## First Order Equation

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

MaryBanh_2K
Posts: 101
Joined: Wed Sep 18, 2019 12:21 am

### First Order Equation

When solving for k, do both of these derivations mean the same thing: (ln([Ao]/[[A]))/t AND (ln[A]/[Ao])/-t? The book uses the first equation but I'm not sure if the two equations mean the same thing because I am getting a different answer.

Brian_Ho_2B
Posts: 221
Joined: Fri Aug 09, 2019 12:16 am

### Re: First Order Equation

Based on logarithmic properties, ln([Ao]/[A])/t is equal to ln([A]/[Ao])/-t. The negative sign applied to a logarithm takes the inverse (^-1) of whatever is in the logarithm.

Kishan Shah 2G
Posts: 132
Joined: Thu Jul 11, 2019 12:15 am

### Re: First Order Equation

Yes there are multiple ways of expressing the equation. The first one is the linearized form and thats the one that Lavelle put on the Kinetics syllabus. I prefer to use that one since its easy to graph and see the linear line.

Posts: 113
Joined: Thu Jul 11, 2019 12:16 am

### Re: First Order Equation

MaryBanh_2K wrote:When solving for k, do both of these derivations mean the same thing: (ln([Ao]/[[A]))/t AND (ln[A]/[Ao])/-t? The book uses the first equation but I'm not sure if the two equations mean the same thing because I am getting a different answer.

Yes, they’re equivalent expressions.