## the slope

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Ilan Shavolian 1K
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?

Nhan Nguyen 2F
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)

Cam Bear 2F
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.

Christina Cen 2J
Posts: 53
Joined: Sat Jul 22, 2017 3:01 am

### Re: the slope

0 and first order: -k
second order: k

Matthew 1C
Posts: 47
Joined: Fri Sep 29, 2017 7:06 am

### Re: the slope

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

Ilan Shavolian 1K
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

lizzygaines1D
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### Re: the slope

Hope this can clarify everything.
Attachments

Michelle Lee 2E
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?

DamianW
Posts: 35
Joined: Fri Sep 29, 2017 7:06 am

### Re: the slope

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

Meredith Steinberg 2E
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.

Joanne Guan 1B
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.

AnuPanneerselvam1H
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

Jimmy Zhang Dis 1K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am