## homework problem #35

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

Sonia Perez 1G
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### homework problem #35

The half life for the second-order reaction of a substance A is 50.5 s when [A]0 = 84 M. Calculate the time needed for the concentration of A to decrease to (a) one-sixteenth, (b) one-fourth, (c) one-fifth of its original value.

For part a, if we are given the half life why are we not able to multiply 50.5 seconds by four half lives? Isn't 1/16 the same as (1/2)(1/2)(1/2)(1/2)?

Chem_Mod
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### Re: homework problem #35

Sonia Perez 1G wrote:The half life for the second-order reaction of a substance A is 50.5 s when [A]0 = 84 M. Calculate the time needed for the concentration of A to decrease to (a) one-sixteenth, (b) one-fourth, (c) one-fifth of its original value.

For part a, if we are given the half life why are we not able to multiply 50.5 seconds by four half lives? Isn't 1/16 the same as (1/2)(1/2)(1/2)(1/2)?

Cause the half life for the second-order reaction is not a constant. it's t1/2=1/(k*[A]initial). Where the [A]initial here is changing for the question. To calculate the time from [A]0 to [A]0/2, you should use [A]0 as [A]initial. When calculate time from [A]0/2 to [A]0/4, you should use [A]0/2 as the [A]initial

Ariana de Souza 4C
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### Re: homework problem #35

So it's because the equation for the second order half lives include the concentration of the initial substance? Then the reason we can multiply half lives for problem 15.27 is because, there, it's talking about a first order reaction, which only concerns the rate constant?
For reference, 15.27 : "A substance A decomposes in ta first order reaction and its half life is 355 s. how much time must elapse for the concentration A to decrease to a) 1/8th b) 1/4th (and it goes on like that)"

Alyssa_Hsu_2K
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### Re: homework problem #35

Because second order reactions include initial concentration as part of the calculation for half life, each individual half life steps are going to be different. As a result, you can not treat it as a constant the way you can when it comes to first order reactions.

Julie Barreto 3A
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Joined: Wed Sep 21, 2016 2:56 pm

### Re: homework problem #35

for question 39 i understand using the second order integrated rate law but what does the question mean when it asks for the rate law expressed in terms of the loss of A??