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Topic: Reaction Orders  (Read 5003 times)

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Offline ILoveISO

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Reaction Orders
« on: March 01, 2010, 06:34:23 PM »
x y and k, do they always have to be a WHOLE number like 1, 2? For example I'm trying to find x so I'm using this equation

rate1 = k[s2o8]^x[I-]^y
______________________
rate2 = k[s2o8]^x[I-]^y

Cancelled the K and I's to get

rate1 = [S2O8]^x
_______________
rate2 = [S2O8]^x

x = 0.72 is what I got is this possible or am I supposed to get near 1 or 2 etc?

Offline Teeny

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Re: Reaction Orders
« Reply #1 on: March 01, 2010, 06:57:50 PM »
I would just round up to 1.

Offline cliverlong

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Re: Reaction Orders
« Reply #2 on: March 01, 2010, 07:02:13 PM »
x y and k, do they always have to be a WHOLE number like 1, 2?
It appears orders do not have to be integers - but generally are:

http://en.wikipedia.org/wiki/Order_of_reaction (if you trust it)

Quote
For example I'm trying to find x so I'm using this equation

rate1 = k[s2o8]^x[I-]^y
______________________
rate2 = k[s2o8]^x[I-]^y

For which reaction? Where is your chemical equation?
Quote

Cancelled the K and I's to get

rate1 = [S2O8]^x
_______________
rate2 = [S2O8]^x

If (and I only write if) what you have done until now is valid, then why not cancel the terms above? From what you have written they are identical and should cancel, leaving the number 1.
Quote
x = 0.72 is what I got is this possible or am I supposed to get near 1 or 2 etc?
And how did you arrive at that number?
You have this habit of only giving fragments of information in your posts (I have read them before) and even less information on the methods you use - so it becomes almost impossible to work out what you are doing or where you are going wrong.

Also, please use the provided super-script and sub-script options provided on the forum editor. They take barely a second to use. Element symbols are written starting with upper-case.

Which of the following is easier to read?

rate1 = k[s2o8]^x[I-]^y

Rate1 = k[S2O8]x[I-]y

Clive

Offline ILoveISO

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Re: Reaction Orders
« Reply #3 on: March 01, 2010, 07:36:36 PM »
Well I have two equations and I combined them to one S2O8 + 2I --> I2 + 2SO4 and  2S2O8 + I2 --> 2I + S4O6 so I got the ratio 1:2

here is what I plugged into the equation

rate1 = k[S2O8]x[I-]y
______________________________________
rate2 = k[S2O8]2x[I-]2y

5.75x10^-6 = [0.383]x
____________________________
1.50x10^-5 = [0.500]2x

x = 1.38

I used Logs a = bx

Offline ILoveISO

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Re: Reaction Orders
« Reply #4 on: March 01, 2010, 08:24:31 PM »
Is what I did correct? Also how would I solve for K, not sure how to isolate it since it's on top and bottom...

Offline cliverlong

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Re: Reaction Orders
« Reply #5 on: March 02, 2010, 10:31:36 AM »
Well I have two equations and I combined them to one S2O8 + 2I --> I2 + 2SO4 and  2S2O8 + I2 --> 2I + S4O6 so I got the ratio 1:2
I find it extremely difficult to read what you have typed because of your refusal to use sub-scripts and superscripts  provided by the editor when you type in polyatomic ions. Also you casually ignore charges on ions

Do you mean by your equations the following

S2O82- + 2I- -> I2 + 2SO4 2-

2S2O82- + I2 --> 2I- + S4O6 2-

If that is what you meant, then what doesn't make sense to me is in the first equation

S2O82- is oxidizing I- to I2

Whereas in your second equation

S2O82- is reducing  I2 to I-

and I don't think both can happen without something else going on.
Quote
so I got the ratio 1:2
I don't understand what you mean by that at all.
Maybe if you clarify the earlier questions the meaning of the ratio statement will become clear.
Remember, rate equation indices have very little to do with the stoichiometry of the chemical equation that relates to them
Quote
here is what I plugged into the equation

rate1 = k[S2O8]x[I-]y
______________________________________
rate2 = k[S2O8]2x[I-]2y

5.75x10^-6 = [0.383]x
____________________________
1.50x10^-5 = [0.500]2x

x = 1.38

I used Logs a = bx
I don't understand what you mean by the subscripts in

[S2O8]2x
and
[I-]2y

I don't understand where you get your data from.

Clive


Offline cliverlong

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Re: Reaction Orders
« Reply #6 on: March 03, 2010, 10:20:38 AM »
Have a look at the following thread

http://www.chemicalforums.com/index.php?topic=40038.0

are those the reactions your wish to discuss?

Clive

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