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4. A third reaction mixture was made up in the following way: 5 mL 4.0 M acetone + 20 mL 1.0 M HCl + 5 mL 0.0050 M 1₂ + 20 mL H₂O If the reaction is zero order in I₂, how long would it take for the I₂ color to disappear at the temperature of the reaction mixture in Problem 3? seconds

4. A third reaction mixture was made up in the following way: 5 mL 4.0 M acetone + 20 mL 1.0 M HCl + 5 mL 0.0050 M 1₂ + 20 mL H₂O If the reaction is zero order in I₂, how long would it take for the I₂ color to disappear at the temperature of the reaction mixture in Problem 3? seconds

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can I please get some help solving questions 3 (c) and question 4? Thank you. 

Advance Study Assignment: The lodination of Acetone 1. In a reaction involving the iodination of acetone, the following volumes were used to make up the reac- tion mixture: mol HCl = 1.010L = .01 mer 0502 2.20 M a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = MAX V, where MA is the molarity of A and V is the volume in liters of the solution of A that was used. molacedone = 4.0M 0.005 L=0.02 mol 0.020 moles acetone 5 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I₂ + 25 mL H₂O b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again, moles of A MA= V of soln. in liters C. 0.020-40 M 0.050 L c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H* ion and I, as in the original mixture? reduce the amount of H₂0 by and 5 ML adding 5mL of 4.0 Macetone. 110ML 4.0 M acetone + 10mL 1.0 M HCl + 10mL 0.0050 MI₂ + 20 ML H₂O 2. Using the reaction mixture in Problem 1, a student found that it took 510 seconds for the color of the 1₂ to disappear. a. What was the rate of the reaction? Hint: First find the initial concentration of I₂ in the reaction mix- ture, [1₂]o. Then use Equation 5. [1₂] = 0,001 M Mol 1₂.0050m..000=.000050 mol = 0.001 M 050L rate= -A[₂][12] 0.001 m -401] [12] AE E rate= 5-109000196 m/ 510 seconds = K[Acetone] [1₂] [4+] P 0.40 m/sec 20 X 10 m/see rate= b. Given the rate from Part (a), and the initial concentrations of acetone, H* ion, and I₂ in the reaction mixture, write Equation 3 as it would apply to the mixture. What are the unknowns that remain in the equation in Part (b)? k m M acetone 2.0X10 2.0 X16-bM/sec = K[020] [0,0010] "B.267P M/sec n P (continued on following page)
Transcribed Image Text:Advance Study Assignment: The lodination of Acetone 1. In a reaction involving the iodination of acetone, the following volumes were used to make up the reac- tion mixture: mol HCl = 1.010L = .01 mer 0502 2.20 M a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = MAX V, where MA is the molarity of A and V is the volume in liters of the solution of A that was used. molacedone = 4.0M 0.005 L=0.02 mol 0.020 moles acetone 5 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I₂ + 25 mL H₂O b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again, moles of A MA= V of soln. in liters C. 0.020-40 M 0.050 L c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H* ion and I, as in the original mixture? reduce the amount of H₂0 by and 5 ML adding 5mL of 4.0 Macetone. 110ML 4.0 M acetone + 10mL 1.0 M HCl + 10mL 0.0050 MI₂ + 20 ML H₂O 2. Using the reaction mixture in Problem 1, a student found that it took 510 seconds for the color of the 1₂ to disappear. a. What was the rate of the reaction? Hint: First find the initial concentration of I₂ in the reaction mix- ture, [1₂]o. Then use Equation 5. [1₂] = 0,001 M Mol 1₂.0050m..000=.000050 mol = 0.001 M 050L rate= -A[₂][12] 0.001 m -401] [12] AE E rate= 5-109000196 m/ 510 seconds = K[Acetone] [1₂] [4+] P 0.40 m/sec 20 X 10 m/see rate= b. Given the rate from Part (a), and the initial concentrations of acetone, H* ion, and I₂ in the reaction mixture, write Equation 3 as it would apply to the mixture. What are the unknowns that remain in the equation in Part (b)? k m M acetone 2.0X10 2.0 X16-bM/sec = K[020] [0,0010] "B.267P M/sec n P (continued on following page)
162 Experiment 20 Rates of Chemical Reactions, I. The lodination of Acetone 3. A second reaction mixture was made up in the following way: 5 mL 4.0 M acetone + 20 mL 1.0 M HCl + 10 mL 0.0050 M 1₂ + 15 mL H₂O a. What were the initial concentrations of acetone, H ion, and I₂ in the reaction mixture? Molac 4.0 M•.005 L = 0,02 mol = 0.410 M [accion] 05L Mol 4₁ = 1.0.020=02 MolI₂ = 60100050 = 0.00005 = .001 M M; [1₂] .0010 M [acetone] 0.40 M; [H+] 40 b. It took 250 seconds for the 1₂ color to disappear from the reaction mixture when it occurred at the same temperature as the reaction in Problem 2. What was the rate of the reaction? Fate == A[1₂] 001 M At = 0000040 = 4.0 x10-6 M/sec 250 Sec rate = -4.6x10-6 rate = 4. A third reaction mixture was made up in the following way: Write Equation 3 as it would apply to the second reaction mixture: K[Acetone] [1₂] "H 41.0 x 10-M/sec = .407.0010] [40] c. Divide the equation in Part (b) by the equation in Problem 2(b). The resulting equation should have the ratio of the two rates on the left side and a ratio of H* concentrations raised to the p power on the right. Write the resulting equation and solve for the value of p, the order of the reaction with respect to H. (Round off the value of p to the nearest integer.) Rate 3b 41,0X10 m/sec_ *[0.40] [0.001] [040] Rate 26 k[0.020][0.0010] (0.40 2.0 x10-6 sec 2= P=. 105 = 40M 5 mL 4.0 M acetone + 20 mL 1.0 M HCl + 5 mL 0.0050 M I₂ + 20 mL H₂O M/sec If the reaction is zero order in I₂, how long would it take for the I₂ color to disappear at the temperature of the reaction mixture in Problem 3? seconds
Transcribed Image Text:162 Experiment 20 Rates of Chemical Reactions, I. The lodination of Acetone 3. A second reaction mixture was made up in the following way: 5 mL 4.0 M acetone + 20 mL 1.0 M HCl + 10 mL 0.0050 M 1₂ + 15 mL H₂O a. What were the initial concentrations of acetone, H ion, and I₂ in the reaction mixture? Molac 4.0 M•.005 L = 0,02 mol = 0.410 M [accion] 05L Mol 4₁ = 1.0.020=02 MolI₂ = 60100050 = 0.00005 = .001 M M; [1₂] .0010 M [acetone] 0.40 M; [H+] 40 b. It took 250 seconds for the 1₂ color to disappear from the reaction mixture when it occurred at the same temperature as the reaction in Problem 2. What was the rate of the reaction? Fate == A[1₂] 001 M At = 0000040 = 4.0 x10-6 M/sec 250 Sec rate = -4.6x10-6 rate = 4. A third reaction mixture was made up in the following way: Write Equation 3 as it would apply to the second reaction mixture: K[Acetone] [1₂] "H 41.0 x 10-M/sec = .407.0010] [40] c. Divide the equation in Part (b) by the equation in Problem 2(b). The resulting equation should have the ratio of the two rates on the left side and a ratio of H* concentrations raised to the p power on the right. Write the resulting equation and solve for the value of p, the order of the reaction with respect to H. (Round off the value of p to the nearest integer.) Rate 3b 41,0X10 m/sec_ *[0.40] [0.001] [040] Rate 26 k[0.020][0.0010] (0.40 2.0 x10-6 sec 2= P=. 105 = 40M 5 mL 4.0 M acetone + 20 mL 1.0 M HCl + 5 mL 0.0050 M I₂ + 20 mL H₂O M/sec If the reaction is zero order in I₂, how long would it take for the I₂ color to disappear at the temperature of the reaction mixture in Problem 3? seconds
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