linear graph


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josmit_1D
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

linear graph

Postby josmit_1D » Mon Mar 09, 2020 3:27 pm

for a second order reaction, what equation gives a linear plot?

Samuel Tzeng 1B
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

Re: linear graph

Postby Samuel Tzeng 1B » Mon Mar 09, 2020 3:46 pm

1/[A] vs time

Hiba Alnajjar_2C
Posts: 108
Joined: Fri Aug 09, 2019 12:17 am

Re: linear graph

Postby Hiba Alnajjar_2C » Mon Mar 09, 2020 3:46 pm

You would get a linear plot when you graph 1/[A] vs. time. This makes a bit more sense when you look at the integrated rate law for a second order reaction: 1/[A]=kt+1/[A]o

(In this plot, k would be the slope).

005162520
Posts: 101
Joined: Tue Sep 24, 2019 12:17 am

Re: linear graph

Postby 005162520 » Wed Mar 11, 2020 1:13 am

how would we determine if it is a positive or negative slope? Or do we just have to memorize it ?

Sofia Barker 2C
Posts: 101
Joined: Wed Sep 18, 2019 12:21 am

Re: linear graph

Postby Sofia Barker 2C » Wed Mar 11, 2020 12:29 pm

005162520 wrote:how would we determine if it is a positive or negative slope? Or do we just have to memorize it ?


If given a set of [A] values vs time, you can determine the slope by taking the inverse of each [A] value, which gives the 1/[A] value, and see if it increases or decreases over time. That way you can see if the graph's slope will be positive (increasing values) or negative (decreasing values).

Nathan Nakaguchi 1G
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Joined: Wed Sep 18, 2019 12:22 am
Been upvoted: 1 time

Re: linear graph

Postby Nathan Nakaguchi 1G » Wed Mar 11, 2020 12:33 pm

A linear plot is in the form y=mx+b and for second order 1/[A] = kt + 1/[A]0 so the plot is 1/[A] versus time.

ABombino_2J
Posts: 102
Joined: Thu Jul 11, 2019 12:15 am

Re: linear graph

Postby ABombino_2J » Wed Mar 11, 2020 12:41 pm

To see if the slope is positive or negative you take the value of k from the equation.

Maika Ngoie 1B
Posts: 97
Joined: Fri Aug 02, 2019 12:16 am

Re: linear graph

Postby Maika Ngoie 1B » Wed Mar 11, 2020 4:58 pm

For a second order reaction the graph of ln[A] vs time will yield a straight line


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