- #1
erik-the-red
- 89
- 1
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?
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?