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### 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?

### Re: the slope

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)

### Re: the slope

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.

### Re: the slope

Posted: Sat Mar 10, 2018 9:22 pm
0 and first order: -k
second order: k

### Re: the slope

Posted: Sat Mar 10, 2018 9:43 pm
In regards to the graphs, how do you know what the y intercepts are?

### Re: the slope

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

### Re: the slope

Posted: Sat Mar 10, 2018 10:25 pm
Hope this can clarify everything.

### Re: the slope

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?

### Re: the slope

Posted: Sun Mar 11, 2018 5:10 pm
I donâ€™t think lavelle would ever switch them on our test

### Re: the slope

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.

### Re: the slope

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.

### Re: the slope

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

### Re: the slope

Posted: Sun Mar 11, 2018 11:54 pm