Zeroeth, First, Second meaning


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Siddiq 1E
Posts: 106
Joined: Fri Aug 09, 2019 12:15 am

Zeroeth, First, Second meaning

Postby Siddiq 1E » Mon Mar 09, 2020 4:10 pm

Hi, I was working on HW question 7A.7 and it made me realize I don't really understand what it means if a reaction is first order? I know how to tell based on the graph etc, but what does it mean in terms of units and reaction speed etc?

Chem_Mod
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Re: Zeroeth, First, Second meaning

Postby Chem_Mod » Mon Mar 09, 2020 4:13 pm

First order just means that the reaction rate is dependent on the concentration of only one reactant in a 1 to 1 relationship.

rate = k [A]1

Paul Hage 2G
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Re: Zeroeth, First, Second meaning

Postby Paul Hage 2G » Mon Mar 09, 2020 4:30 pm

Also, for a first-order reaction, ln[A]=-kt+ln[A]0

Leonardo Le Merle 1D
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Joined: Wed Feb 27, 2019 12:16 am

Re: Zeroeth, First, Second meaning

Postby Leonardo Le Merle 1D » Thu Mar 12, 2020 11:05 am

Also, note that just because a reactant is first order, doesn't necessarily mean the reaction itself is as well. If multiple reactants are first order, the reaction is the sum of those reactants (So if N and O were both first order in forming NO, the reaction is second order).

Tanmay Singhal 1H
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Re: Zeroeth, First, Second meaning

Postby Tanmay Singhal 1H » Thu Mar 12, 2020 11:06 am

first order means that rate of reaction is proportional to concentration of reacting substance

Myka G 1l
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Re: Zeroeth, First, Second meaning

Postby Myka G 1l » Thu Mar 12, 2020 11:07 am

In a first order reaction, the rate of reaction is directly proportional to the concentration of a single reacting substance. Whereas in a 2nd order reaction the rate depends on the concentration of 2 reacting first order substances or a single reactant with a squared concentration. Because it depends on a single substance, the graph of a first order reaction is a straight line. For a zeroth order reaction the rate is solely dependent on the rate constant.

Shutong Hou_1F
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Re: Zeroeth, First, Second meaning

Postby Shutong Hou_1F » Fri Mar 13, 2020 11:24 am

The units of k:
first-order: s^-1, second-order: (s^-1)*(mol^-1)*(L), zero-order: (mol)(L^-1)*(s^-1);


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