Deriving the rate laws for first and second order reactions

[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?

 

[gravenewworld]

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

 

[GCT]

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 going to write the rate equation in terms of a reactant/product with a unique coefficient, they'll be multiples to account for.

 

[erik-the-red]

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 :)