## Using ln[A]

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Cynthia Ulloa
Posts: 52
Joined: Fri Feb 23, 2018 3:02 am

### Using ln[A]

Can someone explain to me why we would use ln[A] instead of [A] like we do for the rest of the orders?

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

### Re: Using ln[A]

You would use ln[A] for first order reactions. For these reactions, rate=k[A]=-1/a * d[A]/dt. If you set a=1 and then you separate the variables, you get k*dt=-d[A]/[A] or (-1/[A])*d[A]. If you integrate both sides, you would get kt=-ln[A], or -kt=ln[A], as the integral of (-1/[A])*d[A] is ln[A]. This is why you use the ln[A] for first order reactions.

gwynlu1L
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am

### Re: Using ln[A]

as a general idea, you use [A] for zero order, ln[A] for first order, and 1/[A] for second order