## Understanding the graph of First Order Integrated Rate Law

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

JuliaPark2H
Posts: 19
Joined: Fri Sep 25, 2015 3:00 am

### Understanding the graph of First Order Integrated Rate Law

How can the graph of the first order integrated rate law be interpreted? I understand that for the graph, ln[A] is graphed against time, and the negative slope is k, but the ln function graphed against time confuses me-- the concentration for the graph isn't one-to-one, as [A] doesn't decrease at a constant rate as time progresses. The ln[x] graph starts from -infinite and increases, but slows down as x continues to increase, so what does the linearity of the ln[A] vs time tell us?

Chem_Mod
Posts: 19511
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 880 times

### Re: Understanding the graph of First Order Integrated Rate L

A first order reaction follows exponential decay (for the reactant):

[A] = [A]0*e-kt

If we take ln of both sides we recover a linear equation if we plot ln[A] vs t. It is in fact the exact same equation, just manipulated mathematically. So if plotting ln[A] vs t is linear, we KNOW that plotting [A] vs t gives exponential decay curve, as dictated for 1st order