clarisse Posted January 18, 2006 Share Posted January 18, 2006 Right now I'm in my school trying to measure the rate of reaction of propanone and iodine... I've added sulphuric acid as a catalyst and even though I'm waiting patiently for something to happen the colorimeter keeps recording the same absorbance... (I've been running my first experiment with no success for 45 minutes already..) I have to do 5 different experiments varying the molarity of iodine and propanone like this: iodine propane 0.01 0.01 0.01 0.02 0.01 0.03 0.02 0.01 0.03 0.01 Am I waiting in vain? what do I have to do to get a measurable rate for this reaction? Link to comment Share on other sites More sharing options...
woelen Posted January 18, 2006 Share Posted January 18, 2006 Right now I'm in my school trying to measure the rate of reaction of propanone and iodine... I've added sulphuric acid as a catalyst and even though I'm waiting patiently for something to happen the colorimeter keeps recording the same absorbance... (I've been running my first experiment with no success for 45 minutes already..)I have to do 5 different experiments varying the molarity of iodine and propanone like this: iodine propane 0.01 0.01 0.01 0.02 0.01 0.03 0.02 0.01 0.03 0.01 Am I waiting in vain? what do I have to do to get a measurable rate for this reaction? What reaction do you expect between the iodine and the propanone? I think you need a LOT of patience for this reaction . Propanone (=acetone) and iodine do react, but only at high pH. Then you get the so-called haloform reaction, yielding iodide and iodoform, CHI3. The haloform reaction is quite fast. Link to comment Share on other sites More sharing options...
RyanJ Posted January 18, 2006 Share Posted January 18, 2006 What reaction do you expect between the iodine and the propanone? I think you need a LOT of patience for this reaction . Propanone (=acetone) and iodine do react' date=' but only at high pH. Then you get the so-called haloform reaction, yielding iodide and iodoform, CHI3. The haloform reaction is quite fast.[/quote'] Would increasing the temperature of the reactants increase the rate of reaction? Cheers, Ryan Jones Link to comment Share on other sites More sharing options...
woelen Posted January 19, 2006 Share Posted January 19, 2006 I don't know of any fast reaction between iodine and propanone at low pH. Maybe that boiling gives some reaction, but still I don't expect any. Increase the pH to 13 or 14 and then you'll see a fast reaction, yellow iodoform, CHI3, is formed in that reaction, which quickly settles at the bottom and can easily be isolated. Link to comment Share on other sites More sharing options...
RyanJ Posted January 19, 2006 Share Posted January 19, 2006 So this is a Ph dependent reaction? Thats interesting don't see many of these Cheers, Ryan Jones Link to comment Share on other sites More sharing options...
clarisse Posted January 20, 2006 Author Share Posted January 20, 2006 What reaction do you expect between the iodine and the propanone? I think you need a LOT of patience for this reaction . Aim: Measure the rate at which iodine and propanone react to form iodopropanone by colorimetric analysis and calculate the order of the reaction with respect to each of the reactants and the rate constant. Chemical Equation: CH3COCH3 (Aq) + I2 (Aq) → CH3COCH2I (Aq) +H+ (Aq) + I-(Aq) This is the reaction I believed would happen. So by using Sulphuric Acid I was actually making the reaction go slower? I'll try raising the pH then and I will also see if increasing the temperature helps. Link to comment Share on other sites More sharing options...
woelen Posted January 20, 2006 Share Posted January 20, 2006 But keep in mind, that at high pH a totally different reaction occurs. One of the products is CHI3, the other is acetate ion. Look at "haloform reaction" in google. This will certain help you. Link to comment Share on other sites More sharing options...
clarisse Posted January 22, 2006 Author Share Posted January 22, 2006 Thankyou very much woelen for your help! Link to comment Share on other sites More sharing options...
Tartaglia Posted January 25, 2006 Share Posted January 25, 2006 The reaction is not really totally different at high and low PH because the first intermediate formed in the haloform reaction is CH3C(O)CH2I. The mechanism is also broadly similar in that at high PH the species CH3C(O-)=CH2 atacks I2 to give CH3C(O)CH2I and at low pH the vinyl alcohol CH3C(OH)=CH2 attacks I2 to give [CH3C(=OH+)CH2I} which then elininates H+ to give CH3C(O)CH2I. At low pH the rate determining step will be the formation of CH3C(OH)=CH2 which involves H+ and CH3C(O)CH3 and not I2. Therefore the reaction is relatively slow and the order wrt I2 is zero Link to comment Share on other sites More sharing options...
grooverider5 Posted January 26, 2006 Share Posted January 26, 2006 im doin the experiment that you talk about, iodine, propanone and sulphric acid. the reaqction is slow if the concs are similar so you have to have excess propanone and sulphuric acid. i used purple as my filter and 10cm of 0.0025 cubed of iodine with 5cm each of 1mol sulpuric acid and 1mol propanone. this reaction with these volumes and concs occurs rapidly with absorbance decreasing by 0.02 every 15 seconds Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now