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

1. May 4, 2015

### mooncrater

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$\frac{dc}{dt}=-kc^n$

3. 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 devide then there is no $C_0$ in that. So why is (C) correct? (B) should be correct. Where am I wrong?

Last edited: May 5, 2015
2. May 5, 2015

### Staff: Mentor

I am not convinced that is correct.

What is $t_{\frac 3 4}$?

3. May 5, 2015

### mooncrater

The time in which 3/4th of the reactant is used up. Is it wrong?

4. May 5, 2015

### Staff: Mentor

Used, or left? I am honestly not sure. Perhaps that's because I am still sleepy.

5. May 5, 2015

### epenguin

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.

6. May 19, 2015

### James Pelezo

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:

For a 1st order reaction

For a 2nd order reaction:

Last edited: May 19, 2015