Question

The acid-catalyzed reaction of an organic compound propanone(CH3COCH3) with iodine may be represented by the equation

CH3COOH + I2----->CH2ICOCH3+ (H+ve charge)+ (I-charge)

other conditions remaining same, what will be the initial rate of reaction if the concentrations of propanone,iodine and acid,all are doubled?

Answer 1

@summerforever21

Answer 2

I can't figure this out, i have tried but i keeping getting syntax errors but maybe he can help you, he's good at chemistry ----> aaronq

Answer 3

@aaronq

Answer 4

3 questions below your is Ivanthebear click him then look for a skelton cat picture and that aaronq

Answer 5

there maybe he can help you

Answer 6

I hope

Answer 7

yeah i tried sorry I didnt get any good results for you

Answer 8

Its alright...atleast you tried :)

Answer 9

thank you, good luck too you :D

Answer 10

sorry, i disabled the notifications, so when you type my name it doesn't alert me.. anyway lets take alook

Answer 11

thank you for helping I tried but kept getting errors and i know your good so i thought about you helping him

Answer 12

i noticed that the reaction is written backwards, it should be:

CH2ICOCH3 + H^+ + I^- -> CH3COOH + I2

HI is the acid

the rate is always dependent on the "rate determining step", and unfortunately you can't figure that out without more data, but based on what is given i would say that the rate would quadruple (increase by 4).

rate = [CH2ICOCH3][HI]

rate = [1][1] =1
rate = [2][2] = 4
rate =[4][4]=16

Answer 13

maybe that why I was having trouble cus the reaction was written backwards

Answer 14

ryaan .-.

Answer 15

I'd also like to point out that this looks like an \(S_{n2}\) reaction so I'm going to support @aaronq this is a second order reaction and is consistent with his argument that propanone conc. doubles, and rate quadruples. If you remember, 2nd order rate reaction definition.