Graph of ln[A] against time


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Dakota_Campbell_1C
Posts: 51
Joined: Fri Sep 28, 2018 12:15 am

Graph of ln[A] against time

Postby Dakota_Campbell_1C » Thu Mar 07, 2019 11:45 pm

In lecture Lavelle had a graph that had ln[A] on the Y axis and time on the X axis. I was confused to see that the plot for the graph was linear. Why is it linear if the graph of ln is exponential?

Sam Joslyn 1G
Posts: 34
Joined: Fri Sep 28, 2018 12:20 am

Re: Graph of ln[A] against time

Postby Sam Joslyn 1G » Fri Mar 08, 2019 12:04 am

This graph is referring to a first-order reaction. The plot of ln[A] vs. time is linear because first-order processes see the concentration of [A] undergo exponential decay; therefore, a graph of ln[A] for a first-order process would theoretically be linear with a slope=-k.

Danny Elias Dis 1E
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

Re: Graph of ln[A] against time

Postby Danny Elias Dis 1E » Fri Mar 08, 2019 9:29 am

Look at it this way: if you were to graph the points without ln, the curve would be exponential. So, in order to make the graph linear, you must take the ln of the data. The ln is changing the curve into a line.

Nico Edgar 4L
Posts: 32
Joined: Fri Sep 28, 2018 12:19 am

Re: Graph of ln[A] against time

Postby Nico Edgar 4L » Fri Mar 08, 2019 11:12 am

This graph is linear if it is a first order reaction, as in that case the reaction rate is in the form of a linear equation or line.

Sophie Roberts 1E
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

Re: Graph of ln[A] against time

Postby Sophie Roberts 1E » Fri Mar 08, 2019 7:46 pm

If the graph of [A] vs time is linear then it's a Zero Order reaction.
If the graph of ln[A] vs time is linear then it's a First Order reaction.
If the graph of 1/[A] vs time is linear then it's a Second Order reaction.


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