15.19c

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Ryan Neis 2L
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

15.19c

Postby Ryan Neis 2L » Thu Mar 08, 2018 5:16 pm

For this portion of the problem, I solved for the rate constant by plugging in the values into the rate law and converting from mmol to mol at the very end, however doing so gave me a completely different answer. So I tried converting all the values from mmol to mol before plugging them into the rate law and it gave me the correct answer. My question is why do you need to convert the values from mmol to mol before you plug them into the rate law equation to get the right rate constant value?

Harrison Wang 1H
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: 15.19c

Postby Harrison Wang 1H » Thu Mar 08, 2018 5:31 pm

Rate is given in units of mol/(Ls), and as such when doing calculations involving the rate law, all your values should be in moles.

Shane Simon 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

Re: 15.19c

Postby Shane Simon 2K » Thu Mar 08, 2018 9:32 pm

You could have gotten the answer wrong because of the units. For part c of the question, if you left the units as mmol, the answer would have had mmol^-4 in the units. Since the mmol has an exponent, you would have to multiply the answer by 1000^4 in order to convert it into mol^-4. If you just converted it by multiplying it by 1000 it would have given you a very off answer. Basically, converting mmol to mol happens just by multiplying by 1000 but converting mmol^n to mol^n, you must multiply it by 1000^n.


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