## 15.29

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

Palmquist_Sierra_2N
Posts: 27
Joined: Wed Sep 21, 2016 2:56 pm
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### 15.29

For the first order reaction A-->3B+C, when initial concentration of A = 0.015 mol/L, the concentration of B increases to 0.018 mol/L in 3.0 minutes.
a) what is the rate constant for the reaction expressed as the rate of loss of A?
b) how much more time would be needed for the concentration of B to increase to 0.030 mol/L?

I don't understand the solution manual could someone help explain. Thanks so much!

feodora_r_3d
Posts: 16
Joined: Wed Sep 21, 2016 2:59 pm

### Re: 15.29

I think what the solution manual's trying to do is to find the concentration of A after 3 minutes. To do so, we need to use the data that is given. The question gave the concentration of B after 3 minutes and we know that initially, the concentration of B is none since the product has not formed yet. therefore the formation of B will be directly proportional to the loss of A.
Using stoichiometric relations (or coefficient in the reaction) we know that one mole of A would form 3 moles of B. Hence forming 0.018 mol/L of B would mean the loss of 0.018/3= 0.006 mol/L of A.
So after 3 minutes, the concentration of A would be 0.015 mol/L - 0.006 mol/L = 0.009 mol/L
Now that we have the initial concentration of A and the concentration of A after 3 minutes, we could use the formula $k = \frac{1}{t}ln\frac{[A]_{0}}{[A]}$
which gives $k = \frac{1}{3}ln\frac{0.015}{0.009}=0.17$

For the second part of the question, you could use the same method (to find the concentration of A if the concentration of B increases to 0.03 mol/L) and then use the same formula but changing t with k.
$t = \frac{1}{k}ln\frac{[A]_{0}}{[A]}$

Hope this helps!

Nikola_Stojcic_3O
Posts: 11
Joined: Wed Sep 21, 2016 2:55 pm

### Re: 15.29

So for part B, would you again convert the concentration of B increased into concentration A through stoichiometry and then plug that back into the rate law?

Janice Kim 3I
Posts: 33
Joined: Wed Sep 21, 2016 2:56 pm

### Re: 15.29

B is a product so the concentration of B increasing means that more product is being formed. Since .018 M of B is formed, using stoichiometric calculations, 0.018M B(1 mol A/3 mol B) = .006M A is lost because reactant concentration lessens as products are formed and product concentration is increased. Subtract this .006M value from the initial value given (in M) to get the concentration of A after 3.0 minutes. Now that you have [A]t, [A]0, and t(time), you can plug these numbers into the first order integrated rate law and solve for the rate constant.