Solving the Order of Reactants: BrO3-, I-, H+

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AI Thread Summary
The discussion focuses on calculating the order of reactants in a chemical reaction involving BrO3-, I-, and H+. The user is confused about obtaining different orders from their calculations and seeks clarification on the correct approach. They provided experimental data and calculations, but some values, like 7.6E-7, are questioned for their origin. A suggestion is made to use linear regression by graphing log concentration versus log reaction rate to determine the best fit. This method could help clarify the discrepancies in the calculated orders of the reactants.
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Homework Statement



Calculate Order of Each Reactants
I am doing a lab on the reaction rates for k[BrO3-]y[I-]x[H+]z However when I do my calculations based on my data, i get different orders! The teacher said that will happen but how am I supposed to know which is the right one?


The Attempt at a Solution


log(5.3E-7/7.6E-7)/log(0.0017/0.0033)=0.5-I
log(5.3E-7/1.2E-6)/log(0.0067/0.0133)=1- BrO3-
log(5.3E-7/2.8E-6)/log(0.0067/0.0200)=1.5- BrO3-
log(1.2E-6/2.8E-6)/log(0.0133/0.0200)=2- BrO3-
log(5.3E-7/2.02E-6)/log(0.0167/0.0333)=2- H
log(5.3E-7/3.8E-6)/log(0.0167/0.0500)=2-H
log(2.02E-6/3.4E-6)/log(0.0333/0.0500)= 1 H
 
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I can't tell from your math what your numbers actually mean, since none of them have units or descriptions.

Can you list your data for each set of measurements?
 
Exp #---- Time of reaction (s) - -- Reaction rate (M/s)---- Initial Concentrations (M)
Trial 1---Trial 2---Trial 3-- -Average --- [I-] --- [BrO3-] ------[H+]
1- 26.27- -- 27.2 --- 24.5 ---25.99 ---5.3E-7---------0.0017---0.0067 ---0.0167

3 -9.59 --- 7.79 --- 5.68--- 7.69--- 1.8E-6------- 0.0050--- 0.0067--- 0.0167
4- 12.87 --- 11.8--- 10.75--- 11.8---1.2E-6 -------0.0017---0.0133 ---0.0167
5 -4.49---4.40--- 5.90--- 4.90 ---2.8E-6-------- 0.0017--- 0.0200--- 0.0167
6- 6.21---7.56--- 6.72 --- 6.83---2.02E-6---------0.0017---0.0067---0.0333
7- 4.39---4.14 --- 3.74--- 4.09--- 3.4E-6--- ---------0.0017--- 0.0067--- 0.0500for BrO3 i used experiments 1,4,5
for I i used experiment 1 and 3
and for H i used experiments 1,6 , 7

Calculations

log(5.3E-7/7.6E-7)/log(0.0017/0.0033)=1-I
log(5.3E-7/1.2E-6)/log(0.0067/0.0133)=1- BrO3-
log(5.3E-7/2.8E-6)/log(0.0067/0.0200)=1.5- BrO3-
log(1.2E-6/2.8E-6)/log(0.0133/0.0200)=2- BrO3-
log(5.3E-7/2.02E-6)/log(0.0167/0.0333)=2- H
log(5.3E-7/3.8E-6)/log(0.0167/0.0500)=2-H
log(2.02E-6/3.4E-6)/log(0.0333/0.0500)= 1 H
 
disneychannel said:
Calculations

log(5.3E-7/7.6E-7)/log(0.0017/0.0033)=1-I
Where does the 7.6e-7 come from?
log(5.3E-7/1.2E-6)/log(0.0067/0.0133)=1- BrO3-
log(5.3E-7/2.8E-6)/log(0.0067/0.0200)=1.5- BrO3-
log(1.2E-6/2.8E-6)/log(0.0133/0.0200)=2- BrO3-
log(5.3E-7/2.02E-6)/log(0.0167/0.0333)=2- H
log(5.3E-7/3.8E-6)/log(0.0167/0.0500)=2-H
log(2.02E-6/3.4E-6)/log(0.0333/0.0500)= 1 H

Since you have 3 points for both of these, that's the minimum needed for linear regression.

Try graphing log[x] vs log(rate) and find the best straight line to fit the data.
 
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