Question

Derivation of the 0th order integrated rate law.

Answer 1

I just remember that it's rate = k lol

Answer 2

So an example 0th order integrated rate law can be said as
A--> products
where \[rate=k[A]^0\]
rate can be expressed in this manner \( \frac{d[A]}{dt} \) however we know that this is a rate of disappearance of reactant so its correctly expressed as \( \frac{-d[A]}{dt} \)
we can now say \[\frac{ -d[A] }{ dt }=k \] through a simple separation of variables we can say \[\int\limits_{[A]_0}^{[A]_t}d[A]=\int\limits_{0}^{t}-kdt \]
\[([A]_t -[A]_0)=-k(t-0)\]
\[[A]_t -[A]_0=-kt \]
\[[A]_t=-kt+[A]_0\]

Answer 3

tbh idk why i keep this in my head

Answer 4

good work

I haven't look at this stuff since freshman chem tbh

Answer 5

thanks, and as you probably know the idea extends easily to higher order reactions its just a different result of the integration