48 The Effect of a Change in Concentration If the concentration of A increases, the system reacts to consume some of it. If a reactant is added, the equilibrium position shifts to the right. If a product is added, the equilibrium position shifts to the left. If the concentration of B decreases, the system reacts to consume some of it. If a reactant is removed, the equilibrium position shifts to the left. If a product is removed, the equilibrium position shifts to the right. Only substances that appear in the expression for Q can have an effect. A change in concentration has no effect on the value of K. 17-49

Figure 17.7 17-50 The effect of a change in concentration. Table 17.3 The Effect of Added Cl2 on the PCl3-Cl2-PCl5 System Concentration (M) Original equilibrium PCl3(g) + 0.200 Disturbance New initial Change New equilibrium 17-51 0.125

PCl3(g) 0.600 +0.075 0.200 0.200 0.600 -x -x +x 0.200 - x 0.200 - x 0.600 + x (0.637)*

Experimentally determined value * Cl2(g) Figure 17.8 The effect of added Cl2 on the PCl3-Cl2-PCl5 system. PCl3(g) + Cl2(g) PCl5(g) When Cl2 (yellow curve) is added, its concentration increases instantly (vertical part of yellow curve) and then falls gradually as it reacts with PCl3 to form more PCl5. Equilibrium is re-established at new concentrations but with the same value of K.

17-52 Sample Problem 17.12 Predicting the Effect of a Change in Concentration on the Equilibrium Position PROBLEM: To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the contaminant hydrogen sulfide with O2: 2H2S(g) + O2(g) 2S(s) + 2H2O(g) What happens to (a) [H2O] if O2 is added? (b) [H2S] if O2 is added? (c) [O2] if H2S is removed? (d) [H2S] if sulfur is added? PLAN: We write the reaction quotient to see how Qc is affected by each disturbance, relative to Kc. This effect tells us the direction in which the reaction proceeds for the system to reattain equilibrium and how each concentration changes. 17-53

Sample Problem 17.12 SOLUTION: 2 Qc = [H2O] [H2S]2[O2] (a) When O2 is added, Q decreases and the reaction proceeds to the right until Qc = Kc again, so [H2O] increases. (b) When O2 is added, Q decreases and the reaction proceeds to the right until Qc = Kc again, so [H2S] decreases. (c) When H2S is removed, Q increases and the reaction proceeds to the left until Qc = Kc again, so [O2] increases. (d) The concentration of solid S is unchanged as long as some is present, so it does not appear in the reaction quotient. Adding more S has no effect, so [H2S] is unchanged. 17-54 The Effect of a Change in Pressure (Volume) Changes in pressure affect equilibrium systems containing gaseous components. Changing the concentration of a gaseous component

causes the equilibrium to shift accordingly. Adding an inert gas has no effect on the equilibrium position, as long as the volume does not change. This is because all concentrations and partial pressures remain unchanged. Changing the volume of the reaction vessel will cause equilibrium to shift if ngas 0. Changes in pressure (volume) have no effect on the value of K. 17-55 Figure 17.9 17-56 The effect of a change in pressure (volume) on a system at equilibrium. Sample Problem 17.13 Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position

PROBLEM: How would you change the volume of each of the following reactions to increase the yield of the products? (a) CaCO3(s) CaO(s) + CO2(g) (b) S(s) + 3F2(g) SF6(g) (c) Cl2(g) + I2(g) 2ICl(g) PLAN: Whenever gases are present, a change in volume causes a change in concentration. For reactions in which the number of moles of gas changes, a decrease in volume (pressure increase) causes an equilibrium shift to lower the pressure by producing fewer moles of gas. A volume increase (pressure decrease) has the opposite effect. 17-57 Sample Problem 17.13

SOLUTION: (a) CO2 is the only gas present. To increase its yield, we should increase the volume (decrease the pressure). (b) There are more moles of gaseous reactants than products, so we should decrease the volume (increase the pressure) to shift the equilibrium to the right. (c) The number of moles of gas is the same in both the reactants and the products; therefore a change in volume will have no effect. 17-58 The Effect of a Change in Temperature To determine the effect of a change in temperature on equilibrium, heat is considered a component of the system. Heat is a product in an exothermic reaction (Hrxn < 0). Heat is a reactant in an endothermic reaction (Hrxn > 0). An increase in temperature adds heat, which favors the endothermic reaction. A decrease in temperature removes heat, which favors the exothermic reaction. 17-59

Temperature and K The only factor that affects the value of K for a given equilibrium system is temperature. For a reaction with Hrxn > 0, an increase in temperature will cause K to increase. For a reaction with Hrxn < 0, an increase in temperature will cause K to decrease. The vant Hoff equation shows this relationship: ln 17-60 K2 K1 o H rxn = R 1 T2

- 1 T1 R = 8.314 J/molK K1 is the equilibrium constant at T1 Sample Problem 17.14 Predicting the Effect of a Change in Temperature on the Equilibrium Position PROBLEM: How does an increase in temperature affect the equilibrium concentration of the underlined substance and K for each of the following reactions? (a) CaO(s) + H2O(l) Ca(OH)2(aq) H = -82 kJ (b) CaCO3(s) (c) SO2(g) CaO(s) + CO2(g) H = 178 kJ S(s) + O2(g) H = 297 kJ

PLAN: We write each equation to show heat as a reactant or product. The temperature increases when we add heat, so the system shifts to absorb the heat; that is, the endothermic reaction is favored. Thus, K will increase if the forward reaction is endothermic and decrease if it is exothermic. 17-61 Sample Problem 17.14 SOLUTION: (a) CaO(s) + H2O(l) Ca(OH)2(aq) + heat An increase in temperature will shift the reaction to the left, so [Ca(OH)2] and K will decrease. (b) CaCO3(s) + heat CaO(s) + CO2(g) An increase in temperature will shift the reaction to the right, so [CO2] and K will increase. (c) SO2(g) + heat S(s) + O2(g)

An increase in temperature will shift the reaction to the right, so [SO2] will decrease. 17-62 Catalysts and Equilibrium A catalyst speeds up a reaction by lowering its activation energy. A catalyst therefore speeds up the forward and reverse reactions to the same extent. A catalyst causes a reaction to reach equilibrium more quickly, but has no effect on the equilibrium position. 17-63 Table 17.4 Effects of Various Disturbances on a System at Equilibrium 17-64 Sample Problem 17.15 Determining Equilibrium Parameters from

Molecular PROBLEM: For the reaction X(g) + Scenes Y2(g) XY(g) + Y(g) H > 0 the following molecular scenes depict different reaction mixtures. (X = green, Y = purple): (a) If K = 2 at the temperature of the reaction, which scene represents the mixture at equilibrium? (b) Will the reaction mixtures in the other two scenes proceed toward reactant or toward products to reach equilibrium? (c) For the mixture at equilibrium, how will a rise in temperature affect [Y2]? 17-65 Sample Problem 17.15 PLAN: To select the scene that represents the reaction at equilibrium, we first write the expression for Q. We count the particles for each scene and calculate the value of Q. The scene that gives Q = K represents equilibrium. For the other

two scenes, we compare Q to K and determine the direction of reaction. SOLUTION: For the reaction, we have Q = For scene 1: Qc = 5x3 = 15 1x1 For scene 3: Qc = 17-66 [XY][Y] [X][Y2] For scene 2: Qc = 3x1 = 3x3 4x2 =2

2x2 Sample Problem 17.15 Q = 15 Q = 2.0 Q = 0.33 (a) In scene 2, Q = K, so it represents the system at equilibrium. (b) In scene 1, Q > K, so the system will proceed toward the reactants to reach equilibrium. In scene 3, Q < K, so the system will proceed toward the products. (c) If H > 0, heat is a reactant (endothermic). A rise in temperature will favor the products and [Y2] will decrease as the system shifts toward the products. 17-67 The Synthesis of Ammonia Ammonia is synthesized industrially via the Haber process: N2(g) + 3H2(g)

2NH3(g) Hrxn = -91.8 kJ There are three ways to maximize the yield of NH3: Decrease [NH3] by removing NH3 as it forms. Decrease the volume (increase the pressure). Decrease the temperature. 17-68 Table 17.5 Effect of Temperature on Kc for Ammonia Synthesis N2(g) + 3H2(g) 2NH3(g) Hrxn = -91.8 kJ T (K) Kc 200.

7.17x1015 300. 2.69x108 400. 3.94x104 500. 1.72x102 600. 4.53x100 700. 2.96x10-1 800.

3.96x10-2 An increase in temperature causes the equilibrium to shift towards the reactants, since the reaction is exothermic. 17-69 Figure 17.11 Percent yield of ammonia vs. temperature at five different pressures. 17-70 At very high P and low T (top left), the yield is high, but the rate is low. Industrial conditions (circle) are between 200 and 300 atm at about 400C. Figure 17.12 Key stages in the Haber process for synthesizing ammonia. 17-71 Chemical Connections

Figure B17.1 The biosynthesis of isoleucine from threonine. Isoleucine is synthesized from threonine in a sequence of five enzyme-catalyzed reactions. Once enough isoleucine is present, its concentration builds up and inhibits threonine dehydratase, the first enzyme in the pathway. This process is called end-product feedback inhibition. 17-72 Chemical Connections Figure B17.2 17-73 The effect of inhibitor binding on the shape of an active site.

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