Question

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 0.0010 M I2 (mL )	Volume of 0.050 M HCl (mL )	Volume of 1.0 M acetone (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 C

What is the rate constant at 42.4 ∘C based on the data collected for trial E?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer: According to the question : Here the balanced equation is shown as

  CH3COCH3 + I2 + H+ ------> CH3COCH2 + HI

Now using the r = k[CH3COCH3]^m [H+]ⁿ [I2]^p

but the rate is still almost equal so order with respect to I2 is 0...as there is no change in rate with change in concentration of I2

rate doubles when conc. of H+ doubles keeping conc. of acetone and I2 constant so order with respect to H+ is 1

and also for the acetone the order is 1 .

Hence the , final equation is equal to : rate = k [acetone] [H+]

So, the value of rate constant at 42.4 ⁰C is

[acetone] initial = 0.2 M

[H+] initial = 0.01 M

so rate= k [acetone] [H+]

5.37 X 10^-6 = k X 0.2 X 0.01

k = 2.68 X 10^-3 ,

and the order of the reaction is 2 so the unit of the rate constant is  L·mol^?1·s^?1.

So , It is all about the given question , thank you :)