chemistry

**euclid2** · December 17th 2008, 05:00 PM

$-\frac{\delta[A]}{\deltat}=r=k[A]$ 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)

**o_O** · December 17th 2008, 05:12 PM

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.

**euclid2** · December 17th 2008, 05:19 PM

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;

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

**o_O** · December 17th 2008, 05:21 PM

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

**euclid2** · December 17th 2008, 05:26 PM

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

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