CHEM Dr. Babb s Sections Lecture Problem Sheets

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1 CHEM Dr. Babb s Sections Lecture Problem Sheets Kinetics: Integrated Form of Rate Law 61. Give the integrated form of a zeroth order reaction. Define the half-life and find the halflife for a general zeroth order reaction. Is the half life constant? For a zeroth order reaction, what type of plot will yield a straight line? 62. Give the integrated form of a first order reaction. Find the half-life for a general first order reaction. Is the half-life constant? For a first order reaction, what type of plot will yield a straight line? A. A certain first order reaction has the rate law Rate = k[a] with k= sec -1. If the initial concentration of A is 0.75 M, what will be the concentration of A after 1 minute? What is the half-life for this reaction? How much time will it take for 75% of A to react? How much A will be left after the passage of three half-lives? What is the initial rate of the reaction? 63. Give the integrated form of a second order reaction. Find the half-life for a general second order reaction. Is the half-life constant? For a second order reaction, what type of plot will yield a straight line? 64. The following data were collected for the reaction: 2 HI(g) > H 2 (g) + I 2 (g) at 580K. Time (min) [HI] M M M M M Determine the reaction order and form of the rate law. (HINT: Plot [HI] vs. t, ln[hi] vs. t and 1/[HI] vs. t.) Calculate the numerical values of the specific rate constant and the halflife of the reaction. After how much time will only 20% of the original amount of HI remain? Kinetics: Elementary Reactions, Mechanisms, Catalysis and Arrhenius Equation 65. Define the terms: elementary step and reaction mechanism. Can the rate law be written directly for an elementary reaction? 66. A reaction mechanism that occurs in polluted air is the following: NO 2 + O 3 > NO 3 + O 2 SLOW NO 3 + NO 2 > N 2 O 5 FAST Write the net reaction for this process, define the molecularity of each elementary step and write the rate law for this process. What is happening to the NO 3? 67. A reaction mechanism for the destruction of ozone, O 3, in the stratosphere is: O 3 > O 2 + O FAST O 3 + O > 2 O 2 SLOW Write the net reaction, define the molecularity of each elementary step and write the rate law for this process. Identify the intermediate. 68. A reaction mechanism for the destruction of ozone by high flying aircraft is the following: O 3 > O 2 + O

2 NO + O 3 > NO 2 + O 2 NO 2 + O > NO + O 2 Write the net reaction and define the molecularity of each elementary step. Identify any reaction intermediates and catalysts. 69. Define the following terms: transition state and activation energy. On a molecular level, how does an increase in temperature lead to an increase in reaction rate? Give the Arrhenius equation. Show how the Arrhenius equation can be used to calculate the activation energy and pre-exponential factor. 70. The specific rate constant for the formation of HI was measured at two different temperatures. Calculate the numerical values of the activation energy and preexponential factor. Calculate the value of the specific rate constant at 427 C. H 2 + I 2 > 2 HI k (1/M-s) Temp (K) The following rate constants were found for the decomposition of N 2 O 5 at four different temperatures. k (1/s) Temp ( C) Make a plot of ln(k) vs. 1/T. Calculate the activation energy and pre-exponential factor from the plot. What is the value of the specific rate constant at C? 72. How does a catalyst increase the rate of a reaction? Is the reaction mechanism the same in the presence of a catalyst? Chemical Equilibrium: Introduction to K p and K c 73. Define the following terms: equilibrium, equilibrium constant expression, and equilibrium constant. 74. Do all reactions go 100 % to completion (such that all reactants react to form products)? Are the concentrations of reactants and products changing at equilibrium? In a chemical reaction, once equilibrium is established it is referred to as a dynamic equilibrium. What does this mean? 75. At equilibrium, how are the rates of the forward and reverse reactions related? 76. For the general elementary reaction below, write the rate law for the forward reaction (Forward Rate=?) and the rate law for the reverse reaction (Reverse Rate=?). Derive the equilibrium constant (K) and equilibrium constant expression (K=?) by setting the Forward Rate = Reverse Rate. aa + bb < > cc + dd 77. Does the numerical value of the equilibrium constant K depend on temperature? 78. Consider the equation below. What does the value of K c tell you about the relative values of k f and k r? What does the value of K c tell you about the concentrations of products and reactants at equilibrium? COCl 2 (g) < > CO(g) + Cl 2 (g) K c =

3 79. Consider the equation below. What does the value of K c tell you about the relative values of k f and k r? What does the value of K c tell you about the concentrations of products and reactants at equilibrium? PCl 3 (g) + Cl 2 (g) < > PCl 5 (g) K c = When K c is large (K>10 +3 ), how do the concentrations of reactants and products compare? When K c is small (K<10-3 ), how do the concentrations of reactants and products compare? When K c is intermediate (10-3 <K<10 +3 ), how do the concentrations of reactants and products compare? 81. Write the equilibrium constant expressions for the chemical reactions below. A. 4 NH 3 (g) + 5 O 2 (g) < > 4 NO (g) + 6 H 2 O(g) K c =? B. 3 Fe(s) + 4 H 2 O(g) < > Fe 3 O 4 (s) + 4 H 2 (g) K c =? C. COCl 2 (g) < > CO(g) + Cl 2 (g) K c =? D. PCl 3 (g) + Cl 2 (g) < > PCl 5 (g) K c =? E. NH 3 (aq) + H 2 O(l) < > NH + 4 (aq) + OH - (aq) K c =? F. H 2 O(s) < > H 2 O(g) K c =? G. 2 NaHCO 3 (s) < > Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) K c =? 82. Are pure solids, pure liquids and solvents included in the equilibrium constant expression? Why not? 83. Given the chemical reaction below, calculate the numerical value of the equilibrium constant for the reactions in A, B, C and D. N 2 (g) + 3 H 2 (g) < > 2 NH 3 (g) K c = at 25 C A. 2 NH 3 (g) < > N 2 (g) + 3 H 2 (g) K c =? B. 3 N 2 (g) + 9 H 2 (g) < > 6 NH 3 (g) K c =? C. ½ N 2 (g) + 3/2 H 2 (g) < > NH 3 (g) K c =? D. NH 3 (g) < > ½ N 2 (g) + 3/2 H 2 (g) K c =? 84. Write the equilibrium constant expression (K p ) in terms of partial pressures for the general reaction: aa(g) + bb(g) > cc(g) + dd(g) where a, b, c, d are coefficients 85. Write the equilibrium constant expressions (K p ) for the reactions below: A. 2 SO 2 (g) + O 2 (g) < > 2 SO 3 (g) B. N 2 (g) + 3 H 2 (g) < > 2 NH 3 (g) C. 4 NH 3 (g) + 5 O 2 (g) < > 4 NO (g) + 6 H 2 O(g) 86. For the reaction in 85A, K c = at a temperature of 727 C. What is the numerical value of K p? 87. What is the general mathematical equation that can be used to calculate K c from K p? K c from K p? 88. Calculate the equilibrium constant in terms of partial pressures (K p ) for the reaction below: CO(g) + 3 H 2 (g) < > CH 4 (g) + H 2 O(g) K c = 3.91 at 927 C 89. Calculate the equilibrium constant in terms of concentrations (K c ) for the reaction below: CO(g) + H 2 O(g) < > CO 2 (g) + H 2 (g) K p = 0.58 at 727 C Under what conditions will K c and K p be equal (have the same value)?

4 Chemical Equilibrium: Reaction Quotient and Concentration Problems 90. What is the form of the reaction quotient Q c? What concentrations are used to calculate the numerical value of the reaction quotient? 91. Consider the reaction below with the given initial concentrations of products and reactants. Is the reaction already at equilibrium? If not, which way will it proceed to attain equilibrium? (HINT: Calculate Q c and compare its value to K c ) CO(g) + 3 H 2 (g) < > CH 4 (g) + H 2 O(g) K c = 3.91 at 927 C Initial: M M M M 92. If Q c is greater than K c, which way will the reaction proceed in order to attain equilibrium (left to right or right to left)? If Q c is less than K c, which way will the reaction proceed in order to attain equilibrium (left to right or right to left)? If Q c is equal to K c, which way will the reaction proceed in order to attain equilibrium (left to right or right to left)? 93. A 10.0 L vessel contains mol CO 2 and mol of CO. Some solid carbon C(s) is added and temperature is raised to 1000 C. Will more CO(g) form? The reaction is shown below. (HINT: Calculate Q c and compare it to K c ) CO 2 (g) + C(s) < > 2 CO(g) K c =1.17 at 1000 C 94. Consider the reaction given below 2 NOBr(g) < > 2 NO(g) + Br 2 (g) K c = at 24 C For each of the following compositions, describe in which direction the reaction will go to reach equilibrium. A. [NOBr]= M C. [NOBr]=0.500 M [NO]= M [NO]= M [Br 2 ]= M [Br 2 ]= M B. [NOBr]=0.115 M [NO]= M [Br 2 ]= M 95. Calculate the missing concentrations and the numerical value of K c for the following: A. 2 ICl(g) < > I 2 (g) + Cl 2 (g) K c =? Initial: M 0.22M Change: At Eq: 0.09M B. H 2 (g) + I 2 (g) < > 2 HI(g) K c =? Initial: M Change: M At Eq: C. 3 A + 2 B < > C + 2 D K c =? Initial: 6.00M 2.00M 1.00M 0 Change: 0.50M At Eq: 96. If 0.50 mol PCl 5 is injected into a 2.0L vessel, what will be the equilibrium concentrations of all species? PCl 5 (g) < > PCl 3 (g) + Cl 2 (g) K c =1.8 at 250 C

5 97. Initially a mixture contains mol each of N 2 and O 2 in an 8.00 L vessel. Calculate the equilibrium concentration of gaseous NO. N 2 (g) + O 2 (g) < > 2 NO(g) K c = Consider the reaction below which is already at equilibrium. Calculate the partial pressures of NO 2 and N 2 O 4 when the total pressure is increased to 20 atm. 2 NO 2 (g) < > N 2 O 4 (g) At Eq: 0.25 atm 0.25 atm Chemical Equilibrium: Le Chatelier s Principle 99. State Le Chatelier s Principle. List the types of stresses that can be applied to a system at equilibrium Why does the addition of a catalyst to an equilibrium mixture not change the equilibrium? Why does an increase of pressure due to addition of an inert gas not change the equilibrium concentrations of reactants and products? 101. If the temperature is changed, does the value of K C change? Why? 102. Consider the reaction below 2 ICl(g) < > I 2 (g) + Cl 2 (g) K c =0.12 At Eq: 0.26M 0.09M 0.09M What will happen (in which direction will the reaction proceed) if the following changes occur? A M ICl is added B M I 2 is added C. All of the Cl 2 is removed 103. Consider the reaction below which is at equilibrium BaSO 4 (s) < > BaO(s) + SO 3 (g) What will happen if more solid BaSO 4 is added? more solid BaO is added? 104. For a reaction at equilibrium, the pressure is increased by decreasing the volume. Will the equilibrium concentrations of species be altered? Is the numerical value of K c or K p changed? 105. Consider the reaction below which has already attained equilibrium N 2 O(g) + NO 2 (g) < > 3 NO(g) K c = at 200 C Which way will the reaction proceed to reach equilibrium if the pressure is increased by decreasing the volume? 106. For the reaction CO(g) + H 2 O(g) < > H 2 (g) + CO 2 (g) Will a decrease in pressure by increasing volume alter the equilibrium concentrations of all species? 107. A certain reaction is exothermic ()H= negative). What happens to the numerical value of K c if the temperature is increased? if the temperature is decreased? 108. A certain reaction is endothermic ()H= positive). What happens to the numerical value of K c if the temperature is increased? if the temperature is decreased? 109. Consider the reaction below which has already attained equilibrium N 2 H 4 (g) + H 2 (g) < > 2 NH 3 (g) )H= kj In which direction will the reaction proceed if the temperature is increased? 110. Consider the reaction below which has already attained equilibrium N 2 (g) + O 2 (g) < > 2 NO(g) )H=+181 kj In which direction will the reaction proceed if the temperature is increased?

6 CHEM 116-Dr. Babb s Section Answer Key to Exam II Lecture Problem Sheet 61. Integrated form of zeroth order rate law: [A] t = -kt + [A] 0 Half-life is the time it takes for half of original amount of reactant to react such that only half remains. For zeroth order reaction, half-life equation is J = [A] 0 /2k and half-life is not constant but directly proportional to initial concentration of reactant (J%[A] 0 or as [A] 0 8, J8). For zeroth order reaction, a plot of [A] t vs. t will yield a straight line. 62. Integrated form of first order rate law: ln[a] t = -kt + ln[a] 0 For first order reaction, half-life equation is J = ln2/k and half-life is a constant and not dependent on initial concentration of reactant. For first order reaction, a plot of ln[a] t vs. t will yield a straight line. A. [A] after 1 minute is 0.50 M; Half-life (J)= sec; it will take sec for 75% of A to react; after three half-lives M of A remains; the initial rate of the reaction is M/sec. 63. Integrated form of second order rate law: 1/[A] t = kt + 1/[A] 0 For second order reaction, half-life equation is J = 1/k[A] 0 and half-life is not constant but inversely proportional to initial concentration of reactant (J%1/[A] 0 or as [A] 0 8, J9). For first order reaction, a plot of 1/[A] t vs. t will yield a straight line. 64. Reaction is second order in [HI] and Rate Law is: Rate = k[hi] 2 ; k = M -1 min -1 ; J=0.696 min; it will take 2.8 min for 20% of the original amount of HI to remain. 66. Net Rxn: 2 NO 2 + O 3 > N 2 O 5 + O 2 ; Both steps are bimolecular; Rate = k[no 2 ][O 3 ]; The NO 3 is an intermediate. 67. Net Rxn: 2 O 3 > 3 O 2 ; First elementary step is unimolecular while second is bimolecular; Rate = k[o 3 ][O]; The O is an intermediate. 68. Net Rxn: 2 O 3 > 3 O 2 ; First elementary step is unimolecular while second and third are bimolecular; the O and NO 2 are intermediates; the NO is a catalyst. 70. Activation Energy (E A )= J/mol; Pre-exponential factor (A) = M -1 sec -1 ; Specific rate constant (k) at 427 C = M -1 sec Activation Energy (E A )= J/mol; Pre-exponential factor (A) = M -1 sec -1 ; Specific rate constant (k) at 427 C = 0.17 sec Numerical value of K depends only on T but does not depend on initial concentrations of reactants or products. 78. Since K C for this reaction is less than 1 and K C =k f /k r, then k r > k f. Since K C =[Products]/[Reactants] and for this reaction K C <1, then [Reactants]>[Products]. 79. Since K C for this reaction is greater than 1 and K C =k f /k r, then k f > k r. Since K C =[Products]/[Reactants] and for this reaction K C >1, then [Products]>[Reactants]. 80. When K C > 10 3, then [Products] >> [Reactants] and reaction proceeds essentially 100% to completion. When K C < 10-3, then [Reactants] >> [Products] and reaction proceeds hardly at all to completion. When 10-3 < K C < 10 3, then [Products] ~ [Reactants] and appreciable amounts (experimentally measurable quantities) of both reactants and products are present. 81. A. K c = [NO] 4 [H 2 O] 6 /[NH 3 ] 4 [O 2 ] 5 B. K c = [H 2 ] 4 /[H 2 O] 4

7 C. K c = [CO][Cl 2 ]/[COCl 2 ] D. K c = [PCl 5 ]/[PCl 3 ][Cl 2 ] E. K c = [NH + 4 ][OH - ]/[NH 3 ] F. K c = [H 2 O] G. K c = [CO 2 ] 82. Concentrations for pure solids, pure liquids and the solvent are not included in the equilibrium constant expression because these concentrations depend only on density and don t vary (i.e. are constant). These constant concentrations are already included into the numerical value of K. 83. A B C D A. K p = (p SO3 ) 2 /(p SO2 ) 2 (p O2 ) B. K p = (p NH3 ) 2 /(p H2 ) 3 (p N2 ) C. K p = (p NO ) 4 (p H2O ) 6 /(p NH3 ) 4 (p O2 ) K p =K c (RT) )n or K c =K p (RT) -)n ; Whenever )n gas = 0, then K c =K p. 90. Reaction quotient (Q C ) has same form as K C except non-equilibrium or initial concentrations are used to calculate Q C. 91. Q C =6.25 and Q c > K c. Thus this reaction is not at equilibrium and will proceed from right to left in order to reach equilibrium. 92. If Q c > K c, reaction proceeds from right to left in order to reach eq. If Q c < K c, reaction proceeds from left to right in order to reach eq. If Q c = K c, reaction is at equilibrium and will not react in a net way in either direction. 93. Q C =0.202 and Q c < K c. Thus this reaction is not at equilibrium and will proceed from left to right in order to reach equilibrium and more gaseous CO will be formed. 94. A. Q C = and Q c > K c. Thus this reaction is not at equilibrium and will proceed from right to left in order to reach equilibrium. B. Q C = and Q c = K c. Thus this reaction is at equilibrium and will not react in a net way in either direction. C. Q C = and Q c < K c. Thus this reaction is not at equilibrium and will proceed from left to right in order to reach equilibrium. 95. A. 2 ICl(g) < > I 2 (g) + Cl 2 (g) K c =0.1 Initial: M 0.22M Change: +0.26M -0.13M -0.13M At Eq: 0.26M 0.09 M 0.09M B. H 2 (g) + I 2 (g) < > 2 HI(g) K c =137 Initial: M Change: M M M At Eq: 0.365M 0.365M 4.270M

8 C. 3 A + 2 B < > C + 2 D K c = Initial: 6.00M 2.00M 1.00M 0 Change: -0.75M -0.50M +0.25M 0.50M At Eq: 5.25M 1.50M 1.25M 0.50M 96. [Cl 2 ] = [PCl 3 ] = 0.22 M; [PCl 5 ]=0.03 M M 98. p NO2 =2.1 atm; p N2O4 =18 atm 102. A. reaction proceeds from left to right in order to regain eq. B. reaction proceeds from right to left in order to regain eq. C. reaction proceeds from left to right in order to regain eq Addition of more solid BaSO 4 or BaO will have no effect on the eq. because concentrations of pure solids are not included in the eq. constant expression If total pressure is altered, the numerical value of K is not changed. The only factor influencing the value of K is temperature. However, a change in the total pressure does change the partial pressures of gaseous species and may remove the reaction from eq as P increases V decreases and reaction will proceed toward side of equation that occupies less volume. This means that reaction will proceed toward side of equation that has less moles of gas. In this instance, reaction will proceed toward the reactants in order to regain eq as P decreases due to V increase reaction will proceed toward side of equation that occupies more volume (i.e. toward side of equation with more moles of gas). However, in this case both sides of equation have same # moles of gas and thus occupy same volume. As a result, the eq. of this reaction is unaffected by changes in P due to V change as T increases, K C decreases or as T decreases, K C increases for an exothermic reaction as T increases, K C increases or as T decreases, K C decreases for an endothermic reaction as T increases, Heat increases so heat on the product side is increased and reaction will proceed from right to left in order to regain eq as T increases, Heat increases so heat on reactant side is increased and reaction will proceed from left to right in order to regain eq.