H.W assignment multiple choice with show work for answers

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Chemical Equilibrium Chapter 14 1 Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. Equilibrium Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: • the rates of the forward and reverse reactions are equal and • the concentrations of the reactants and products remain constant Physical equilibrium H2 O 𝑙 ⇄ H2 O 𝑔 Chemical equilibrium N2 O4 𝑔 ⇄ 2NO2 𝑔 2 Equilibrium of N2 O4 N2 O4 𝑔 ⇄ 2NO2 𝑔 Copyright © McGraw-Hill Education. Permission required for reproduction or display equilibrium equilibrium Start with N2 O4 Start with NO2 & N2 O4 equilibrium Start with NO2 Copyright © McGraw-Hill Education. Permission required for reproduction or display 3 Equilibrium of N2 O4 (1) N2 O4 𝑔 ⇄ 2NO2 𝑔 NO2 2 𝐾= = 4.63 × 10−3 N2 O4 𝑎A + 𝑏B ⇄ 𝑐C + 𝑑D C 𝐾= A 𝑐 D𝑑 𝑎 B 𝑏 Law of Mass Action Copyright © McGraw-Hill Education. Permission required for reproduction or display 4 Copyright © McGraw-Hill Education. Permission required for reproduction or display Equilibrium of N2 O4 (2) constant 5 Relationship of Equilibrium and K C 𝐾= A 𝑐 D𝑑 𝑎 B 𝑏 𝑎A + 𝑏B ⇄ 𝑐C + 𝑑D Equilibrium Will 𝐾≫1 Lie to the right Favor products 𝐾≪1 Lie to the left Favor reactants Copyright © McGraw-Hill Education. Permission required for reproduction or display Copyright © McGraw-Hill Education. Permission required for reproduction or display 6 Example 14.1 Write expressions for 𝐾c , and 𝐾𝑝 if applicable, for the following reversible reactions at equilibrium: (a) HF 𝑎𝑞 + H2 O 𝑙 ⇌ H3 O+ 𝑎𝑞 + F − 𝑎𝑞 (b) 2NO 𝑔 + O2 𝑔 ⇌ 2NO2 𝑔 (c) CH3 COOH a𝑞 + C2 H5 OH a𝑞 ⇌ CH3 COOC2 H5 a𝑞 + H2 O 𝑙 9 Example 14.1 (1) Strategy Keep in mind the following facts: (1) the 𝐾𝑝 expression applies only to gaseous reactions and (2) the concentration of solvent (usually water) does not appear in the equilibrium constant expression. 10 Example 14.1 (2) Solution (a) Because there are no gases present, 𝐾𝑝 does not apply and we have only 𝐾c . + − H O F 3 𝐾c′ = HF H2 O HF is a weak acid, so that the amount of water consumed in acid ionizations is negligible compared with the total amount of water present as solvent. Thus, we can rewrite the equilibrium constant as 𝐇𝟑 𝐎+ 𝐅 − 𝑲c = 𝐇𝐅 11 Example 14.1 (3) (b) 𝑲c = 𝐍𝐎2 2 𝐍𝐎 2 𝐎𝟐 𝑲𝑝 = 𝑷2No2 𝑷2NO 𝑷o2 (c) The equilibrium constant 𝐾c′ is given by 𝐾c′ CH3 COOC2 H5 H2 O = CH3 COOH C2 H5 OH Because the water produced in the reaction is negligible compared with the water solvent, the concentration of water does not change. Thus, we can write the new equilibrium constant as 𝐂𝐇3 𝐂𝐎𝐎𝐂2 𝐇5 𝑲c = 𝐂𝐇3 𝐂𝐎𝐎𝐇 𝐂2 𝐇5 𝐎𝐇 12 Example 14.2 The following equilibrium process has been studied at 230°C: 2NO 𝑔 + O2 𝑔 ⇌ 2NO2 𝑔 In one experiment, the concentrations of the reacting species at equilibrium are found to be NO = 0.0542 M, O2 = 0.127 M, and NO2 = 15.5 M. Calculate the equilibrium constant 𝐾c of the reaction at this temperature. 13 Example 14.2 (1) Strategy The concentrations given are equilibrium concentrations. They have units of mol/L, so we can calculate the equilibrium constant 𝐾c using the law of mass action [Equation (14.2)]. Solution The equilibrium constant is given by NO2 2 𝐾c = NO 2 O2 Substituting the concentrations, we find that 15.5 2 𝟓 𝐾c = = 𝟔. 𝟒𝟒 × 𝟏𝟎 0.0542 2 0.127 14 Heterogeneous Equilibrium Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO3 𝑠 ⇄ CaO 𝑠 + CO2 𝑔 𝐾c′ CaO CO2 = CaCO3 𝐾𝑐 = CO2 = 𝐾c′ x CaCO3 CaO CaCO3 =constant CaO =constant 𝐾𝑝 = 𝑃co2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 22 Heterogeneous Equilibrium (1) CaCO3 𝑠 ⇄ CaO 𝑠 + CO2 𝑔 Copyright © McGraw-Hill Education. Permission required for reproduction or display 𝑃CO2 = 𝐾𝑝 𝑃CO2 does not depend on the amount of CaCO3 or CaO 23 Example 14.6 Consider the following heterogeneous equilibrium: CaCO3 𝑠 ⇌ CaO 𝑠 + CO2 𝑔 At 800° C, the pressure of CO2 is 0.236 atm. Calculate (a) 𝐾𝑝 and (b) 𝐾c for the reaction at this temperature. 28 Example 14.6 (1) Strategy Remember that pure solids do not appear in the equilibrium constant expression. The relationship between 𝐾𝑝 and 𝐾c is given by Equation (14.5). Solution (a) Using Equation (14.8) we write 𝐾𝑃 = 𝑃CO2 = 𝟎. 𝟐𝟑𝟔 29 Example 14.6 (2) (b) From Equation (14.5), we know 𝐾𝑃 = 𝐾c 0.0821𝑇 ∆𝑛 In this case, 𝑇 = 800 + 273 = 1073 K and ∆𝑛 = 1, so we substitute these values in the equation and obtain 0.236 = 𝐾c 0.0821 × 1073 𝐾c = 𝟐. 𝟔𝟖 × 𝟏𝟎−𝟑 30 Equilibrium Constants A+B⇄C+D 𝐾𝑐′ C+D⇄E+F 𝐾𝑐′′ A+B⇄E+F 𝐾𝑐 𝐾c′ C D = A B 𝐾𝑐′′ E F = C D E F 𝐾c = A B 𝐾𝑐 = 𝐾𝑐′ × 𝐾𝑐′′ If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 31 Equilibrium Constants (1) N2 O4 𝑔 ⇄ 2NO2 𝑔 NO2 2 𝐾= = 4.63 × 10−3 N2 O4 2NO2 𝑔 ⇄ N2 O4 𝑔 𝐾′ N2 O4 1 = = = 216 2 NO2 𝐾 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 32 Example 14.7 The reaction for the production of ammonia can be written in a number of ways: (a) N2 𝑔 + 3H2 𝑔 ⇌ 2NH3 𝑔 (b) 1 N 2 2 (c) 1 N 3 2 𝑔 3 + H2 2 𝑔 ⇌ NH3 𝑔 𝑔 + H2 𝑔 ⇌ 2 NH3 3 𝑔 Write the equilibrium constant expression for each formulation. (Express the concentrations of the reacting species in mol/L.) (d) How are the equilibrium constants related to one another? 33 Example 14.7 (1) Strategy We are given three different expressions for the same reacting system. Remember that the equilibrium constant expression depends on how the equation is balanced, that is, on the stoichiometric coefficients used in the equation. 34 Example 14.7 (2) Solution (a) (b) (c) 𝐾a = NH3 2 N2 H2 3 𝐾b = 𝐾c = NH3 1 N2 2 3 H2 2 2 NH3 3 1 N2 3 H2 35 Example 14.7 (3) (d) 𝐾a = 𝐾b2 𝐾a = 𝐾c3 3 2 𝐾b2 = 𝐾c3 or 𝐾b = 𝐾c 36 Chemical Kinetics and Chemical Equilibrium kf ratef = 𝑘f A B kr rater = 𝑘r AB2 ⎯⎯⎯ → AB2 A+2B ⎯ 2 Equilibrium ratef = rater 𝑘f A B 2 = 𝑘r AB2 𝑘𝑓 AB2 = 𝐾𝑐 = 𝑘𝑟 A B2 38 Reaction Quotient The reaction quotient 𝑸𝐜 is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant 𝐾𝑐 expression. IF • 𝑄𝑐 < 𝐾𝑐 system proceeds from left to right to reach equilibrium • 𝑄𝑐 = 𝐾𝑐 the system is at equilibrium • 𝑄𝑐 > 𝐾𝑐 system proceeds from right to left to reach equilibrium 39 Example 14.8 Strategy We are given the initial amounts of the gases (in moles) in a vessel of known volume (in liters), so we can calculate their molar concentrations and hence the reaction quotient Q c . How does a comparison of 𝑄c with 𝐾c enable us to determine if the system is at equilibrium or, if not, in which direction will the net reaction proceed to reach equilibrium? 40 Example 14.8 (1) At the start of a reaction, there are 0.249 mol N2 , 3.21 × 10−2 mol H2 , and 6.42 × 10−4 mol NH3 in a 3.50 − L reaction vessel at 375°C. If the equilibrium constant 𝐾c for the reaction N2 𝑔 + 3H2 𝑔 ⇌ 2NH3 𝑔 is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed. 41 Example 14.8 (2) Solution The initial concentrations of the reacting species are N2 H2 0 0 NH3 0.249mol = = 0.0711 𝑀 3.50L 3.21 × 10−2 mol = = 9.17 × 10−3 𝑀 3.50L 0 6.42 × 10−4 mol = = 1.83 × 10−4 𝑀 3.50L 42 Example 14.8 (3) Next we write NH3 20 𝑄c = N2 0 H2 3 0 1.83 × 10−4 2 = 0.0711 9.17 × 10−3 3 = 0.611 Because Qc is smaller than 𝐾c (1.2), the system is not at equilibrium. The net result will be an increase in the concentration of NH3 and a decrease in the concentrations of N2 and H2 . That is, the net reaction will proceed from left to right until equilibrium is reached. 43 Calculating Equilibrium Concentrations 1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3. Having solved for x, calculate the equilibrium concentrations of all species. 44 Example 14.9 A mixture of 0.500 mol H2 and 0.500 mol l2 was placed in a 1.00 − L stainless-steel flask at 430° C. The equilibrium constant 𝐾c for the reaction H2 𝑔 + l2 𝑔 ⇌ 2Hl 𝑔 is 54.3 at this temperature. Calculate the concentrations of H2 , l2 , and HI at equilibrium 45 Example 14.9 (1) Strategy We are given the initial amounts of the gases (in moles) in a vessel of known volume (in liters), so we can calculate their molar concentrations. Because initially no HI was present, the system could not be at equilibrium. Therefore, some H2 would react with the same amount of l2 (why?) to form HI until equilibrium was established. 46 Example 14.9 (2) Solution We follow the preceding procedure to calculate the equilibrium concentrations. Step 1: The stoichiometry of the reaction is 1 mol H2 reacting with 1 mol l2 to yield 2 mol HI. Let x be the depletion in concentration (mol/L) of H2 and l2 at equilibrium. It follows that the equilibrium concentration of HI must be 2𝑥. We summarize the changes in concentrations as follows: Initial (M): Change (M): Equilibrium (M): H2 + 0.500 l2 0.500 ⇌ 2HI 0.000 −𝑥 −𝑥 +2𝑥 0.500 − 𝑥 0.500 − 𝑥 2𝑥 47 Example 14.9 (3) Step 2: The equilibrium constant is given by HI 2 𝐾c = H2 I2 Substituting, we get 2𝑥 2 54.3 = 0.500 − 𝑥 0.500 − 𝑥 Taking the square root of both sides, we get 2𝑥 7.37 = 0.500 − 𝑥 𝑥 = 0.393 𝑀 48 Example 14.9 (4) Step 3: At equilibrium, the concentrations are H2 = 0.500 − 0.393 𝑀 = 𝟎. 𝟏𝟎𝟕𝑴 l2 = 0.500 − 0.393 𝑀 = 𝟎. 𝟏𝟎𝟕𝑴 Hl = 2 × 0.393𝑀 = 𝟎. 𝟕𝟖𝟔𝑴 Check You can check your answers by calculating 𝐾c using the equilibrium concentrations. Remember that 𝐾c is a constant for a particular reaction at a given temperature. 49 Example 14.10 For the same reaction and temperature as in Example 14.9, H2 𝑔 + l2 𝑔 ⇌ 2HI 𝑔 , suppose that the initial concentrations of H2 , l2 , and HI are 0.00623 𝑀, 0.00414 𝑀, and 0.0224 𝑀, respectively. Calculate the concentrations of these species at equilibrium. 50 Example 14.10 (1) Strategy From the initial concentrations we can calculate the reaction quotient 𝑄c to see if the system is at equilibrium or, if not, in which direction the net reaction will proceed to reach equilibrium. A comparison of 𝑄c with 𝐾c also enables us to determine if there will be a depletion in H2 and l2 or HI as equilibrium is established. 51 Example 14.10 (2) Solution First we calculate 𝑄c as follows: 𝑄c = HI H2 2 0 0 I2 0 0.0224 2 = = 19.5 0.00623 0.00414 Because 𝑄c (19.5) is smaller than 𝐾c (54.3), we conclude that the net reaction will proceed from left to right until equilibrium is reached (see Figure 14.4); that is, there will be a depletion of H2 and l2 and a gain in HI. 52 Example 14.10 (3) Step 1: Let x be the depletion in concentration (mol/L) of H2 and I2 at equilibrium. From the stoichiometry of the reaction it follows that the increase in concentration for HI must be 2𝑥. Next we write Initial (M): Change (M): Equilibrium (M): H2 + I2 ⇌ 2HI 0.00623 0.00414 0.0224 −𝑥 −𝑥 +2𝑥 0.00623 − 𝑥 0.00414 − 𝑥 0.0224 + 2𝑥 53 Example 14.10 (4) Step 2: The equilibrium constant is HI 2 𝐾c = H2 l2 Substituting, we get 0.0224 + 2x 2 54.3 = 0.00623 − 𝑥 0.00414 − 𝑥 It is not possible to solve this equation by the square root shortcut, as the starting concentrations [H2 ] and [I2 ] are unequal. Instead, we must first carry out the multiplications 54.3 2.58 × 10−5 − 0.0104𝑥 + 𝑥 2 = 5.02 × 10−4 + 0.0896𝑥 + 4𝑥 2 54 Example 14.10 (5) Collecting terms, we get 50. 3𝑥 2 − 0.654𝑥 + 8.98 × 10−4 = 0 This is a quadratic equation of the form a𝑥 2 + 𝑏𝑥 + 𝑐 = 0. The solution for a quadratic equation (see Appendix 4) is −𝑏 ± 𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 Here we have a = 50.3, 𝑏 = −0.654, and 𝑐 = 8.98 × 10−4 , so that 𝑥= −0.654 2 − 4 50.3 8.98 × 10−4 2 × 50.3 𝑥 = 0.0114 𝑀 or 𝑥 = 0.00156 𝑀 0.654 ± 55 Example 14.10 (6) The first solution is physically impossible because the amounts of H2 and l2 reacted would be more than those originally present. The second solution gives the correct answer. Note that in solving quadratic equations of this type, one answer is always physically impossible, so choosing a value for 𝑥 is easy. Step 3: At equilibrium, the concentrations are H2 = 0.00623 − 0.00156 𝑀 = 𝟎. 𝟎𝟎𝟒𝟔𝟕 𝑴 I2 = 0.00414 − 0.00156 𝑀 − 𝟎. 𝟎𝟎𝟐𝟓𝟖 𝑴 Hl = 0.0224 + 2 × 0.00156 𝑀 = 𝟎. 𝟎𝟐𝟓𝟓 𝑴 56 Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Copyright © McGraw-Hill Education. Permission required for reproduction or display • Changes in Concentration N2 𝑔 + 3H2 𝑔 ⇄ 2NH3 𝑔 Equilibrium shifts left to offset stress Add NH3 58 Le Châtelier’s Principle (1) • Changes in Concentration continued 𝑎A + 𝑏B ⇄ 𝑐C + 𝑑D Change Increase concentration of product(s) Decrease concentration of product(s) Shifts the Equilibrium left right Increase concentration of reactant(s) Decrease concentration of reactant(s) right left 59 Example 14.11 At 720° C, the equilibrium constant 𝐾c for the reaction N2 𝑔 + 3H2 𝑔 ⇌ 2NH3 𝑔 is 2.37 × 10−3 . In a certain experiment, the equilibrium concentrations are N2 = 0.683 𝑀, H2 = 8.80 𝑀, and NH3 = 1.05 𝑀. Suppose some NH3 is added to the mixture so that its concentration is increased to 3.65 M. (a) Use Le Châtelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium. (b) Confirm your prediction by calculating the reaction quotient 𝑄c and comparing its value with 𝐾c . 60 Example 14.11 (1) Strategy (a) What is the stress applied to the system? How does the system adjust to offset the stress? (b) At the instant when some NH3 is added, the system is no longer at equilibrium. How do we calculate the 𝑄c for the reaction at this point? How does a comparison of 𝑄c with 𝐾c tell us the direction of the net reaction to reach equilibrium. 61 Example 14.11 (2) Solution (a) The stress applied to the system is the addition of NH3 . To offset this stress, some NH3 reacts to produce N2 and H2 until a new equilibrium is established. The net reaction therefore shifts from right to left; that is, N2 𝑔 + 3H2 𝑔 ⟵ 2NH3 𝑔 62 Example 14.11 (3) (b) At the instant when some of the NH3 is added, the system is no longer at equilibrium. The reaction quotient is given by NH3 20 𝑄𝒄 = N2 0 H2 3 0 2 3.65 = 0.683 8.80 3 = 2.86 × 10−2 Because this value is greater than 2.37 × 10−3 , the net reaction shifts from right to left until 𝑄c equals 𝐾c . 63 Example 14.11 (4) Figure 14.8 shows qualitatively the changes in concentrations of the reacting species. 64 Le Châtelier’s Principle (2) • Changes in Volume and Pressure A 𝑔 +B 𝑔 ⇄C 𝑔 Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas 65 Example 14.12 Consider the following equilibrium systems: (a) 2PbS 𝑠 + 3O2 𝑔 ⇌ 2PbO 𝑠 + 2SO2 𝑔 (b) PCl5 𝑔 ⇌ PCl3 𝑔 + Cl2 𝑔 (c) H2 𝑔 + CO2 𝑔 ⇌ H2 O 𝑔 + CO 𝑔 Predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature. 66 Example 14.12 (1) Strategy A change in pressure can affect only the volume of a gas, but not that of a solid because solids (and liquids) are much less compressible. The stress applied is an increase in pressure. According to Le Châtelier’s principle, the system will adjust to partially offset this stress. In other words, the system will adjust to decrease the pressure. This can be achieved by shifting to the side of the equation that has fewer moles of gas. Recall that pressure is directly proportional to moles of gas: 𝑃𝑉 = 𝑛𝑅𝑇 so 𝑃 ∝ 𝑛. 67 Example 14.12 (2) Solution (a) Consider only the gaseous molecules. In the balanced equation, there are 3 moles of gaseous reactants and 2 moles of gaseous products. Therefore, the net reaction will shift toward the products (to the right) when the pressure is increased. (b) The number of moles of products is 2 and that of reactants is 1; therefore, the net reaction will shift to the left, toward the reactant. (c) The number of moles of products is equal to the number of moles of reactants, so a change in pressure has no effect on the equilibrium. 68 Le Châtelier’s Principle (3) • Changes in Temperature Change Exothermic Rx Endothermic Rx Increase temperature K decreases K increases Decrease temperature K increases K decreases N2 O4 𝑔 ⇄ 2NO2 𝑔 ∆𝐻 ° = 58.0 kJΤmol 69 Le Châtelier’s Principle (4) • Adding a Catalyst • does not change K • does not shift the position of an equilibrium system • system will reach equilibrium sooner Copyright © McGraw-Hill Education. Permission required for reproduction or display Copyright © McGraw-Hill Education. Permission required for reproduction or display Catalyst lowers 𝐸a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. 70 Le Châtelier’s Principle - Summary Change Shift Equilibrium Concentration yes Change Equilibrium Constant no Pressure yes* no Volume yes* no yes yes no no Temperature Catalyst *Dependent on relative moles of gaseous reactants and products 71 Chapter 14 HW On a piece(s) of paper, write your name and the chapter number. For each slide, write the question number and the letter corresponding to your answer choice. On certain problems the required written work must be shown for full credit. * The homework assignment is worth a total of 16 pts. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Question 1 The reaction N2O4 (g) ↔ 2 NO2 (g) is endothermic. Which of the following is true? (Classify the reaction as homogeneous or heterogeneous) A) Decreasing the pressure drives the reaction to the right. B) Increasing the temperature drives the reaction to the right. C) Decreasing the volume drives the reaction to the left. D) All of the above. Question 2 If an equilibrium for a reaction is said to lie to the right then the equilibrium constant (K)… (Write the equilibrium expression for Question 1) A) is a negative value. B) is a positive value less than one. C) is a positive value greater than one. D) is equal to one. Question 3 The value of the equilibrium constant is not affected by the concentrations of… (Provide an example chemical equation and the expression for K) A) pure solids and liquids. B) gases. C) aqueous solutions. D) all of the above. Question 4 The value of an equilibrium constant (K) depends on… (In your own words, write a definition for K) A) the speed of the reaction. B) the coefficients of the balanced chemical equation. C) the starting concentrations of the reactants. D) all of the above. Question 5 For a particular set of concentrations, if the value of Q is much greater than the value of the equilibrium constant, K, for a reaction, then… (In your own words, write a definition for Q) A) the initial reaction rate will be fast, but slows down near equilibrium. B) the reaction will proceed to the right, but the rate cannot be determined from the Q or K. C) the reaction will proceed to the left, but the rate cannot be determined from the Q or K. D) the reaction will proceed to make more products if the value of K is much greater than one. Question 6 Consider the reaction of hydrogen gas and oxygen gas to produce liquid water. Which of the following is true? (Write the equilibrium expression of the reaction) A) The position of the equilibrium is unaffected by the addition of liquid water. B) Removing some of the hydrogen gas has no effect on the position of the equilibrium. C) Reducing the volume of the reaction vessel has no effect on the position of the equilibrium. D) None of the above are true. Question 7 Which of the following is (are) unaffected by reversing the direction in which a chemical equation is written? A) The equilibrium concentrations. B) The equilibrium constant value. C) The mass action expression. D) All of the above. Question 8 The reaction of hydrogen gas with nitrogen gas to produce ammonia gas is exothermic. Which of the following is true? (Write the equilibrium expression A) Increasing the pressure of the reaction system will increase production of ammonia. B) Condensing the ammonia gas produced into liquid will increase the production of ammonia. C) Decreasing the temperature of the reaction will increase the production of ammonia. D) All of the above. Question 9 Consider the chemical change AB. After the reaction starts, but before the system reaches equilibrium, the only process that takes places is A being converted into B. A) True B) False Question 10 Which of the following statements is true of a catalyst? (Give an example of a catalyst and the corresponding reaction it affects) A) Its concentration stays constant throughout the reaction B) it increases the rate of a reaction C) it provides a new pathway for the reaction D) all of the above E) none of the above Question 11 For reactions in gas phase, a relationship can be established between Kc and Kp. For which of the following processes will the two constants have the same numerical value? (Show a calculation that supports your answer) A) formation of water vapor from hydrogen and oxygen. B) formation of carbon dioxide from carbon monoxide and oxygen. C) synthesis of ammonia from hydrogen and nitrogen. D) decomposition of hydrogen iodide into hydrogen and iodine gas. E) conversion of nitrogen dioxide into dinitrogen tetroxide. Question 12 Calculate the equilibrium constant for the process AB, given the following equilibrium concentrations: [N2O4] = 0.0427 M ; [NO2] = 0.0141 M (Show all calculation steps) A) 6.60 x10-1 B) 215 C) 0.00466 D) 0.330 E) 3.03 Question 13 Consider the following reaction: 4 HCl (g) + O2 (g) ↔ 2 H2O (g) + 2 Cl2 (g) with DH = -114.4 kJ One way to decrease the amount of chlorine present at equilibrium is… (Explain why your answer affects the equilibrium of the reaction) A) Removing the water vapor while it is being formed. B) Raising the temperature. C) Adding more hydrogen chloride. D) Decreasing the volume of the reaction vessel. E) None of the above. Question 14 The concentrations of pure solids are normally excluded in an equilibrium constant expression because… A) These substances do not participate in the processes that are at equilibrium. B) At the time the equilibrium is established, their absolute amounts have not varied from the original amounts. C) The concentration of a pure solid remains almost constant at a high value throughout the reaction. D) They do not play a vital role in the rate-determining step of the reaction. E) None of the above. Question 15 At equilibrium… (Using your answer, write a definition for equilibrium) A) the rate constants for the forward and reverse reactions are equal. B) the reaction stops. C) all reactants have been converted to products. D) the forward and reverse rates are equal. E) none of the above. Question 16 Consider the following reaction: CaCO3 (s) ↔ CO2 (g) + CaO (s) What will happen to the system if more CaCO3 is added? Explain how this condition affects the K value (concentration). A) Nothing B) The concentration of CO2 will increase. C) less CaO will be produced. D) the pressure will increase. E) the amount of CO2 will decrease. Question 17 Consider the following reaction: 2 H2 (g) + 2 Cl2 (g) ↔ 2 HCl (g) + ∆ (heat) What will happen to the system if the volume of the reaction vessel is decreased? Explain how this condition affects the K values (concentration and pressure) as well as if the value of one constant will change more. A) more HCl will dissociate B) amounts of H2 and Cl2 will decrease C) less heat is produced D) all of the above E) none of the above
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