## Problem 7B.1 7th edition

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

Ashley McClearnen 1B
Posts: 59
Joined: Fri Sep 28, 2018 12:26 am

### Problem 7B.1 7th edition

Why does the solution manual write the 1st order integrated rate law as ln[A]/[A]0 = -kt instead of ln[A] = -kt+[A]0? Do you get the same answer if you write it either way?

Jessica Castro 2H
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

### Re: Problem 7B.1 7th edition

For first order reactions, the equation for the plot is ln[A] = -kt + ln[A]0. If you solve for kt you would get kt = ln[A]0 - ln[A]. We know that for subtraction with ln, lnx -lny = ln(x/y). Therefore, kt = ln[A]0 - ln[A] = ln([A]0/[A]).

There should be an ln next to the [A]0, but both equations should get you the same answer.