Name Oicou moura Name NCOS Section CH UaL-0Oi feb a,3010 Advance Study Assignmen
ID: 982242 • Letter: N
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Name Oicou moura Name NCOS Section CH UaL-0Oi feb a,3010 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- on mixture: 5 mL. 4.0 M acetone + 10 mL. 1.0 M HCI+10 mL.0.0050 M I,+ 25 mL. H,O How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = M4 × V, where MA is the molarity of A and Vis the volume in liters of the solution of A that was used a. N W 0.03 moles acetone A+ o03 mous 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 V of soln. in liters 0.4 _M acetone 0.0 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? c. Sumlaer of mous oF A Holari CHa) ouumg Using the reaction mixture in Problemi,a studentfound that itt ok5ionun disappear 2. Using the reaction mixture in Problem 1, a student found that it took 510 seconds for Ihe color of the 1, to 2. neco r T T What was the rate of the reaction? Hint: First find the initial concentration of I2 in the reaction mix- ture, [Ilo- Then use Equation 5 a. rate = M/sec b. Given the rate from Part (a), and the initial concentrations of acetone, H' ion, and I, in the reaction mixture, write Equation 3 as it would apply to the mixture. rate = c. What are the unknowns that remain in the equation in Part (b)? (continued on following page)Explanation / Answer
2 a) rate of reaction = - d [I2] / dt = - 0.0050 / 510 = 9.8039 * 10-6M/sec
b ) rate =d [I2] / dt= - d [acetone ] / dt= - d [hcl] / dt
3a) no of moles = molarity (moles/lit )* volume in litre
acetone = 4 * 0.005 hcl = 1 * 0.02 I2 = 0.005 * 0.01
= 0.02 moles = 0.02 moles = 0.00005 moles
3b ) rate of reaction = - d [I2] / dt = - 0.0050 / 250 = 2 * 10-5 M/sec
rate =-d [I2] / dt= - d [acetone ] / dt= - d [hcl] / dt
4 ) rate= k [a]0 =-d [I2] / dt =
2 * 10-5 M sec-1 = -d 0.005 M / dt
dt = 0.005/ 2 * 10-5 = 0.0025 * 105 =250 sec
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