Reaction rate: first or second order?

start by reading through my answer here

http://answers.yahoo.com/question/index;_ylt=Aul8r...

and here

http://answers.yahoo.com/question/index;_ylt=Aj7Mn...

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there are generally 2 types of reaction order problems you're going to see in chemistry class.

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Type 1... see the first link.

the INTEGRATED rate equations are

.. (1)... [A] = -kt + [Ao].... . ... ..zero order in A

.. (2)... ln[A] = -kt + ln[Ao].... . first order in A

.. (3)... 1/[A] = +kt + 1/[Ao].... .second order in A

meaning that exponent on

.. rate = k x [A]^n

is 0, 1, or 2...

keep in mind, the "order" of A can also be 3 or 4, it can be -, and it can even be a fractional number

in each of those equations..

.. k is a constant

.. [Ao] is a constant.. think about it.. we start with a fixed concentration

.. . ... .of A and measure the concentration of A over time.

.. .. .. .the fixed value we start with is [Ao]

.. [A] is the concentration of component A and it varies with time

so.. each of those equations is a line of the form y = mx + b, with different things for "y" and "m" and "b"

for a zero order reaction,

.. a plot of [A] vs t will give a line with slope = -k and intercept = [Ao]

for a first order reaction

.. a plot of ln[A] vs t will give a line with slope = -k and intercept = ln[Ao]

for a second order reaction... (1/[A]) = +kt + 1/[Ao]..

.. so a plot of 1/[A] vs t will give a line with slope = +k and intcp = 1/[Ao]

so..

.. we take the data put it in EXCEL

.. we make 4 columns... time, [A], ln[A], 1/[Ao]

.. we make 3 plots.. [A] vs t, ln[A] vs t, 1/[A] vs t

and then..

.. if the [A] vs t plot is linear, the reaction is ZERO order in A..

... .. and rate = k x [A]^0 ===> rate = k.. is our NON-integrated rate equation

.. if the [A] vs t plot is NOT linear, we look at the ln[A] vs t plot

.. . .if that is linear, the reaction is 1st order in A..

... ..and rate = k x [A] is our non-integrated rate equation

.. if neither the [A] vs t nor the ln[A] vs t plots are linear

.......and IF the 1/[A] vs t is linear, the rxn is 2nd order in A

... .. and rate = k x [A]² is our non-integrated rate equation.

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type 2 problem...

you have a table of concentration vs RATE like this

.. A + B ---> AB

run #... . [A].... .. [B]...... .rate

.. 1.. . ...0.100.. 0.100.. 3x10^-4

.. 2.. . ...0.100.. 0.200.. 6x10^-4

.. 3.. . ...0.200.. 0.100.. 1.2x10^-3

you compare change in concentrations and change in rates for the different runs. in this example

.. [B] doubled from 1 to 2 and rate doubled while [A] was constant

.. [A] doubled from 1 to 3 and the rate quadrupled while [B] was constant

so the NON-INTEGRATED rate equation goes like this

.. rate = k x [A]^n x [B]^m

and from the table of data we know that n=2 and m=1

.. rate = k x [A]² x [B]¹

because

.. doubling [A] quadruples rate.. so the exponent on A must be "2"

.. doubling [B] doubles the rate so the exponent on B must be "1"

and once you have the orders of the reactants, you can plug data from any run to calculate k. in the above example

.. k = rate / ([A]² x [B]¹)

if you plug in the date from say trial #3, you calculate k like

.. k = 1.2e-3 / ((0.2)² x (0.100)) = ?

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do you have a specific problem you're working on?