the slope
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 58
- Joined: Fri Sep 29, 2017 7:03 am
the slope
can someone tell me what the slope is for the three orders of the reactions. Is it not just k for all of them?
-
- Posts: 52
- Joined: Thu Jul 13, 2017 3:00 am
- Been upvoted: 1 time
Re: the slope
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)
-
- Posts: 60
- Joined: Thu Jul 27, 2017 3:01 am
Re: the slope
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.
-
- Posts: 53
- Joined: Sat Jul 22, 2017 3:01 am
-
- Posts: 47
- Joined: Fri Sep 29, 2017 7:06 am
-
- Posts: 58
- Joined: Fri Sep 29, 2017 7:03 am
Re: the slope
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
-
- Posts: 52
- Joined: Thu Jul 13, 2017 3:00 am
-
- Posts: 64
- Joined: Thu Jul 27, 2017 3:01 am
Re: the slope
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?
-
- Posts: 50
- Joined: Thu Jul 13, 2017 3:00 am
Re: the slope
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.
-
- Posts: 30
- Joined: Sat Jul 22, 2017 3:01 am
Re: the slope
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]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.
-
- Posts: 52
- Joined: Fri Sep 29, 2017 7:07 am
Re: the slope
negative k for zero and first order reaction rates and positive k for second order reaction rates
Return to “Second Order Reactions”
Who is online
Users browsing this forum: No registered users and 5 guests