Half life

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Half life

Postby DamianW » Sun Mar 11, 2018 5:20 pm

for a second order reaction, when given something like 1/16 why do you flip it in the equation and write 16-1?

Alyssa Pelak 1J
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Re: Half life

Postby Alyssa Pelak 1J » Sun Mar 11, 2018 5:58 pm

you would use the integrated rate law for this question.


You know that 1/[A] is 1/16[A]0 so you sub that into the equation:

In order to remove the fraction in the denominator you get

Hope this helps!

Joanne Guan 1B
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Re: Half life

Postby Joanne Guan 1B » Sun Mar 11, 2018 10:14 pm

If you're talking about finding the time for when the concentration is 1/16 of the initial concentration, you would use the integrated second-order rate law which is:

1/{A] = 1/[A]0 + kt
[A] = (1/16)[A]0

Subbing that into the 1/[A] part would give you 1/((1/16)[A]0), and if you reorganize it, you have 16[A]0.

Salman Azfar 1K
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Re: Half life

Postby Salman Azfar 1K » Mon Mar 12, 2018 10:14 am

The key is that when you put them into the formula (integrated second order rate law) I believe you end up using reciprocals due to the way the law is formatted. Just remember that half life problems for second order problems usually require use of the rate law; they aren't usually as simple as adding together multiple half lives for first order.

Timothy Kim 1B
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Re: Half life

Postby Timothy Kim 1B » Mon Mar 12, 2018 4:08 pm

Since it is a fraction, the value would appear flipped when it is used in the second-order reaction. take a look at the respective equations and you will see that the values correspond to each other.

Christina Bedrosian 1B
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Re: Half life

Postby Christina Bedrosian 1B » Tue Mar 13, 2018 11:16 am

it is 1/(1/16) making it 16/1 (it flips); just plug it into the equation and you should be fine

Jennie Fox 1D
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Joined: Sat Jul 22, 2017 3:01 am

Re: Half life

Postby Jennie Fox 1D » Tue Mar 13, 2018 4:23 pm

If you plug it in to the equation, you should get the correct answer. 1/(1/16) is 16.

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