Example 15.2

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Sabrina Fardeheb 2B
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Joined: Sat Jul 22, 2017 3:00 am
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Example 15.2

Postby Sabrina Fardeheb 2B » Sun Mar 04, 2018 5:01 pm

Can someone please explain the explanations on page 622 for me? I understand the definitions and the rate laws of each reaction order, but I don't know how to determine the order if given a table of concentration values.

Vasiliki G Dis1C
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

Re: Example 15.2

Postby Vasiliki G Dis1C » Sun Mar 04, 2018 5:24 pm

To find the order given a table of values, you want to find where only the concentration of one reactant is changing. If the other concentrations are constant, then you know that they have not affected the rate, and that the change in concentration of the one reactant is the only influence on the rate. For example, for BrO3-, in experiments 1 and 2, the concentration changes form 0.10 to 0.20, but the concentration of the other reactants stays the same. You can set up a rate law equation for experiment 1 to be 1.2= (0.1)^n(0.1)^m(0.1)^l and for experiment 2 to be 2.4=(0.2)^n(0.1)^m(0.1)^l. If you divide these 2 equations by each other, the (0.1)^m(0.1)^l cancel, and you get 2=2^n. Therefore n=1, and the order for BrO3- is 1. You can do the same thing for Br- using experiments 1 and 3, and use experiments for 2 and 4 for H3O+. For the H3O+, because you get 2.3=(1.5)^l, you have to take use ln to find the value of l. You get ln(2.3)=l(ln(1.5), and therefore l=ln(2.3)/ln(1.5), which is 2. The explanation skips setting up the different rate laws for the experiments and skips to the step where everything has been canceled out, but the full way to calculate individual orders is to set up the full rate law and then cancel out the constant terms. Hope this helps!

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