First order vs zero-order

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Bryce Ramirez 1J
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Joined: Sat Aug 24, 2019 12:16 am

First order vs zero-order

Postby Bryce Ramirez 1J » Sat Mar 14, 2020 11:04 pm

For the graphs of first order and zero order reactions, why are they both exactly the same? The first order reaction has the Ln of A while the zero order is just A. The slopes for both of the graphs are -k, but how can they be the same when the equations are significantly different?

Tracy Tolentino_2E
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Re: First order vs zero-order

Postby Tracy Tolentino_2E » Sat Mar 14, 2020 11:25 pm

They are not the same precisely because of the y-axis. Because they are the integrated rate laws, they are supposed to be linear. They have to follow the y = mx+b format. As a result, the slope = -k, but the y axis are different: first order is ln[A] and zero order [A]. You could tell which order reaction we are looking at based on the axis.

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Re: First order vs zero-order

Postby gconcha » Sun Mar 15, 2020 4:42 am

integrated rate laws are designed to express the rate linearly. Always. I'm sure the graphs don't look exactly the same, but if they're both linear and have the same sign of their slope, then that might be why they look so similar. Remember to always take into account what the y-axis is labeled in ANY graph that you read. I'm sure if you re-plotted the first order rxn's data on the graph with [A] as the y-axis that it would look significantly different.

Connor Chappell 2B
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Re: First order vs zero-order

Postby Connor Chappell 2B » Sun Mar 15, 2020 5:52 am

The first order and zero-order integrated rate laws are designed to create a linear plot, using a different y-axis depending on the order of the reaction. While the two graphs can look similar in that they are both decreasing lines, the clue to what order the reaction is is in the y-axis. If it is first order, ln(A) plotted on the y-axis against time is liner, and the slope is -k. A zero-order reaction plots [A] vs time, and the slope is -k.

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Re: First order vs zero-order

Postby BeylemZ-1B » Sun Mar 15, 2020 7:06 am

to ease some stress before the final, know that we will most likely not be asked to graph these ourselves by hand, but maybe only asked to interpret graphs. KEY IS TO REMEMBER:
zeroth: [A]t vs t (-k)
first order: ln[A]t vs t (-k)
second order: 1/[A]t vs t (k)

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