the slope

[Ilan Shavolian 1K]

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]

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]

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]

0 and first order: -k
second order: k

[Matthew 1C]

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

[Ilan Shavolian 1K]

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]

Hope this can clarify everything.

[Michelle Lee 2E]

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]

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

[Meredith Steinberg 2E]

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]

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]

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

[Jimmy Zhang Dis 1K]