## 15.23 (c)

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

Sarahi Garza 3k
Posts: 16
Joined: Wed Sep 21, 2016 2:57 pm

### 15.23 (c)

Why is it that to find the final concentration of A (to later find k), we must subtract the initial concentration of A to 2 times the final concentration of B?

Jose_Arambulo_2I
Posts: 35
Joined: Wed Sep 21, 2016 2:59 pm

### Re: 15.23 (c)

This is because the rate that B increases is equal to the rate that A decreases. I currently don't have the book on me, but it's probably doubled due to the rxn's stoichiometric coefficients

Sarahi Garza 3k
Posts: 16
Joined: Wed Sep 21, 2016 2:57 pm

### Re: 15.23 (c)

Thank you! Yeah, the stochiometric coefficient for A is 2.

Aishwarya_Natarajan_2F
Posts: 11
Joined: Mon Jul 11, 2016 3:00 am

### Re: 15.23 (c)

In this question when you use the unique rate equation for the reaction you know that the rate of increase in products in the rate of decrease in reactants, and therefore:

-(1/2)d[A]/dt = d[b]/dt

Using this we can find the rate of decrease in A by multiplying the rate of increase of B by 2. Then this gives us the rate of decrease in A which we can subtract from the A initial to give us [A] after 115 s which we can substitute into the integrated rate law to solve for k.

ln[0.153 + x] - ln[0.153] = -k(115 s)
with x being d[A]/dt.

using this equation, we get k= 5.11 x 10^-3