Rate Law Slopes


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Aiden Metzner 2C
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Joined: Wed Sep 18, 2019 12:21 am

Rate Law Slopes

Postby Aiden Metzner 2C » Thu Mar 05, 2020 11:00 pm

In lecture it seemed like professor said that k is always the slope of the graph of the rate. Is this true for zero order, first order, and second order reactions. Or is it only true for zero order?

Katie Bart 1I
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Joined: Sat Aug 24, 2019 12:16 am

Re: Rate Law Slopes

Postby Katie Bart 1I » Thu Mar 05, 2020 11:05 pm

The second order graph has a slope of k, and the zero order graph has a slope of -k. For the first order, sometimes the graph is exponential, but the slope is equal to k if ln(A) is on the y axis.

Jessica Li 4F
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Joined: Fri Aug 09, 2019 12:16 am

Re: Rate Law Slopes

Postby Jessica Li 4F » Fri Mar 06, 2020 9:50 am

For a zero order reaction, if you just plot [A] over time, it should be a straight line, and -k will be the slope. For other order reactions, the graphs will not be straight lines, so you cannot determine k (the slope) unless you take either the ln[A] of the reaction for a first order reaction, in which k will be the slope, or 1/[A] for a second order reaction.

This helps you tell what order a reaction is.

Mariah
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Joined: Fri Aug 02, 2019 12:16 am

Re: Rate Law Slopes

Postby Mariah » Fri Mar 06, 2020 11:08 am

Aiden Metzner 2C wrote:In lecture it seemed like professor said that k is always the slope of the graph of the rate. Is this true for zero order, first order, and second order reactions. Or is it only true for zero order?


There is a great diagram in the book explaining this, I believe it is on page 608.


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