## First oder graphs

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

Jasmin Argueta 1K
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am

### First oder graphs

During lecture Lavelle derived the first order reaction and showed us the graph that correlated. I am a bit confused because the graph for a first order reaction is liner, although he said ln[A] on the y axis is exponential. Why is the graph not exponential if it includes ln?

Chem_Mod
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### Re: First oder graphs

So f(x) = ln(x) is itself exponential. However, we make ln[A] the x-axis. Because the x-axis and y-axis are measured with the same units, the graph is linear.

Andrew Bennecke
Posts: 62
Joined: Fri Sep 28, 2018 12:15 am

### Re: First oder graphs

Using the equation ln[A]=-kt+ln[A]o, we see how it has the same format as y=mx + b, the skeleton equation for a linear equation. In this case, y=ln[A], m=-k, x=t, and b=ln[A]o. Therefore, we can graph ln[A] in terms of t, where ln[A]o is the initial concentration of [A] at t=o. ln[A] does not take t as an input, so we don't graph it as a logarithmic function, but as a linear one instead.