## 15.35a

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

Ethan-Van To Dis2L
Posts: 50
Joined: Fri Sep 29, 2017 7:05 am

### 15.35a

So I found the k value (0.02357), and used the integrated rate law for second order reactions, but my answer was 757.6 second but the answer was 7.4 x 10^2. Would this just be a rounding error on my part, or is the book wrong?

Jared Smith 1E
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.35a

I did the same thing and recalculated it with 0.02357 rounded to 0.024 like it does in the answer key, and it gives 744. I think it's just a rounding error on our part.

Jack Papciak 2F
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am

### Re: 15.35a

The book rounded the k value to 0.024L/molxs so that's why it's a slightly different answer.

Alejandra Rios 1L
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am
Been upvoted: 1 time

### Re: 15.35a

This should be a rounding error, if you were to use 0.024 as the k value you would get 744.05 as your final answer which is equivalent to 7.4 x 10^2, however if you were to leave the whole decimal, this would give you a value of 757.62. I think both answers would be correct as long as you provided the work.