# 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?

Apr 9, 2015

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

$\text{2600 ppm = 2600 mg/L}$ of $N {H}_{4}^{+}$
"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. $N {H}_{4}^{-}$? I've never heard of ${H}^{-}$ ever wanting to react with such a good nucleophile as $N {H}_{3}$ (though its pKa is much higher than that of $N {H}_{3}$, being way over 35 as the conjugate base of ${H}_{2}$, with the pKa of $N {H}_{3}$ being 36...). I'm going to assume it's $N {H}_{4}^{+}$, 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.

$\setminus m a t h b f \left(r \left(t\right) = \frac{d \left[N {O}_{3}^{-}\right]}{\mathrm{dt}} = {k}_{1} \left[N {H}_{4}^{+}\right]\right)$

So, you can see that the rate law says the particular mechanistic step is first order with respect to $N {H}_{4}^{+}$. 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 $N {H}_{4}^{+}$ is positive.

Then, notice how the number of moles of $N {H}_{4}^{+}$ are equal to that of $N {O}_{3}^{-}$. Since we are looking at the half-life of $N {H}_{4}^{+}$, its change will be negative after the initial formation is over (going off of the first equation):

$- \frac{d \left[N {H}_{4}^{+}\right]}{\mathrm{dt}} = {k}_{1} \left[N {H}_{4}^{+}\right]$

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

$- \frac{1}{\left[N {H}_{4}^{+}\right]} d \left[N {H}_{4}^{+}\right]$ = ${k}_{1} \mathrm{dt}$

Follow up with an integral.

${\int}_{{\left[N {H}_{4}^{+}\right]}_{0}}^{\left[N {H}_{4}^{+}\right]} \frac{1}{\left[N {H}_{4}^{+}\right]} d \left[N {H}_{4}^{+}\right]$ = ${\int}_{{t}_{0}}^{t} - {k}_{1} \mathrm{dt}$

$\to \ln \left[N {H}_{4}^{+}\right] - \ln {\left[N {H}_{4}^{+}\right]}_{0} = - {k}_{1} \left(t - {t}_{0}\right)$
$\to \ln \left[N {H}_{4}^{+}\right] - \ln {\left[N {H}_{4}^{+}\right]}_{0} = - {k}_{1} \left(t - 0\right)$
$\to \ln \left[N {H}_{4}^{+}\right] - \ln {\left[N {H}_{4}^{+}\right]}_{0} = - {k}_{1} t$

$\to$ $y = m x + b$ form:

$\textcolor{g r e e n}{\ln \left[N {H}_{4}^{+}\right] = - {k}_{1} t + \ln {\left[N {H}_{4}^{+}\right]}_{0}}$

where the slope is -k and the y-intercept is $\ln {\left[N {H}_{4}^{+}\right]}_{0}$, 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:

$\ln \setminus \frac{\left[N {H}_{4}^{+}\right]}{{\left[N {H}_{4}^{+}\right]}_{0}} = \ln \left(\setminus \frac{\frac{1}{2}}{1}\right) = - {k}_{1} \cdot {t}_{\frac{1}{2}}$
so...

$\ln 2 = {k}_{1} \cdot {t}_{\frac{1}{2}}$

The final result is:

$\textcolor{b l u e}{{t}_{\frac{1}{2}} = \frac{\ln 2}{k} _ 1}$

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