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### Overall reaction order

Posted: Fri Feb 28, 2020 8:01 pm
What do you set rate 1/rate 2 equal to when trying to find m or n. In the example from class, they set it equal to 2 so that it was 2^m= 2 and m was just 1. I’m confused how to know what to set 2^m equal to.

### Re: Overall reaction order

Posted: Fri Feb 28, 2020 9:45 pm
We set rate 1/rate 2 equal to 2 in the example because the ratio of the initial rates between experiment 1 and experiment 2 is 2 and the units cancel out. And the initial rates of [NH4+] doesn't change, so, after applying equation: rate = k[NH4+]^n*[NO2-]^m, we are left with the ratio between initial concentrations of [NO2-] and thus we are calculating m, rather than n, using experiments 1 and 2.

### Re: Overall reaction order

Posted: Sat Feb 29, 2020 10:33 am
The key point is that the rate is made equal to the rate law equation with the concentrations inputted into their respective spots. Then, after division, the rate side is equal to 2 and the rate law side is equal to 2^m.

### Re: Overall reaction order

Posted: Sun Mar 01, 2020 9:41 pm
In his example during class, he divides (2.7*10^-7 mol/Ls = k(0.1M)^n (0.01M)^m) by (1.35*10^-7 mol/Ls = k(0.1M)^n (0.005M)^m).
This gives an overall equation of 2 mol/Ls = 2^m, since the two k(0.1M)^n terms cancel out.
He isn't setting rate 1/ rate 2 equal to a specific value, he is simply dividing the two equations for rate to isolate the variable m. So his technique is to divide equations.

### Re: Overall reaction order

Posted: Sun Mar 01, 2020 10:40 pm
Rate 2 is divided by Rate 1. Since the concentration of [NH4+] stays the same in reactions 1 and 2, these values and their units cancel out. k also cancels out when the two rates are divided. So, the only components left in the calculation are initial rate 2/initial rate 1 (which is equal to 2) and the ratio of the initial concentrations of [NO2-] (which is equal to 2) raised to the power of m. So, 2 = 2^m and you can then solve for m.