##\frac {t_{1/2}}{t_{3/4}}## for an nth order reaction

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SUMMARY

The discussion centers on the relationship between the half-life ratio ##\frac {t_{1/2}}{t_{3/4}}## for nth order reactions, specifically when n is not equal to 1. The conclusion drawn is that the correct answer to the question posed is (C), indicating that both initial concentration and the order of the reaction (n) influence this ratio. The mathematical derivation confirms that the concentration term cancels out, leaving n as a determining factor. This highlights the unique behavior of half-lives in reactions of different orders.

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Homework Statement


The question:
For an nth order reaction ##\frac {t_{1/2}}{t_{3/4}}## depends on (n isn't equal to 1) :
(A) initial concentration only
(B) 'n' only
(C) initial concentration and 'n' both
(D) sometimes 'n' and sometimes initial concentration.

Homework Equations


##\frac{dc}{dt}=-kc^n##

The Attempt at a Solution


Integrating this equation:
(from ##C_0## to ##C## and ##0## to ##t##)
##\frac {1}{n-1}## ##[\frac {1}{C^{n-1}}-\frac {1}{(C_{0})^{n-1}}]=kt##
So when ##t=t_{1/2}## then ##C=C_0/2## and when ##t=t_{3/4}## then ##C=C_0/4##
And when we put this in the equation and divide then there is no ##C_0## in that. So why is (C) correct? (B) should be correct. Where am I wrong?
 
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mooncrater said:
when ##t=t_{3/4}## then ##C=C_0/4##

I am not convinced that is correct.

What is t_{\frac 3 4}?
 
Borek said:
I am not convinced that is correct.

What is t_{\frac 3 4}?
The time in which 3/4th of the reactant is used up. Is it wrong?
 
Used, or left? I am honestly not sure. Perhaps that's because I am still sleepy.
 
So far as I can see you are not wrong and there is a cancellation such that only n appears in the ratio.

Where are you getting these problems from? It would be useful if your personal info included where you are.
 
Borek said:
I am not convinced that is correct.

What is t_{\frac 3 4}?
I've seen reports on decay times referred to as t3/4 that refer to the time for completion of two half-lives. That is, t1/2(1) + t1/2(2) = t3/4, time required to consume 3/4 of the initial amount of reagent. However, for C = Co/4 which is the sum of two half-lives, but holds only for 1st order reactions. (for example, radioactive decay is 1st order) For n ≠ 1 each subsequent t1/2 calculated compresses or expands depending on the order of reaction being studied. Here are some trends of half-lives relative to order of reaction. For 'dependence on initial concentration', and in the light of the fact that none of the half-life equations contain a concentration term, one would have to assume (for the sake of answering the above question) that to consider a time as t1/2, one has to assume (arbitrarily) that an initial concentration value must exist so as to give meaning to the term half-life, which, I must admit, is a stretch, but is the only logical inference I could relate to the question given.

For a zero-order reaction:
zero_order_half-life.jpg


For a 1st order reaction
first_order_half-life.jpg


For a 2nd order reaction:
Screen_shot_2011-03-14_at_2.02.16_PM.png
 
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