15.23


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Rucha Kulkarni 2A
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15.23

Postby Rucha Kulkarni 2A » Sun Mar 04, 2018 10:06 pm

Does anyone know how to solve question 15.23?

Thu Uyen Tran 1B
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Re: 15.23

Postby Thu Uyen Tran 1B » Sun Mar 04, 2018 10:15 pm

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
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Re: 15.23

Postby Joshua Hughes 1L » Sun Mar 04, 2018 10:16 pm

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
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Re: 15.23

Postby Joshua Hughes 1L » Sun Mar 04, 2018 10:22 pm

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
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Re: 15.23

Postby Joshua Hughes 1L » Sun Mar 04, 2018 10:25 pm

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
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Re: 15.23

Postby Gurkriti Ahluwalia 1K » Tue Mar 06, 2018 12:01 am

how do you get rid of the negative??

Akash_Kapoor_1L
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Re: 15.23

Postby Akash_Kapoor_1L » Tue Mar 06, 2018 4:24 pm

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


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