Question

How is the half-life of a first-order reaction affected by change in concentration?

Answer 1

It's not. The half-life of a reactant is intrinsic to the identity of the reactant, which has its own rate constant.

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A first-order reaction rate law, for the reaction

`A -> B`,

would be written as:

`bb(r(t) = k[A] = -(d[A])/(dt))`

where `r(t)` is the rate of reaction in `"M/s"`, `k` is the rate constant in `"s"^(-1)`, and `(d[A])/(dt)` is the change in the concentration `[A]` of `A` over the change in time `t` (otherwise known as the rate of disappearance).

When you separate variables by multiplying by `-(dt)/([A])` on both sides, you'd get:

`-kdt = 1/([A])d[A]`

To observe the process over time, we will be integrating from `0` to `t` on the left and `[A]_0` (initial concentration) to `[A]` (current concentration) on the right.

We get:

`-int_(0)^(t) kdt = int_([A]_0)^([A]) 1/([A])d[A]`

`-kt = ln[A] - ln[A]_0`

`ln(([A])/([A]_0)) = -kt`

which is just a rearrangement of the integrated rate law for a first-order reaction.

Now, if we say that we want to know how much time it takes for `[A]` to be `([A]_0)/2`, we are asking for the half-life, `t_"1/2"`.

Therefore:

`ln((1/2cancel([A]_0))/(cancel([A]_0))) = -kt_"1/2"`

Since `ln(1/2) = ln(2^(-1)) = -ln2`, we have:

`-ln2 = -kt_"1/2"`

`ln2 = kt_"1/2"`

`=> color(blue)(t_"1/2" = ln2/k)`

So, the half-life depends only on the rate constant for the first-order reaction that is occurring.