Question

Using the equation `(d[NO_3^(-)])/(dt) = k_1[NH_4^(-)]` calculate the predicted half-life of `"NH"_4^(-)` if the initial concentration is 2600 ppm?

Answer 1

First off, just to keep this in "common" units, I'm going to convert the 2600 ppm to molarity.

`"2600 ppm = 2600 mg/L"` of `NH_4^+`
`"2600 mg/L" * "1 g"/"1000 mg" * ("1 mol "NH_4^(+))/("14.007 g"+4*1.007"9 g") = \mathbf("0.1441 M")`

So, I feel like the question is off. `NH_4^-`? I've never heard of `H^-` ever wanting to react with such a good nucleophile as `NH_3` (though its pKa is much higher than that of `NH_3`, being way over 35 as the conjugate base of `H_2`, with the pKa of `NH_3` being 36...). I'm going to assume it's `NH_4^+`, though it doesn't really matter in the long run.

First, it helps to write the rate law of the reaction to get some context, if nothing else.

`\mathbf(r(t) = (d[NO_3^-])/dt = k_1[NH_4^+])`

So, you can see that the rate law says the particular mechanistic step is first order with respect to `NH_4^+`. So, what you can do is derive the first order integrated rate law, and graph this. Additionally, you can go further and make a few reasonable assumptions, getting a useful expression out of it. You need the rate constant before you can do anything else.

But first, realize that it helps to have a full reaction context. This is probably fine:

`mathbf(HNO_3 + NH_3 rightleftharpoons NO_3^(-) + NH_4^(+)`

where the equilibrium lies towards the products for the most part. In this case, the change in `NH_4^(+)` is positive.

Then, notice how the number of moles of `NH_4^+` are equal to that of `NO_3^-`. Since we are looking at the half-life of `NH_4^(+)`, its change will be negative after the initial formation is over (going off of the first equation):

`-(d[NH_4^+])/dt = k_1[NH_4^+]`

Then, rearrange this (separation of variables) to be like so:

`-1/([NH_4^+])d[NH_4^+]` = `k_1dt`

Follow up with an integral.

`int_([NH_4^+]_0)^([NH_4^+])1/([NH_4^+])d[NH_4^+]` = `int_(t_0)^t -k_1dt`

`-> ln[NH_4^+] - ln[NH_4^+]_0 = -k_1(t-t_0)`
`-> ln[NH_4^+] - ln[NH_4^+]_0 = -k_1(t-0)`
`-> ln[NH_4^+] - ln[NH_4^+]_0 = -k_1t`

`->` `y = mx + b` form:

`color(green)(ln[NH_4^+] = -k_1t + ln[NH_4^+]_0)`

where the slope is -k and the y-intercept is `ln[NH_4^+]_0`, the natural log of the initial concentration. When you assume that the initial concentration is `100%` and the final is `50%` (for the half life), you can condense it down to:

`ln\frac{[NH_4^+]}{[NH_4^+]_0} = ln (\frac{1/2}{1}) = -k_1*t_(1/2)`
so...

`ln 2 = k_1*t_(1/2)`

The final result is:

`color(blue)(t_(1/2) = (ln2)/k_1)`

So if you have k, you can get the half life.