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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 |
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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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