# Order Rate

## 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

## The Attempt at a Solution

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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--- 0.0500

for 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

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.