ASA-iodination

ASA-iodination - moles H + = (1.0 mol/L) x 0.010 L = 0.010...

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ASA for Expt. 14 1. a. 0.010 L x (4.0 mol/L) = 0.040 moles acetone b. M = 0.040 mol/0.050 L = 0.80 M acetone c. Use 20 mL of 4.0 M acetone. Mix with 10 mL of 0.10 M HCl, 10 mL of 0.0050 M I 2 and 10 mL water. The total volume will be 50 mL. 2. a. moles I 2 = (0.0050 mol/L) x 0.010 L = 5.0 x 10 -5 mol [I 2 ] 0 = 5.0 x 10 -5 mol/0.050 L = 0.0010 M rate = 0.0010 M/250 s = 4.0 x 10 -6 mol/L-s b. 4.0 x 10 -6 mol/L-s = k[acetone] m [I 2 ] n [H + ] p c. unknowns are k, m, n, p 3. a. moles acetone = (4.0 mol/L) x 0.010 L = 0.040 mol [acetone] 0 = 0.040 mol/0.050 L = 1.6 M moles I 2 = (0.0050 mol/L) x 0.010 L = 5.0 x 10 -5 mol [I 2 ] 0 = 5.0 x 10 -5 mol /0.050 L = 0.0010 M
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Unformatted text preview: moles H + = (1.0 mol/L) x 0.010 L = 0.010 mol [H + ] = 0.010 mol /0.050 L = 0.20 M b. rate = [I 2 ] /t = (0.0010 mol/L)/120 s = 8.3 x 10-6 mol/L-s 8.3 x 10-6 = k(1.60) m (0.20) p (0.0010) n c. 8.3 x 10-6 /4.0 x 10-6 = k(1.60/0.80) m (0.20/0.20) p (0.0010/0.0010) n 2.1 = 2 m ; m ≈ 1 4. This mixture contains twice as much I 2 as the one in #3. If the reaction is 0-order in I 2 , the rate of the reaction does not depend on [I 2 ]. If there is twice as much I 2 , it will take twice as long for the color to disappear at the same rate . So, it will take 240 seconds....
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This lab report was uploaded on 04/08/2008 for the course CHEM 128 taught by Professor All during the Spring '08 term at UConn.

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