zero order slopes


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Simrina Desar Dis 1H
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

zero order slopes

Postby Simrina Desar Dis 1H » Sun Feb 25, 2018 2:23 pm

do zero order reactions have a constant slope, a slope of zero, or a varying slope throughout the entire reaction?

Sean Monji 2B
Posts: 66
Joined: Fri Sep 29, 2017 7:06 am

Re: zero order slopes

Postby Sean Monji 2B » Sun Feb 25, 2018 3:06 pm

Since zero order reactions occur at a constant rate independent of the concentration, the rate of the reaction will not change with time or concentration as long as the reaction can continue. Therefore, the slope should be zero.

Justin Folk 3I
Posts: 43
Joined: Wed Sep 21, 2016 2:56 pm

Re: zero order slopes

Postby Justin Folk 3I » Sun Feb 25, 2018 4:24 pm

Yes, slope should be zero. It's the derivative of a constant, which makes it zero.

Veronica Rasmusen 2B
Posts: 49
Joined: Sat Jul 22, 2017 3:01 am

Re: zero order slopes

Postby Veronica Rasmusen 2B » Tue Feb 27, 2018 3:00 pm

If you are graphing rate vs. time the slope will be zero, however if you are graphing [A] vs. time there will be a slope = -k

Irma Ramos 2I
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

Re: zero order slopes

Postby Irma Ramos 2I » Sat Mar 03, 2018 10:35 am

Can someone explain why the slope is negative for a zero order reaction when we are plotting [A] vs. time?

Gwyneth Huynh 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

Re: zero order slopes

Postby Gwyneth Huynh 1J » Sat Mar 03, 2018 11:32 am

The graph [A] vs. time has a negative slope because the concentration of reactants (A) decreases as the reaction progresses. For zero order reactions, the concentration doesn't affect the rate and [A] = -kt + [A]initial, so the slope = -k. Also, the k value should be positive.


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