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### 7th edition 7A. 17

Posted: Sun Mar 10, 2019 3:30 pm
For part c) determine the value of the rate constant, I get 2.85 as the answer for K not 2.85 x 10^12. How do you get the correct answer and why are the units L^4 x mmol^4 x s^-1 as shown in the back of the book?

### Re: 7th edition 7A. 17

Posted: Sun Mar 10, 2019 6:23 pm
The solutions manual's answer is in terms of moles instead of mmol (1 mmol = 1x10^-3 mol). They converted all the units from mmol to mol before solving for the units of k. Since the whole problem is in terms of mmol, I think you are okay leaving the answer in terms of mmol as well.

### Re: 7th edition 7A. 17

Posted: Sun Mar 10, 2019 7:08 pm
If you turn the concentrations given in mmol units to mole units, you will get the right answer. For example, on the top row turn each of the 1.25 mmol to 1.25*10^-3 mol. The reaction is of the fifth order. (1.25*10^3)^5 is equal to about 3.05*10^-15.

When you set up the equality k(3.05*10^-15)=8.7*10^-3, you will get k=2.85*10^12 L^4*mol^-4*s^-1. (I have displayed the rate constant here as it is displayed in the solutions manual.)

The units are as shown because you are trying to get k into units that will lead to the actual rate being in terms of mol*L^-1*s^-1. Page 595 in the seventh edition explains this relationship between the rate and the rate constant.

### Re: 7th edition 7A. 17

Posted: Sat Mar 16, 2019 2:20 am
Oh ok, so is the answer key just wrong then? (They calculated it in terms of moles but wrote it down as mmol)