In: Chemistry
The iodination of acetone, CH3COCH3, in aqueous solution is catalyzed by the hydrogen ion H+:
I2+CH3COCH3H+?HI+CH3COCH2I
The reaction can be followed visually by adding starch. The purple starch-iodine complex forms immediately, then disappears when the iodine has been consumed. The kinetics of the reaction may be determined by recording the time required for the color to disappear, which is the time required for the iodine, I2, to be consumed.
The following data were acquired in an iodination experiment involving acetone. All reaction times are in terms of the rate of disappearance of I2.
Trial | Volume of 0.0010M I2 (mL) |
Volume of 0.050 M HCl (mL) |
Volume of 1.0 Macetone (mL) |
Volume of water (mL) |
Temperature (?C) |
Reaction time (s) |
A | 5.0 | 10.0 | 10.0 | 25.0 | 25.0 | 130 |
B | 10.0 | 10.0 | 10.0 | 20.0 | 25.0 | 249 |
C | 10.0 | 20.0 | 10.0 | 10.0 | 25.0 | 128 |
D | 10.0 | 10.0 | 20.0 | 10.0 | 25.0 | 131 |
E | 10.0 | 10.0 | 10.0 | 20.0 | 42.4 | 38 |
part A
In the general rate law
rate=k[I2]^X[H+]^Y[CH3COCH3]^Z
what are the values of X, Y, and Z?
I2 + CH3COCH3 (H+) ---> HI + CH3COCH2I
Trial A
Concentration of I2 = moles of I2/solution volume = (0.001*5 )/( 5+10+10+25 ) = 0.0001M
Rate of reaction = rA = (change in concentration)/time = 0.0001/130 = 0.77*10^(-6)M
Trial B
Concentration of acetone = moles of H+/solution volume = (1.0*10 )/( 10+10+10+20) = 0.2 M
Concentration of H+ = moles of H+/solution volume = (0.05*10 )/( 10+10+10+20) = 0.01M
Concentration of I2 = moles of I2/solution volume = (0.001*10 )/( 10+10+10+20) = 0.0002M
Rate of reaction = rB = (change in concentration)/time = 0.0002/249 = 0.80*10^(-6)M
rate=k[I2]^X[H+]^Y[CH3COCH3]^Z
Only I2 changes between A and B
rA/rB = (.0001/.0002)^x = .77/.80 =0.96 = approximately 1
So x= 0
Trial C
Concentration of H+ = moles of H+/solution volume = (0.05*20 )/( 10+10+10+20) = 0.02M
Concentration of I2 = moles of I2/solution volume = (0.001*10 )/( 10+10+10+20) = 0.0002M
Rate of reaction = rC = (change in concentration)/time = 0.0002/128 = 1.56*10^(-6)M
rC and rB, only H+ changes
rC/rB = (.02/.01)^y = 1.56/.80 = 1.95 approximately 2
y = 1
Trial D
Concentration of acetone = moles of H+/solution volume = (1.0*20 )/( 10+10+10+20) = 0.4 M
Concentration of I2 = moles of I2/solution volume = (0.001*10 )/( 10+10+10+20) = 0.0002M
Rate of reaction = rD = (change in concentration)/time = 0.0002/131 = 1.53*10^(-6)M
rD and rB, only acetone changes
rD/rB = (.4/.2)^z = 1.53/.80 = 1.9 approximately 2
So z= 1
X=0, Y=1, Z=1