CHEMISTRY 253

Spring, 2015 - Dixon

Homework Set 1.2 Turn In Problem Solutions

 

Problems:

1-6.  Not all XO molecules such as NO2 survive long enough to react with oxygen atoms; some are photochemically decomposed to X and atomic oxygen, which then reacts with O to re-form ozone.  Write out the three steps (including one for ozone destruction) for this process and add them together to deduce the net reaction.  Does this sequence destroy ozone overall, or is it a null cycle, which is defined as one that involves a sequence of steps with no chemical change overall?

                        Step 1) NO2 + hn → NO + O

                        Step 2) O + O2 + M  → O3 + M

                        Step 3) NO + O3 → NO2 + O2

                        Net rxn: each reactant can be cancelled out by products, so this is a null             cycle.

 

2-1.  A minor route for ozone destruction in the ozone hole involves Mechanism II with bromine as X’ and chlorine as X (or vice-versa).  The ClO and BrO free radical molecules produced in these processes then collide with each other and rearrange their atoms to eventually yield O2 and atomic chlorine and bromine.  Write out the mechanism for this process, and add up the steps to determine the overall reaction.

                        Step 1) Br +  O3 → BrO + O2

                        Step 2) Cl +  O3 → ClO + O2

                        Step 3) BrO + ClO → BrOOCl

                        Step 4) BrOOCl + hn → Br + Cl + O2

            Net rxn: 2O → 3O2

 

Additional Problems:

1-7  At an altitude of about 35 km, the average concentrations of O* and of CH4 are approximately 100 and 1 x 1011 molecules cm-3, respectively, and the rate constant k for the reaction between them is approximately 3 x 10-10 cm3 molec-1 s-1.  Calculate the rate of destruction of methane in molecules per s per cm3 and in g per year per cm3 under these conditions.

rate = k[O*][CH4] = (3 x 10-10 cm3 molec-1 s-1)(100 molecules cm-3)(1 x 1011 molecules cm-3)

rate = 3 x 103 molecules cm-3 s-1

rate = (3 x 103 molecules cm-3 s-1)(3600 s/hr)(24 hr/d)(365.25 d/yr)(1 mol/6.02 x 1023 molec)(16.0 g/mol)

rate = 2.52 x 10-12 g cm-3 s-1

 

2-6  When Mechanism II for ozone destruction operates with X = Cl and X’ = Br, the radicals ClO and BrO react together to reform atomic Cl and Br.  A fraction of the latter process proceeds by the intermediate formation of BrCl, which undergoes photolysis in daylight.  At night, however, all the bromine eventually ends up as BrCl, which does not decompose and restart the mechanism until dawn.  Deduce why all the bromine exists as BrCl at night, even though only a fraction of the ClO with BrO collisions yield this product.

 

BrCl is relatively stable if there is no photolysis (e.g. at night), so once formed, it will stay in that form.  During daylight hours, most of stratospheric Br is in the form of BrO or Br with little present in reservoir species.  Thus, Br will quickly react with ozone to form BrO, and BrO can eventually react with ClO to form the dimer.  The dimer will either produce Br or BrCl, with Br atoms eventually re-reacting to form BrO and re-reacting with ClO.  [Note: this is assuming that insignificant quantities of BrONO2 and HBr form]