The iodination of acetone, CH3COCH3, in aqueous solution is catalyzed by the hyd

Question

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.

part A

In the general rate law

rate=k[I2]^X[H+]^Y[CH3COCH3]^Z

what are the values of X, Y, and Z?

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

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.