15.19(c)

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Haocheng Zhang 2A
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

15.19(c)

Postby Haocheng Zhang 2A » Sat Mar 03, 2018 10:07 pm

For homework 15.19 part c, my answer is 2.85(8.7=k*1.25*(1.25^2)*(1.25^2)), but the solution manual says the answer is 2.85*10^12. I notice that the unit of initial rate is (mmol A)*L^-1*s^-1, so (mmol A) means mmol*10^12?

Vivian Nguyen
Posts: 66
Joined: Fri Sep 29, 2017 7:04 am
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Re: 15.19(c)

Postby Vivian Nguyen » Sat Mar 03, 2018 10:29 pm

mmol means millimoles, so x10^-3 moles.
However, because they converted all the values to moles, this gave them the 10^-12.
so in terms of experiment 1
8.7x10^-3/(1.25x10^-3)(1.25x10^-3)^2(1.25x10^-3)^2

which would give you
=8.7x10^-3/3.05x10^-15
=2.85x10^12

hope that makes sense! :)

Haocheng Zhang 2A
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

Re: 15.19(c)

Postby Haocheng Zhang 2A » Sun Mar 04, 2018 2:31 pm

But in the solution manual, the unit of k is L^4*mmol^-4*s^-1, so I think we don't have to convert mmol to mol. Is the solution manual wrong?


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