The Student Room Group

Chemistry;finding rate, to find k, to find Ea!

Possibly yet another stupid chemistry question!

I'm trying to find the activation enthalpy using the annoying (I don't do maths!) Arrhenius equation. Therefore I need the rate constant at my 5 different temperatures. The stupid thing is I changed my method after finding the rate equation and rate constant at 298K. Instead of using the initial rates method I decided I would use the reciprocal (sp?) of time instead, as I only recorded the time taken for my solution to reach 50% transmission on the colorimeter rather than taking a reading every 30seconds for 6mins.

Basically I want to know how to calculate the rate using my new method!
I think it's something like:

rate = concentration of I2 needed for 50% transmission/time taken for 50% transmission to me reached

therefore as concentration of I2 needed will be the same each time;

rate is proportional to 1/time taken

(does proportional mean same as equals?!)

Is this right? What are my units? (concentration is in moldm-3, time is in mins but easily convertable into seconds) Is it moldm-3s-1?

I need the rate to substitue it into my rate equation as follows;

rate = k [propanone] [acid catalyst]

and I need the k for the Arrhenius equation to plot 1/T (T=temp in Kelvin) agains log k! (Don't get me started on logarithms...I know to press 'log' on my calculator but otherwise don't really understand all of that yet!)

I feel silly for asking but I'm not sure & its driving me nuts! Thanks
Reply 1
Katie J
Possibly yet another stupid chemistry question!

I'm trying to find the activation enthalpy using the annoying (I don't do maths!) Arrhenius equation. Therefore I need the rate constant at my 5 different temperatures. The stupid thing is I changed my method after finding the rate equation and rate constant at 298K. Instead of using the initial rates method I decided I would use the reciprocal (sp?) of time instead, as I only recorded the time taken for my solution to reach 50% transmission on the colorimeter rather than taking a reading every 30seconds for 6mins.

Basically I want to know how to calculate the rate using my new method!
I think it's something like:

rate = concentration of I2 needed for 50% transmission/time taken for 50% transmission to me reached

therefore as concentration of I2 needed will be the same each time;

rate is proportional to 1/time taken

(does proportional mean same as equals?!)

Is this right? What are my units? (concentration is in moldm-3, time is in mins but easily convertable into seconds) Is it moldm-3s-1?

I need the rate to substitue it into my rate equation as follows;

rate = k [propanone] [acid catalyst]

and I need the k for the Arrhenius equation to plot 1/T (T=temp in Kelvin) agains log k! (Don't get me started on logarithms...I know to press 'log' on my calculator but otherwise don't really understand all of that yet!)

I feel silly for asking but I'm not sure & its driving me nuts! Thanks


Can you get your point across clearer please.

And proportional is not the same as equals.
Reply 2
Katie J
Possibly yet another stupid chemistry question!

I'm trying to find the activation enthalpy using the annoying (I don't do maths!) Arrhenius equation. Therefore I need the rate constant at my 5 different temperatures. The stupid thing is I changed my method after finding the rate equation and rate constant at 298K. Instead of using the initial rates method I decided I would use the reciprocal (sp?) of time instead, as I only recorded the time taken for my solution to reach 50% transmission on the colorimeter rather than taking a reading every 30seconds for 6mins.

Basically I want to know how to calculate the rate using my new method!
I think it's something like:

rate = concentration of I2 needed for 50% transmission/time taken for 50% transmission to me reached

therefore as concentration of I2 needed will be the same each time;

rate is proportional to 1/time taken

(does proportional mean same as equals?!) - no: "rate is proportional to 1/time" means rate equals 1/time multiplied by a constant

Is this right? What are my units? (concentration is in moldm-3, time is in mins but easily convertable into seconds) Is it moldm-3s-1? yeah, I think so; yes, the units are mol dm^-3 s^-1

I need the rate to substitue it into my rate equation as follows;

rate = k [propanone] [acid catalyst]

and I need the k for the Arrhenius equation to plot 1/T (T=temp in Kelvin) agains log k! (Don't get me started on logarithms...I know to press 'log' on my calculator but otherwise don't really understand all of that yet!) - want an explanation? I still find them difficult to get my head round when I'm working with them but the principle is relatively easy:smile:

I feel silly for asking but I'm not sure & its driving me nuts! Thanks


Hope I didn't miss anything there... also hope it's right!! Hoping a physical scientist will have a look to check it, lol...
Reply 3
Thanks! :smile:

Yup, just what I feared...proportional bit! How do I find the constant then?
All I really want is to calculate the rate at each temperature so I can substitue it into the rate equation to find k. But I can't work out how using the results I have! I think it's a case of being blinded by too much exposure to it; I'll leave it for a bit and go back & I hope it will be obvious!

Before, I found the gradient to work out rate (I2 conc/time) but now I don't have the time, just the known I2 conc to have a 50% transmission and the overall time taken to reach this transmission. I need to get rate using just this data!

Rate = conc/time...it can't be that simple as its not give as '=' in my book, only 'proportional to'...which is ultimately my problem!



Logarithms; I sort of understand them...it's like standard form 'to the power of' bit! They make very big/small numbers easier to work with and I know when to press 'log', which is a great improvement on yesterday evening when I was moaning to my wonderful maths A-level friends! I think of them in horribly over simplistic way; like the Richter Scale...each one greater than the last on scale mulitplying sort of way (lol, that makes sense to me, even if no one else!)

(I'm not good with numbers & I'm terrible at explaining things, sorry!!)
Reply 4
I think rate does equal change in concentration over time.. it might be proprtional to 1/time because the constant may be the concentration of I2; alternatively I may be barking up entirely the wrong tree (forgive me, I'm a biologist). Chemists! Damn you lazy buggers, come and help this girl before I give her all the wrong answers!!

Logarithms; I sort of understand them...it's like standard form 'to the power of' bit! They make very big/small numbers easier to work with and I know when to press 'log', which is a great improvement on yesterday evening when I was moaning to my wonderful maths A-level friends! I think of them in horribly over simplistic way; like the Richter Scale...each one greater than the last on scale mulitplying sort of way (lol, that makes sense to me, even if no one else!)


Hmm, well at the risk of confusing you... Think of powers. 10^2 = 100. So log-to-base-10 of 100 = 2 (damn the lack of subscripts!) [*from now on when I say log I mean log-to-base-10*] log1 = 0 because 10^0 =1. If you've ever come across it, 'ln' means log-to-base-e (e being a horribly complicated but very useful number for calculus); ln1 is also zero because, as GCSE maths should teach everyone I *think*, any number-to-the-power-zero is 1.

Basically yeah, the log just makes large numbers easier to deal with because it tells you what power 10 has to be raised to to make that number. So log6 = 0.778 - try typing in 10^0.778 on your calculator and you should get around 6 (it won't be bang on as I've rounded the number).

I guess you don't really need to know about logs exactly but I think it's useful to have some idea of what they are...
Reply 5
MadNatSci
I think rate does equal change in concentration over time.. it might be proprtional to 1/time because the constant may be the concentration of I2; alternatively I may be barking up entirely the wrong tree (forgive me, I'm a biologist). Chemists! Damn you lazy buggers, come and help this girl before I give her all the wrong answers!!



Hmm, well at the risk of confusing you... Think of powers. 10^2 = 100. So log-to-base-10 of 100 = 2 (damn the lack of subscripts!) [*from now on when I say log I mean log-to-base-10*] log1 = 0 because 10^0 =1. If you've ever come across it, 'ln' means log-to-base-e (e being a horribly complicated but very useful number for calculus); ln1 is also zero because, as GCSE maths should teach everyone I *think*, any number-to-the-power-zero is 1.

Basically yeah, the log just makes large numbers easier to deal with because it tells you what power 10 has to be raised to to make that number. So log6 = 0.778 - try typing in 10^0.778 on your calculator and you should get around 6 (it won't be bang on as I've rounded the number).

I guess you don't really need to know about logs exactly but I think it's useful to have some idea of what they are...


Yeah, that makes sense!! Thanks so much! I think you're right about the constant being [I2], it would make more sense that way anyway! I checked with a friend who's also doing kinetics, she's stuck too with the Arrhenius stuff! She's getting a higher Ea with a catalyst than without!
Thanks for the Log stuff as well, it does make more sense...glad I was right that it's linked to the standard form powers thing! :biggrin:
Thanks again! :smile:
Reply 6
Katie J
Yeah, that makes sense!! Thanks so much! I think you're right about the constant being [I2], it would make more sense that way anyway! I checked with a friend who's also doing kinetics, she's stuck too with the Arrhenius stuff! She's getting a higher Ea with a catalyst than without!
Thanks for the Log stuff as well, it does make more sense...glad I was right that it's linked to the standard form powers thing! :biggrin:
Thanks again! :smile:



No problem :smile: Hope the log stuff is helpful and glad the stuff I said was kinda right! Good luck with your coursework :smile: