## 15.23

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

Rucha Kulkarni 2A
Posts: 43
Joined: Fri Sep 29, 2017 7:05 am

### 15.23

Does anyone know how to solve question 15.23?

Thu Uyen Tran 1B
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: 15.23

Use the half life equation for first order reactions, t1/2=0.693/k (you can find the derivation on page 628). Using this equation and the information given, we can find the rate constant, k. The t1/2 is the time it takes for the concentration of A to decrease to half its original concentration, which is 1000 sec.

Joshua Hughes 1L
Posts: 81
Joined: Sat Jul 22, 2017 3:01 am
Been upvoted: 3 times

### Re: 15.23

For part A of your question, you can use the half-life for first-order reaction equation to solve k. k = .693/1000 second = about 6.93 x 10^-4 /s

Joshua Hughes 1L
Posts: 81
Joined: Sat Jul 22, 2017 3:01 am
Been upvoted: 3 times

### Re: 15.23

For part, c use the equation ln [At]/[Ao] = -kt and rearrange to set equal to K. Then plug in given values of [Ao] and the time(115s). You also need to plug in [At] which you get by subtracting the rise in [B](multiplied by mole conversion of 2) from [Ao]

Joshua Hughes 1L
Posts: 81
Joined: Sat Jul 22, 2017 3:01 am
Been upvoted: 3 times

### Re: 15.23

ln [At]/[Ao] = -kt solve for k so it k=(ln([At]/[Ao]))/(-t) or k= ln([Ao]/[At])/t then plug in the given values ([Ao] being the .67 mol/L)

Gurkriti Ahluwalia 1K
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

### Re: 15.23

how do you get rid of the negative??

Akash_Kapoor_1L
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Re: 15.23

Gurkriti Ahluwalia 1K wrote:how do you get rid of the negative??

You get rid of the negative by flipping the natural log... Use the laws of logs and remember that ln[At]/[Ai] = ln[At] - ln[Ai], so if you bring the negative to the other side, you turn that into ln[Ai] - ln[At] = ln[Ai]/[At]

This is why you get: t = 1/k * ln[Ai]/[At] and k = 1/t * ln[Ai]/[At] for first-order reactions

Ai = A @ time 0
At = A @ time t