### the slope

Posted:

**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?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=149&t=29330

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Posted: **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?

Posted: **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)

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)

Posted: **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.

Posted: **Sat Mar 10, 2018 9:22 pm**

0 and first order: -k

second order: k

second order: k

Posted: **Sat Mar 10, 2018 9:43 pm**

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

Posted: **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]

Posted: **Sat Mar 10, 2018 10:25 pm**

Hope this can clarify everything.

Posted: **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?

Posted: **Sun Mar 11, 2018 5:10 pm**

I donâ€™t think lavelle would ever switch them on our test

Posted: **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.

Posted: **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.

The integrated first-order rate law is ln[A] = ln[A]

The integrated second-order rate law is 1/[A] = 1/[A]

Posted: **Sun Mar 11, 2018 11:22 pm**

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

Posted: **Sun Mar 11, 2018 11:54 pm**