
#1
Dec1205, 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? 



#2
Dec1205, 03:53 PM

P: 1,389

At time =0 C=Ao 



#3
Dec1205, 04:05 PM

Sci Advisor
HW Helper
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.




#4
Dec1205, 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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