# Thread: chemistry

1. ## chemistry

$-\frac{\delta[A]}{\deltat}=r=k[A]$ How would i manipulate this using integral calculus to yield the expression In $\frac{([A]_o)}{([A])}= kt$ (This is a question asked out of curiosity)

2. For a first-order reaction, the differential rate is given by: $r = -\frac{d[A]}{dt} = k[A]$

To get its integrated form, separate the variables:
\begin{aligned}-\frac{d[A]}{[A]} & = k dt \\ -{\color{red}\int}\frac{d[A]}{[A]} & = {\color{red}\int} k dt \\ \ln [A] & = -kt + C\end{aligned}

Normally, we're given an initial concentration, i.e. $[A]_0 \ \ @ \ \ t=0$

This enables us to solve for the constant: $\ln [A]_0 = -k(0) + C \ \Leftrightarrow \ C = \ln [A]_0$

I'm sure you can take it from here.

3. Originally Posted by o_O
For a first-order reaction, the differential rate is given by: $r = -\frac{d[A]}{dt} = k[A]$

To get its integrated form, separate the variables:
\begin{aligned}-\frac{d[A]}{[A]} & = k dt \\ -{\color{red}\int}\frac{d[A]}{[A]} & = {\color{red}\int} k dt \\ \ln [A] & = -kt + C\end{aligned}

Normally, we're given an initial concentration, i.e. $[A]_0 \ \ @ \ \ t=0$

This enables us to solve for the constant: $\ln [A]_0 = -k(0) + C \ \Leftrightarrow \ C = \ln [A]_0$

I'm sure you can take it from here.
Thanks very much. For some reason I was expecting a reply from you!

I suppose we can assume then after one half life we can simplify further to;

$In 2 = kt_\frac{1}{2}$

4. This is only applicable to first-order reactions though as I'm sure you will learn soon enough.

5. Originally Posted by o_O
This is only applicable to first-order reactions though as I'm sure you will learn soon enough.
Something to look forward to! Ahh the wonders of chemistry