## Exam 3: Question 6

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

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Angel R Morales Dis1G
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Joined: Sat Jul 22, 2017 3:01 am

### Exam 3: Question 6

How would we determine which rate constant to use given the following facts: 1st order reaction, proceeds for 10 minutes and that only 71.9% of the initial concentration of reactants remain.

Nha Dang 2I
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

### Re: Exam 3: Question 6

You would use ln([A]t/[A]0)=-kt with [A]t=.719[A]0

Then, you would can cancel out [A]0 in ln(.719[A]0/[A]0)=-k(10 minutes) and solve for k from there.

Jonathan Tangonan 1E
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: Exam 3: Question 6

If you were to use the integrated rate law for the constant you could determine it by substituting for the concentration at time t with (.719)(initial concentration) and then isolate k from there.

ln[A]= -kt + ln[Ainitial]

ln(.719[Ainitial]) - ln[A] = -kt

ln[.719[Ainitial]/[A]] = -kt

-ln[.719]/10min = k

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