Example Help

$\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$

Gabriel Ordonez 2K
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Example Help

Can someone walk through a zero order reaction problem step by step? Please include side information, conceptual knowledge, and graphical representations?

Chem_Mod
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Re: Example Help

Zero order does not depend on the concentration of reactants. The equation is [A]= -kt + [A]o. The graph is a negative line because the slope is -k. Zero order are common in enzymes and catalysts. Various UA sessions will cover these, and you can ask TA's for clarification.

Leonardo Le Merle 1D
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Re: Example Help

You can identify a reactant as zero-order if changing its concentration does not affect the rate in any way.

Michael Nguyen 1E
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Re: Example Help

If you see a graph of [A] v. t and it is linear with a negative slope, you know that the reaction is zero-order with respect to that reactanct. The rate constant k is equal to the negative of the slope.

Juana Abana 1G
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Re: Example Help

Michael Nguyen 1E wrote:If you see a graph of [A] v. t and it is linear with a negative slope, you know that the reaction is zero-order with respect to that reactanct. The rate constant k is equal to the negative of the slope.

Thank you this is very helpful.

CalvinTNguyen2D
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Joined: Thu Jul 25, 2019 12:16 am

Re: Example Help

In a zero order reaction, the rate of the reaction does NOT depend of the concentration of the reactant; the rate remains constant throughout the entire reaction.

RobertXu_2J
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Joined: Fri Aug 30, 2019 12:17 am

Re: Example Help

In a zero order reaction, the rate of reaction does not depend on the concentration of the reactants, so the rate of reaction is constant as the concentration decreases.
If you want examples and graphs: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Rates/Zero-Order_Reactions