15.39 6th edition

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Ana Pedreros
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Joined: Fri Sep 28, 2018 12:19 am

15.39 6th edition

Postby Ana Pedreros » Thu Mar 14, 2019 4:42 pm

in this question part b calls for the time required for the concentration of B to increase to 0.19 M, given that k=.0035 1/Mmin in the rate law for the loss .The solution manual has the answer .055(molA)/L= .37[A]0. Can someone explain how they got .37

Jennifer Su 2L
Posts: 47
Joined: Wed Nov 21, 2018 12:20 am

Re: 15.39 6th edition

Postby Jennifer Su 2L » Thu Mar 14, 2019 5:04 pm

So first you want to find the concentration of [A] after the time period. Since they tell us that delta[B]=0.19M, we can multiply it by (1 mol A)/(2 mol B) to get delta[A].

0.19 mol B/L x (1 mol A/2 mol B) = 0.095 M of A

To get the final [A], simply subtract delta[A] from [A]not
[A]final=0.15M-0.095M =0.055M

I believe they simply solved 0.055M= x*[A]not= x*[0.15M], to which they got x=0.37. Not sure why this is important because it's not used in the rest of the problem...

You can then plug this into the second order integrated rate law to find time.
1/[0.055M] = -(0.0035 L/mol*min)*t + 1/[0.15M]
t=3290 mins= 3.3 x 10^3 mins


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