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Q: Atmpospheric chemistry and reaction rates ( No Answer,   1 Comment )
Question  
Subject: Atmpospheric chemistry and reaction rates
Category: Science > Chemistry
Asked by: raad-ga
List Price: $20.00
Posted: 06 Feb 2005 09:28 PST
Expires: 07 Feb 2005 00:42 PST
Question ID: 469881
i have a term paper in atmospheric chemistry regarding the OH radical.
I have encountered a problem prooving an OH concentration equation. My
paper is due this tuesday 2-8-05 @4:00 pm.....i would appreciate an
answer before that time.
Here is my question:
In the use of lab reactors to study atmospheric chemistry. To estimate
the [OH] present in the reactor, one can irradiate a mixture of NOx in
air with trace levels of two hydrocarbons A &B whose OH rate constant
are known. If the reaction with OH is the only significant removal
process for the two hydrocarbons and if ka & kb are the rate constants
for the rxn of OH radicals with hydrocarbons A & B.......I have to
show that the OH concentration is determined by the following
equation:
[OH] = 1/(ka-kb) *  dln([B]/[A]) / dt

Thank you
Answer  
There is no answer at this time.

Comments  
Subject: Re: Atmpospheric chemistry and reaction rates
From: hfshaw-ga on 06 Feb 2005 21:52 PST
 
It the removal reactions are assumed to be first order with respect to
the concentrations of OH, and A or B, then the rates of removal of A
and B can be written as:

d[A]/dt = -ka*[A]*[OH]  and   d[B]/dt = -kb*[B]*[OH]

Divide each equation by the concentration of the trace hydrocarbon involved:

1/[A]*d[A]/dt = -ka*[OH]  and   1/[B]*d[B]/dt = -kb*[OH]

you should know that 1/x*dx/dt = d(ln(x))/dt, right?, so:

d(ln[A])/dt = -ka*[OH]   and   d(ln[B])/dt = -kb*[OH]

Subtracting the first equation (involving A) from the second (involving B) yields:

d(ln[B])/dt - d(ln[A])/dt = (ka - kb)*[OH]

we can bring the d_/dt outside of the logarithms of [A] and [B]:

d(ln[B] - ln[A])/dt = (ka - kb)*[OH]

and, of course, ln(x) - ln(y) = ln(x/y), so:

d(ln([B]/[A]))/dt = (ka - kb)*[OH]

dividing through by (ka - kb) yields the desired result: 

[OH] = (ka - kb)^-1 * d(ln([B]/[A]))/dt

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