CEE 670 Fall 2011

# Kinetics Homework #1

26. A glass vial contains 0.5% potassium by weight.  The natural abundance of 40K, which decays by β-emission is 0.00118%, and its half-life is 1.28x109 years.  If the vial weighs 15g, calculate the expected rate of β- production by the vial in disintegrations per minute or dpm.  (problem 1 in Chapter 2 of Brezonik; pg.102).

Amount present now (Mass or Atoms)

15 g total weight

[40K] = 15g*(0.5/100)*(0.00118/100) = 8.85 x 10-7 g

= 8.85 x 10-7 g ÷  40 g/M = 2.21 x 10-8 M

 1 Point

=  6.028 x 1023 atoms/M x 2.21 x 10-8 M = 1.33 x 1016 atoms

Rate:

T ½ = 1.28 x109 years  = 6.73 x 1014 min

K= ln(2)/ t ½  = 0.693/ 6.73 x 1014 = 1.0296 x 10-15 min-1

Disintegrations =  1.33 x 1016 atoms x 1.0296 x 10-15 min-1

=  13.72 atoms/m or dpm

27. Using the kinetic plot and data for the hydrolysis of Benzyl Chloride as presented in class (refer to posted slides), determine the following:

a.
 1 Point

The expected concentration of benzyl chloride after 72 hours (@25°C), when the initial concentration was 33µg/L.

C = 0.923 µg/L

b.      The expected concentration of benzyl chloride after 72 hours (@25°C), when the initial concentration was 7.9µg/L.

C = 0.221 µg/L

c.       The expected concentration of benzyl chloride after 72 hours (@0.1°C), when the initial concentration was 33µg/L.

C = 29.6 µg/L

28. It has been known for some time that concentrated solutions of aqueous chlorine lose strength with time.  Accompanying this is the accumulation of chlorate ion.  Early chemical studies have determined this to be a second order reaction in hypochlorite.  The stoichiometry is as follows:

Bolyard and co-workers [1992, Env. Sci. Technol. 26(8)1663-1665] found that chlorate is being inadvertently added to drinking water when aqueous chlorine is used as a disinfectant.  Data collected at 14 sites showed that raw water chlorate levels were 0.02 mg/L or below.  However, finished water levels were as high as 0.66 mg/L, apparently due to contamination from the chlorine stocks.  Gordon and Adam [Water Disinfection News, 5:1, Novatek] have subsequently studied this reaction.  They determined that a 2.776 M chlorine stock solution (15.89% FAC, d=1.239 g/mL) has a 61.7 day half-life at 25°C and pH 13.  Based on this information answer the following questions.

a.       What is the 2nd order rate constant for this reaction in units of M-1s-1?

First recognize that all of the chlorine (2.776 M) is  in the form of hypochlorite (OCl-) at pHs greater than the pKa (7.6) and that chlorine is expressed as Cl2 (GFW = 71).  Note also that the parenthetical note “(15.89% FAC, d=1.239 g/mL)” is what I used to calculate the 2.776 M concentration.  You don’t need to use this information in the solution.

 1 Point

For a half-life:

k =2.25 x 10-8 M-1s-1

b.      Calculate the concentration of chlorate in the stock after 40 days holding time at 25°C

First calcuate the final hypochlorite concentration

and

So at 40 days

C = [OCl-] = 1.68 M

And recognize that the chlorate formed is equal to one-third times the hypochlorite lost

[ClO3-] = (Co-C)/3

[ClO3-] = 0.364 M

c.       What is the dosed concentration (mg/L) of chlorate if sufficient stock (aged 40 days at 25°C) is added to a finished drinking water to achieve a free chlorine dose of 2.5 mg/L?

For this you’ll want to calculate the molar ratio of chlorate to hypochlorite in the aged stock.

[ClO3-]/[OCl-] = 0.364M/1.68M = 0.2161 M/M

Now it’s a simple matter of converting the 2.5 mg/L dose of chorine to moles per liter (remember that chlorine is expressed as Cl2 and thus has a GFW of 71) and then convert to mg/L of ClO3 (GFW  = 83.45)

[ClO3-] = 0.635 mg/L

d.      Repeat "c", but assume that the chlorine stock was diluted by a factor of 4 prior to holding for 40 days.  Assume that dilution does not change the value of the rate constant [we will find out later that it will change a bit due to the change in ionic strength].

Again, you need to first calcuate the final hypochlorite concentration in the stock

and

So at 40 days, using the lower initial stock concentration (2.776/4 = 0.694M), we get:

C = [OCl-] = 0.597 M

And recognize that the chlorate formed is equal to one-third times the hypochlorite lost

[ClO3-] = (Co-C)/3

[ClO3-] = 0.03226 M

Now consider the concentration in the dosed water.  For this you’ll want to calculate the molar ratio of chlorate to hypochlorite in the aged stock.

[ClO3-]/[OCl-] = 0.03226M/0.597M = 0.054 M/M

Now it’s a simple matter of converting the 2.5 mg/L dose of chorine to moles per liter (remember that chlorine is expressed as Cl2 and thus has a GFW of 71) and then convert to mg/L of ClO3 (GFW  = 83.45)

[ClO3-] = 0.159 mg/L

e.       Consider an MCLG (maximum contaminant level goal) of 0.2 mg/L for chlorate.  Determine the amount of dilution of the chlorine stock, which will just allow compliance after 40 days holding time at 25°C.

Can be done analytically, or by trial and error.  For the latter, you can go back to the original kinetic model, and re-calculate the concentrations based on different dilution factors.  In excell this is quite easy if you had already set up the problem.  The answer is:

Dilution Factor = 3.2