# 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?

##### 1 Answer

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.

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

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 **half-life** of

#-(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 thehalflife), 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.