## 7B.9

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

Henry Krasner 1C
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

### 7B.9

Before you use the first order integrated rate law, to calculate the concentration of A at the time you want to calculate, you have to do
[A]=[A(initial)]-(mol A/mol B)[B]. What equation is this?

deepto_mizan1H
Posts: 65
Joined: Fri Sep 28, 2018 12:16 am

### Re: 7B.9

Essentially the equation is just finding the change from initial [A(initial)] to find the value of [A] after a point in the reaction. Since we know the initial concentration of the product [B] is 0, the formation concentration of it can be treated as a decomposition of [A] if we multiply by the moles of reactant/product to find how much reactant is decomposing in relation to its formation of product. So, now that we've gotten a change in concentration by how much B has formed, we can subtract it from our initial value to find what [A] itself is.

It's basically utilizing the knowledge that [B] will start at 0 or near 0.

paytonm1H
Posts: 74
Joined: Fri Sep 28, 2018 12:18 am

### Re: 7B.9

This equation is used to find the concentration of A when 0.018 M of B has formed. Essentially, it is just balancing using the stoichiometric coefficients, and it is not an equation we explicitly learned. It makes sense because 3 moles of B forms from 1 mole of A, so it only took 1/3 of 0.018 M of [A] to form 0.018 M [B].

Additionally, the solutions manual does this same procedure for part (b), however in the manual they use 0.030 M [B], but the question asks for the time it takes for an increase of 0.30 M [B]. Is there a conversion we need to do or is this a manual mistake??
Thanks!

Philipp_V_Dis1K
Posts: 32
Joined: Fri Sep 28, 2018 12:20 am

### Re: 7B.9

if you use the formula of 1 over the molar coefficient one can find the change in [a] as a ratio of change of [b]

### Who is online

Users browsing this forum: No registered users and 1 guest