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Deriving the rate laws for first and second order reactions

by erik-the-red
Tags: deriving, laws, order, rate, reactions
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erik-the-red
#1
Dec12-05, 03:14 PM
P: 89
So, the average rate for a reaction of type A --> product is given by [tex]\text{rate} = -\frac{\Delta A}{\Delta t}[/tex]. Also, [tex]\text{rate} = k \cdot \text{A}[/tex].

The instantaneous rate for a reaction of that type is [tex]\lim_{\Delta t\rightarrow\0} -\frac{\Delta A}{\Delta t} = -\frac{dA}{dt}[/tex].

Setting the instantaneous rate for a reaction equal to the second equation, there is [tex]-\frac{dA}{dt} = k \cdot \text{A}[/tex].

Well, this is a very friendly separable differential equation. I get [tex]\ln A = -kt + C[/tex].

How do I get [tex]\ln{\frac{A}{A_o}} = -kt[/tex] from my derivation? Definite integration?
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gravenewworld
#2
Dec12-05, 03:53 PM
P: 1,405
Quote Quote by erik-the-red
So, the average rate for a reaction of type A --> product is given by [tex]\text{rate} = -\frac{\Delta A}{\Delta t}[/tex]. Also, [tex]\text{rate} = k \cdot \text{A}[/tex].
The instantaneous rate for a reaction of that type is [tex]\lim_{\Delta t\rightarrow\0} -\frac{\Delta A}{\Delta t} = -\frac{dA}{dt}[/tex].
Setting the instantaneous rate for a reaction equal to the second equation, there is [tex]-\frac{dA}{dt} = k \cdot \text{A}[/tex].
Well, this is a very friendly separable differential equation. I get [tex]\ln A = -kt + C[/tex].
How do I get [tex]\ln{\frac{A}{A_o}} = -kt[/tex] from my derivation? Definite integration?


At time =0 C=Ao
GCT
#3
Dec12-05, 04:05 PM
Sci Advisor
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P: 1,769
yes, definite integration, you'll merely have one term on the right since initial time is considered zero. Also remember to always derive the equation for a particular reaction....if you're gonna write the rate equation in terms of a reactant/product with a unique coefficient, they'll be multiples to account for.

erik-the-red
#4
Dec12-05, 07:29 PM
P: 89
Deriving the rate laws for first and second order reactions

Thanks! It's like every intro chem book always uses the exact phrase, "Using calculus, we can derive..."

I mean at the intro chem level most students haven't taken calculus, so to show the derivation is unnecessary. I was just curious :)


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