## Half-Lives for First Order Reactions

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

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Abigail Urbina 1K
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Joined: Thu Jul 27, 2017 3:01 am

### Half-Lives for First Order Reactions

Why is it that for a first-order reaction, the half-life of a substance is independent of its initial concentration? Why can the concentration of A be used as its "initial" concentration at any stage of the reaction? Can someone please clarify this?

Jana Sun 1I
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

### Re: Half-Lives for First Order Reactions

To understand why the half-life of first order reactants are independent of the substance's initial concentration, it really helps to look at how we get our half-life equation.

We start with our integrated rate law ln[A] = -kt +ln[A]o. Since the half-life of anything is the time that it takes for half of the substance to decompose, our [A]final is equal to 0.5 of our [A]original. So then when we substitute that in our integrated rate law, we get ln(0.5[A]o) = -k*(t half) + ln[A]o. We then try to isolate t half by first subtracting ln[A]o from ln(0.5[A]o). This will give us the same number no matter what initial concentration we have, so we'll always have -0.693 = -k(t half). Then we can solve for t half so that t half = 0.693/k.

I hope that answered the first part of your question! I'm a little confused about what you mean by your second question though, sorry!

Emily Duggan 1F
Posts: 42
Joined: Fri Sep 29, 2017 7:07 am

### Re: Half-Lives for First Order Reactions

For the half life formula for first order reactions, the initial condition is in the equation but they are canceled to simplify!
For [A] at its half life it is equivalent to 1/2[A]o (half of initial concentration)
So in the integrated rate law, ln 1/2[A]o = -kt1/2 + ln [A]o
simplifying... ln ((1/2[A]o)/[A]o) = -kt1/2
now u can cancel the initial concentrations...
ln 1/2 = -kt1/2
then rearranging to
t1/2 = .693/k
hope this helps!

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