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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: grignardgirl on October 17, 2007, 12:22:18 AM

Title: Trying to convert from Abs/min to M/min
Post by: grignardgirl on October 17, 2007, 12:22:18 AM

Hey everyone!  So I'm trying to convert from Absorbance per minute to Molarity per minute.  I know that I am supposed to use A=ebc but I'm not sure how the dimensional analysis works with this. 

I'm given: .429 Abs/min and e = 2500 1/M x cm 

This is from a Lineburke-weaver plot.  .429 is the Vmax and the Km is .016 M.  Thank you in advance!!

Title: Re: Trying to convert from Abs/min to M/min
Post by: ARGOS++ on October 17, 2007, 12:27:05 PM

Dear GrignardGirl,

Who is “Lineburke”?, - never heard of him/it, and it seems nobody (Google, Yahoo!, MSN, Alta Vista, etc.) else too!

Abs/min indicates a “Single Wavelength measurement” (SWM) for a Reaction, what seems to be not very wisely!

But let’s do the little “brain acrobatics” anyway:
“Abs/min” or more correct ΔAbs/Δt,  like dAbs/dt is a “Reaction Rate”, and if you use the “Special Beer-Lambert Law” (for only a SWM it’s enough) you have to write:

ΔAbs/Δt    ==>   ΔA_λ/Δt = Δ(ε_λ * c * d)/Δt

On the other side and from the definition of ε_λ: 
If you set for your case for A_λ = 2500 AU into the “Special Beer-Lambert Law”: 
A_λ = ε_λ * c * d you will result in:
 

2500 AU = 2500 L mole^-1 cm^-1  *  c * d

Or with simple words:  ε_λ is “only” the theoretical Absorbance (value) for a Solution with Concentration = 1.00 Mole/L measured in a 1.00cm Cuvette with the given ε_λ.

And the final calculation:

2500 AU/min = 1.00 mole/min  ===>  0.429AU/min = ??? mole/min.

(I’m still searching for its real sense.)
 
Could this be of help to you?

Good Luck!
                    ARGOS⁺⁺


P.S.: What should in this case “Vmax” and “Km” tell us?
.