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Author Topic: What is the relationship between rate of reaction and surface area?  (Read 6091 times)

Offline good_day_sunshine_08

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okay in an experiment involving the dissolving of a peppermint at varying water temperatures where the rate law is:
rate(g/sec)= k M2/3
;and in this experiment due to the candy's spherical shape the surface area = 4(pi)r2, what exactly is the algebraic relationship between the rate and the surface area?

is the relationship relatively similar to the similarity between

y=mx+b and ln[A]t= -kt + ln[A]0

or is it more complex? Assistance and elaboration please.
« Last Edit: 30/04/2009 12:01:09 by chris »


 

Offline lightarrow

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okay in an experiment involving the dissolving of a peppermint at varying water temperatures where the rate law is:
rate(g/sec)= k M2/3
;and in this experiment due to the candy's spherical shape the surface area = 4(pi)r2, what exactly is the algebraic relationship between the rate and the surface area?

is the relationship relatively similar to the similarity between

y=mx+b and ln[A]t= -kt + ln[A]0

or is it more complex? Assistance and elaboration please.
M is the mass, I presume. If it is, then you can write M = ρ*V, where ρ = density, V = volume = (4/3)πR3, R = radius.  Area A = 4πR2 --> R2 = A/4π.

 So:

rate = v = kM2/3 = k(ρ*V)2/3 = kρ2/3((4/3)πR3)2/3 = (16/9)1/3 k(πρ)2/3 R2 = (16/9)1/3 k(πρ)2/3 A/4π

= kρ2/3A/(36π)1/3.
 

Offline good_day_sunshine_08

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yes thank you but there's no elaboration
 

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