## K constant

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

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MorganYun1H
Posts: 59
Joined: Thu Jul 13, 2017 3:00 am

### K constant

can someone explain what Lavelle means when he says that a negative k means negative slope, but it is negative because we want a positive k? How do you get a positive k from a negative k?

Julianna Thrasher 1B
Posts: 27
Joined: Thu Jul 27, 2017 3:00 am

### Re: K constant

For a first order reaction, the integrated rate law is: ln[A] = -kt + ln[A]0. If you set ln[A]=y and -k=m you can get a typical slope equation in the form of y=mx+c. Since -k=m then the slope of the curve is negative so a negative k implies a negative slope. A 1st order reaction of ln[A] vs time means the graph will have a slope of -k.
The only way there is a positive k for a first order reaction is for the first order half life equations, Rate = k[A0]e^-kt.
Professor Lavelle discussed this in Lecture 19 on 2/26/18.

Lindsay Kester 2L
Posts: 29
Joined: Thu Jul 27, 2017 3:00 am

### Re: K constant

When we graph K, we make it negative, because the reaction is using up reactants (at least initially). K itself is usually positive.

MorganYun1H
Posts: 59
Joined: Thu Jul 13, 2017 3:00 am

### Re: K constant

So then will we ever see a positive k value?

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