the slope


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Ilan Shavolian 1K
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Joined: Fri Sep 29, 2017 7:03 am

the slope

Postby Ilan Shavolian 1K » Sat Mar 10, 2018 8:39 pm

can someone tell me what the slope is for the three orders of the reactions. Is it not just k for all of them?

Nhan Nguyen 2F
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Re: the slope

Postby Nhan Nguyen 2F » Sat Mar 10, 2018 8:44 pm

zero order= neg slope
1st order= neg slope
2nd order= pos slope
These are taken from the plots of time vs [A], ln[A], 1/[A] (for the 0th, 1st, and 2nd orders respectively)

Cam Bear 2F
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Joined: Thu Jul 27, 2017 3:01 am

Re: the slope

Postby Cam Bear 2F » Sat Mar 10, 2018 8:48 pm

The slope is k. For zero order reactions, you plot [A] against time which gives you a negative slope. Because k is always positive -slope=k. For first order reactions, you plot ln[A] against time which gives you a negative slope. So -slope=k. For second order reactions, you plot 1/[A] against time which gives you a positive slope. So the slope=k.

Christina Cen 2J
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Re: the slope

Postby Christina Cen 2J » Sat Mar 10, 2018 9:22 pm

0 and first order: -k
second order: k

Matthew 1C
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Re: the slope

Postby Matthew 1C » Sat Mar 10, 2018 9:43 pm

In regards to the graphs, how do you know what the y intercepts are?

Ilan Shavolian 1K
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Joined: Fri Sep 29, 2017 7:03 am

Re: the slope

Postby Ilan Shavolian 1K » Sat Mar 10, 2018 10:20 pm

Matthew 1C wrote:In regards to the graphs, how do you know what the y intercepts are?

i think for zero order its [A]0, for first order it's ln([A]0, and for second, its 1/[A]0

lizzygaines1D
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Joined: Thu Jul 13, 2017 3:00 am

Re: the slope

Postby lizzygaines1D » Sat Mar 10, 2018 10:25 pm

Hope this can clarify everything.
Attachments
IMG_0330.JPG

Michelle Lee 2E
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Re: the slope

Postby Michelle Lee 2E » Sat Mar 10, 2018 11:10 pm

Is it possible at all for the slopes of zero and first order to be positive and for the slope of second order to be negative? I think I heard a ta say it could be switched? If so, then in what circumstances?

DamianW
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Re: the slope

Postby DamianW » Sun Mar 11, 2018 5:10 pm

I don’t think lavelle would ever switch them on our test

Meredith Steinberg 2E
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Joined: Thu Jul 13, 2017 3:00 am

Re: the slope

Postby Meredith Steinberg 2E » Sun Mar 11, 2018 7:05 pm

If the slope were to be positive for the first order reaction, we would have to look at -ln(concentration) on the y axis, which doesn't conceptually make sense, so I don't think we would have to calculate a positive slope for equations that are given with a slope w a -k value.

Joanne Guan 1B
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Joined: Sat Jul 22, 2017 3:01 am

Re: the slope

Postby Joanne Guan 1B » Sun Mar 11, 2018 10:20 pm

The integrated zero-order rate law is [A] = -kt so the slope would be -k.
The integrated first-order rate law is ln[A] = ln[A]0 - kt, so the slope would be -k.
The integrated second-order rate law is 1/[A] = 1/[A]0 + kt, so the slope is k.

AnuPanneerselvam1H
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Joined: Fri Sep 29, 2017 7:07 am

Re: the slope

Postby AnuPanneerselvam1H » Sun Mar 11, 2018 11:22 pm

negative k for zero and first order reaction rates and positive k for second order reaction rates

Jimmy Zhang Dis 1K
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Re: the slope

Postby Jimmy Zhang Dis 1K » Sun Mar 11, 2018 11:54 pm

Image


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