M5 Exam

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Jan 9, 2024

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Question 1 (5.): (1): In the reaction of 0.200 M gaseous N205 to yield NO2 gas and 02 gas as shown below: 2 N205 (g) > 4 NO2 (g) + 02 (g) the following data table is obtained: Time (sec) [N205] [02] 0 0.200 M 0 300 0.180 M 0.010 M 600 0.162 M 0.019 M 300 0.146 M 0.027 M 1200 0.132 M 0.034 M 1800 0.110 M 0.045 M 2400 0.096 M 0.052 M 3000 0.092 M 0.054 M (a) Use the [02] data from the table to calculate the average rate over the measured time interval from 0 to 3000 secs. (b) Use the [02] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 300 sec). (c) Use the [02] data from the table to calculate the instantaneous rate late in the reaction (2400 secs to 3000 secs). (d) Explain the relative values of the average rate, early instantaneous rate and late instantaneous rate. Answer: (a) The average rate over the measured time interval from 0 to 3000 secs is: rate = A[02] / At =(0.054 - 0) / 3000 - 0 = 1.80 x 10”-5 mol/L's (b) The instantaneous rate early in the reaction from 0 to 300 secs is: rate = A[02] / At =(0.010-0) /300 - 0 = 3.33 x 107-5 mol/L's (c) The instantaneous rate late in the reaction from 2400 to 3000 secs is: rate = A[02] / At = (0.054 - 0.052) / 3000 - 2400 = 3.33 x 10”-6 mol/L-s (d) It can be seen that the early instantaneous rate is the largest since the concentrations of reactants is highest during the earliest stages of the reaction and the late instantaneous rate is smallest since the concentrations of reactants is lowest during the late stages of the reaction.

(2): In the reaction of gaseous CH3CHO to yield CH4 gas and CO gas as shown below: CH3CHO (g) - CH4 (g) + CO (g) the following data table is obtained: Time (sec) [CH3CHO] 0 0.0500 M 1200 0.0300 M 2000 0.0240 M 6000 0.0120 M 10000 0.0080 M 15000 0.0056 M 20000 0.0043 M (a) Use the [CH3CHO] data from the table to calculate the average rate over the measured time interval from 0 to 20000 secs. (b) Use the [CH3CHO] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 1200 sec). (c) Use the [CH3CHO] data from the table to calculate the instantaneous rate late in the reaction (15000 secs to 20000 secs). (d) Explain the relative values of the average rate, early instantaneous rate and late instantaneous rate. Question 2: (1): The following rate data was obtained for the reaction: 2 ClO2 + 2 OH- - CIO3- + ClO2- + H20 the following data table is obtained: Experiment # [ClO2] [OH-] rate 1 0.060 0.030 2.48 x 107-2 2 0.020 0.030 2.76 x 107-3 3 0.020 0.090 8.28 x 107-3

Determine the reaction order with respect to (1) CIO2 and (2) OH-, (3) write the rate law and then (4) determine the value of the rate constant, k. Answer: Rate = k [CIO2]x [OH-]y If data from experiments 1 and 2 was substituted into the equation we would obtain (2.48 X 107-2)/(2.76 X 10~-3) = (k[.060]*x X [.030]*y)/(k[.020]~x X [.030]*y) and if we cancel all common terms we would obtain (2.48 X 107-2)/(2.76 X 107-3) = ([.060]*x)/([.020]*x) 9=(3)"x (1) which yields x = 2 as the order of reaction with respect to ClO2 If data from experiments 2 and 3 was substituted into the equation we would obtain (2.76 X 107-3)/(8.28 X 107-3) = (k[.020]*x X [.030]*y)/(k[.020]~x X [.090]*y) and if we cancel all common terms we would obtain (2.76 X 107-3)/(8.28 X 107-3) = ([.030]y X [.090]*y) .333=(.333)7y (2) which yields y = 1 as the order of reaction with respect to OH-. (3) the overall rate law can now be written as follows: Rate = k [ClO02]*2 [OH-]*1 (4) and using the data from experiment 1 we can determine the rate constant as follows: 2.48 x 10~-2 = k [0.060]*2 [0.030] k =2.48 x 10"-2 / [0.060]*2 [0.030] = 229.6 (2): The following rate data was obtained for the reaction which takes place in a solution of OH-: ClO-+1- - 10- + CI- the following data table is obtained: Experiment | [I-] [CIO-] [OH-] rate # 1 0.0030 | 0.0010 | 1.00 1.8 x 1074 2 0.0030 | 0.0020 | 1.00 3.6 x 1074 3 0.0060 | 0.0020 | 1.00 7.2x107-4

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4 0.0030 | 0.0010 | 0.50 9.0x10%-5 Determine the reaction order with respect to (1) ClO-, (2) |- and (3) OH-, (4) write the rate law and then (5) determine the value of the rate constant, k. Answer: 1) Rate = k [CIO-]”x [I-]*y [OH-]*z If data from experiments 1 and 2 was substituted into the equation we would obtain: (1.8 X 107-4)/(3.6 X 10”-4)= (k[.0010]*x [.0030]*y [1.00]7z)/ (k[.0020]*x [.0030]7y [1.00]”z) (1.8 X 107-4)/(3.6 X 107-4)=[.0010]*x / [.0020]*x .50=(.5)"x which yields x = 1 as the order of reaction with respect to CIO- If data from experiments 2 and 3 was substituted into the equation we would obtain: (3.6 X 107-4)/(7.2 X 107-4)= (k[.0020]*x [.0030]*y [1.00]”z)/ (k[.0020]*x [.0060]*y [1.00]”z) (3.6 X 107-4)/(7.2 X 107-4)=[.0030]*y/[.0060]"y .50=(.5)"y 2) which yields y = 1 as the order of reaction with respect to I-. If data from experiments 1 and 4 was substituted into the equation we would obtain: (1.8 X 107-4)/(9.0 X 107-5)= (k[.0010]”x [.0030]*y [1.00]”z) / (k[.0010]”x [.0030]*y [.50]7z) (1.8 X 107-4)/(9.0 X 107-5)=[1.00]*z / [.50]*z 2.0=(2.0)*z 3) which yields z = 1 as the order of reaction with respect to OH-. The overall rate law can now be written as follows: 4) Rate = k [CIO-]1 [I-]1 [OH-]1 5) and using the data from experiment 1 we can determine the rate constant as follows: 1.8 x 10-4 = k [0.0010] [0.0030] [1.00] k=1.8 x10-4 /[0.0010] [0.0030] [1.00] = 60

Question 3 (): (1): A sample of wood from an ancient tomb was found to contain 15.7% "*14C content as compared to a present-day sample. The t_1/2 for ~14Cis 5720 yrs. What is the age of the wood? Answer: 0.693 =kt _1/2 0.693 =k (5720 years) k=0.693/5720=1.2115x 1074 In[A] -In[A]*0=-kt In15.7-In100=-(1.2115 x 107-4) t 2.7537 -4.6052 =-(1.2115x 107-4) t t (age)=-1.8515/-1.2115 x 10~-4 = 15,283 years (2): A sample of paper from an ancient scroll was found to contain 39.5% 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 years. What is the age of the scroll? Answer: 0.693=kt 1/2 0.693 =k (5720 years) k=0.693/5720=1.2115x 10*-4 In[A]-In[A]*0=-kt In 39.5 - In 100 = - (1.2115 x 107-4) t 3.6763 -4.6052 =-(1.2115x 107-4) t t=-0.9289/-1.2115 x 107-4 = 7667.4 years Question 4 (5.4): Using the potential energy diagram below, state whether the reaction described by the diagram is endothermic or exothermic and spontaneous or nonspontaneous. Explain your answer.

Transtion State i Eact(small) T Reactants A\ AH=- Energy Products Progress of Reaction — Answer: Smal E_act = spontaneous AH- = exothermic Explanation: If the reaction has a negative heat of reaction (-AH), it will be exothermic. If the reaction has a positive heat of reaction (+AH), it will be endothermic. A reaction with relatively large E_act will be a nonspontaneous. A reaction with relatively small E_act will be a spontaneous. Question 5 (5.5): (1): Calculate the K_c for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H20 and corresponding amounts of CO, H2, and CHA4. CO (g) +3H2(g) &S CH4 (g) + H20 (g) Answer: H20 =0.198 mole / 5.00 L = 0.0396 M CH4 =0.198 mole / 5.00 L = 0.0396 M CO =0.302 mole / 5.00 L = 0.0604 M H2 =0.906 mole /5.00 L=0.1812 M K_c=([CH4][H20]) / ([CO][H2]*3) = ([.0396][.0396]) / ([.0604][.1812]) = 4.36

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(2): Calculate the K_c for the following reaction if an initial reaction mixture of 0.700 mole of NH3 and 0.910 mole of 02 in a 1.00 liter container forms an equilibrium mixture containing 0.230 mole of N2 and corresponding amounts of NH3, 02, and H20. 4 NH3 (g) +3 02 (g) 5 2 N2 (g) + 6 H20 (g) Answer: N2 = 0.230 mole of (as stated) H20 =0.230 x 6/2 mole (6 mole of H20 forms for every 2 mole of N2 that is formed) NH3 = 0.700 - 4/2 x 0.230 mole (4 mole of NH3 reacts for every 2 mole of N2 that is formed) 02 =0.910 - 3/2 x 0.230 mole (3 mole of 02 reacts for every 2 mole of N2 that is formed) Change all amounts to moles/L before entering in Kc expression: N2 =0.230 mole /1.00L=0.230 M H20 =0.690 mole / 1.00 L = 0.690 M NH3 =0.240 mole /1.00 L=0.240 M 02 =0.565mole /1.00L=0.565 M K_c=([N2]~2 [H20]*6) / (INH3]*4 [02]~3) = ([.230]*2 [.690]"6) / ([.240]*4 [.565]"3) = 9.54 Questions 6 (): Explain in detail how a catalyst increases the rate of reaction Answer: A catalyst increases the rate of a chemical reaction by combining with reactants to form an intermediate state achieving lower energy activation. Question 7: Problem to solve: The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants? Be sure to explain your answer. 2H2(g)+S2(g) S 2H2S(g) K _c=9.39 X 10A-5

Answer: The very small K_c indicated that this mixture will be composed of mostly reactants Explanation: If the K_c value is abnormally larger than 1 (example: 4.1 x 1078), the equilibrium mixture will be predominately products. If the K_c value is abnormally smaller than 1 (example: 4.6 x 107-31), the equilibrium mixture will be predominately reactants. If the value of K_c is neither abnormally large nor abnormally small but = 1, the equilibrium mixture will contain a significant concentration of both products and reactants. Question 8 (chapter 5.7): (1): The equilibrium reaction below has the following equilibrium mixture concentrations: SO3 =[0.0894], SO2 = [0.400] and 02 = [0.200] with Kc = 4.00 If the concentration of SO3 at equilibrium is increased to [0.300], how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2503 (g) 5 2S02 (g) + 02 (g) Answer: Kc = 4.00 when SO3 = [0.0894], SO2 = [0.400] and 02 = [0.200] When SO3 = [.300], Q = ([.400]*2 [.200]) / [.300]*2 = .356 The reaction must shift briefly in the forward direction to decrease the [SO3] to come back to equilibrium. This is in agreement with Qc < Kc which also predicts the reaction will proceed to the right. (2): The equilibrium reaction below has the following equilibrium mixture concentrations: SO3 =[0.0894], SO2 = [0.400] and 02 = [0.200] with Kc = 4.00 If the concentration of O2 at equilibrium is decreased to [0.100], how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2503 (g) 52502 (g) + 02 (g) Answer:

Kc = 4.00 when SO3 = [0.0894], SO2 = [0.400] and 02 = [0.200] When SO3 = [.100], Q = ([.400]*2 [.100]) / [.0894]#2 = 2.0 The reaction must shift briefly in the forward direction to increase the [02] to come back to equilibrium. This is in agreement with Qc < Kc which also predicts the reaction will proceed to the right. Questions 9 (5.7): (1): The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is decreased from 4.00 liters to 2.00 liters, how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2503 (g) 5 2S02 (g) +02 (g) Answer: When volume decreases from 4.00 to 2.00, the pressure doubles and the concentration of all gases (SO3, SO2 and 02) doubles so: (at equilibrium) Qc= Kc= ([SO2]72] [02])/[SO3]*2=4.0 (volume halved = pressure doubled = conc doubled) Qc = ([SO2]*2] [202])/[2 SO3]*2= 2Kc The reaction must shift briefly in the direction that decreases the pressure by going toward the side with the lesser moles of gas (reverse direction : 3 moles of gas yields 2 moles of gas) to come back to equilibrium. This is in agreement with Qc > Kc: the reaction will proceed to the left (in the direction of the reactants). (2): The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is increased from 3.00 liters to 9.00 liters, how and for what reason will the equilibrium shift. Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. 2503 (g) 52502 (g) + 02 (g) Answer: When volume increases from 3.00 to 9.00, the pressure is cut to 1/3 of original and the concentration of all gases (SO3, SO2 and 02) decreases to 1/3 so: (at equilibrium) Qc= Kc= ([SO2]*2] [02])/[SO3]*2=4.0 (volume tripled = Pi/3= conc/3) Qc= ([1/3 SO2]*2 [1/3 02])/[1/3 SO3]*2= Kc/3 The reaction must shift briefly in the direction that increases the pressure by going toward the side with the greater moles of gas (forward direction: 2 moles of gas yields 3 moles of gas) to come back to

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equilibrium. This is in agreement with Qc < Kc: the reaction will proceed to the right (in the direction of the products). Questions 10 (5.7): (1): The equilibrium reaction below has the Kc = 0.250 at 25°C. If the temperature of the system at equilibrium is decreased to 0°C, how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change. 2502 (g)+02(g) 52S03(g) AHMN0=-94.1kJ Answer: If the temperature is decreased on this reaction at equilibrium which has a -AH”0, the reaction will briefly shift in the direction that produces some heat [-AH”0 indicates an exothermic forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of SO3 (in the numerator of Kc) and decrease the concentration of SO2 and 02 (in the denominator of Kc) which will increase the value of Kc. (2): The equilibrium reaction below has the Kc = 0.254 at 25°C. If the temperature of the system at equilibrium is decreased to 0°C, how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change. CH4 (g) +H20 (g) - CO (g) + 3H2 (g) AHA0=-+206.2 kJ Answer: If the temperature is decreased on this reaction at equilibrium which has a +AHA0, the reaction will briefly shift in the direction that produces some heat [+AH”0 indicates an endothermic forward reaction] so this reaction will shift in the reverse direction. Simultaneously, this reverse shift will decrease the concentration of CO and H2 and increase concentration of CH4 and H20 which will decrease value of Kc. (3): The equilibrium reaction below has the Kc = 4.00 at 25°C. If the temperature of the system at equilibrium is increased to 100°C, how and for what reason will the equilibrium shift. Also show and explain how and why the Kc value will change. 2503 (g) 5 2502 (g) + 02 (g) AHANO=+94.1 k! Answer:

If the temperature is increased on this reaction at equilibrium which has a +AH”0, the reaction will briefly shift in the direction that uses up some of the added heat [+AH”0 indicates an endothermic (heat absorbing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of SO2 and 02 (in the numerator of Kc) and decrease the concentration of SO3 (in the denominator of Kc) which will increase the value of Kc.

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