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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: tiger1987 on November 04, 2006, 08:30:18 PM

Title: Calculating the Diffusion Coeifficient
Post by: tiger1987 on November 04, 2006, 08:30:18 PM
How would you calculate the diffusion coeifficient if m^(2/3)t^(1/6) was 1.66mg^(2/3)sec^(-1/2), c=2.10mM, n=2, and id=15.2 uA.

Thanks to anyone that can help. I am especially confused by the meaning of m^(2/3)t^(1/6) being 1.66mg^(2/3)sec^(-1/2), what does that simplify as?
Title: Re: Calculating the Diffusion Coeifficient
Post by: Borek on November 04, 2006, 08:54:15 PM
I am assuming you are talking about polarography?
Title: Re: Calculating the Diffusion Coeifficient
Post by: tiger1987 on November 04, 2006, 09:02:34 PM
Yes, this is a polarography problem
Title: Re: Calculating the Diffusion Coeifficient
Post by: Borek on November 05, 2006, 07:13:29 AM
Quote from: tiger1987 on November 04, 2006, 08:30:18 PM
mg^(2/3)sec^(-1/2)

Take a look at Ilkovi? equation. What units m is in? What units m2/3t1/6 is in?
Title: Re: Calculating the Diffusion Coeifficient
Post by: tiger1987 on November 05, 2006, 06:50:37 PM
I know that m=mg/sec and t=sec, but I still dont understand where the exponents on the t and m come from after m^(2/3)t^(1/6) is evaluated.

In other words, how does m^(2/3)t^(1/6) = 1.66 mg^(2/3)sec^(-1/2)
Why doesn't it just equal 1.66(mg/sec)(sec), leaving 1.66mg?
Title: Re: Calculating the Diffusion Coeifficient
Post by: Borek on November 05, 2006, 07:03:06 PM
You are not given m, you are given value of m2/3t1/6 combined, just as a thing to put into the equation:

15.2 = 607*2*D1/2*1.66*2.10

This combined value (containing both speed of mercury flowing and droplet lifetime in correct powers) has units kg2/3s-1/2

(note that equation as shown is correct only in general, I am not sure about mili/micro factors; also note that I have translated Polish terms into English without checking them, so I can be completely off with the established nomenclature).