# Deriving the rate laws for first and second order reactions

by erik-the-red
Tags: deriving, laws, order, rate, reactions
 P: 89 So, the average rate for a reaction of type A --> product is given by $$\text{rate} = -\frac{\Delta A}{\Delta t}$$. Also, $$\text{rate} = k \cdot \text{A}$$. The instantaneous rate for a reaction of that type is $$\lim_{\Delta t\rightarrow\0} -\frac{\Delta A}{\Delta t} = -\frac{dA}{dt}$$. Setting the instantaneous rate for a reaction equal to the second equation, there is $$-\frac{dA}{dt} = k \cdot \text{A}$$. Well, this is a very friendly separable differential equation. I get $$\ln A = -kt + C$$. How do I get $$\ln{\frac{A}{A_o}} = -kt$$ from my derivation? Definite integration?
P: 1,389
 Quote by erik-the-red So, the average rate for a reaction of type A --> product is given by $$\text{rate} = -\frac{\Delta A}{\Delta t}$$. Also, $$\text{rate} = k \cdot \text{A}$$. The instantaneous rate for a reaction of that type is $$\lim_{\Delta t\rightarrow\0} -\frac{\Delta A}{\Delta t} = -\frac{dA}{dt}$$. Setting the instantaneous rate for a reaction equal to the second equation, there is $$-\frac{dA}{dt} = k \cdot \text{A}$$. Well, this is a very friendly separable differential equation. I get $$\ln A = -kt + C$$. How do I get $$\ln{\frac{A}{A_o}} = -kt$$ from my derivation? Definite integration?

At time =0 C=Ao
 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.
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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