second order half life calculation


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505166714
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Joined: Fri Sep 28, 2018 12:17 am

second order half life calculation

Postby 505166714 » Thu Mar 14, 2019 2:01 pm

For exercise 15.35 on the 6th edition book, can I use the half-life and multiply by 4 to find the time for the reactant to become 1/16 of the original amount in a second-order reaction?
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LilyL1C
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Joined: Fri Sep 28, 2018 12:20 am

Re: second order half life calculation

Postby LilyL1C » Thu Mar 14, 2019 5:41 pm

No you cannot. If you do one half life, you'll end up with half of the initial concentration. If you take another half life, you get 1/4 of the initial concentration. Another half life, and it's 1/8. One more half life, and you finally get 1/16. So it was 4 half lives, which is (1/2)^4 = 1/16. You need to take the half life to the 4th power, not multiply by 4.

For part b, you need to square the half life because (1/2)^2 = 1/4.

And for part c, 1/5 is not a multiple of 1/2, so you need to go back to the original 1/[A] = kt + 1/[A(initial)] and plug in [A]=1/5[A(initial)], and solve for an equation.


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