15.23
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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.
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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
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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]
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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)
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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
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