CHEMISTRY 253

Spring, 2015 - Dixon

Homework Set 1.1 (Problems to turn in)

 

Ch. 1                           

Problems: 2, 4, 5

2.  The DHº for the decomposition of ozone into O2 and atomic oxygen is +105kJ mol-1.  What is the longest wavelength of light that could dissociate ozone in this manner? By reference to Figure 1-2, decide the region of sunlight (UV, visible, or infrared) in which this wavelength falls.

Converting this to an energy per molecule,

(105 x 103 J/mol)(1 mol/6.02 x 1023 molecules) = 1.744 x 10-19 J/molec.

Now we can convert this to a wavelength: E = hc/l

l = hc/E = (6.63 x 10-34 J·s)(3.00 x 108 m/s)/1.744 x 10-19 J = 1.14 x 10-6 m = 1140 nm

This corresponds to near infrared light.

 

4.  Given the total concentration of molecules in air decreases with increasing altitude, would you expect the relative concentration of ozone, on the ppb scale to peak at a higher or lower altitude or the same altitude compared to the peak for the absolute concentration of the gas?

The mixing ratio gives the ratio of the concentration of ozone to air.  Since the air concentration decreases with altitude, the peak mixing ratio will occur at higher altitudes than the peak concentration.

 

5.  By reference to the information in Problem 1-2, calculate the longest wavelength of light that decomposes ozone to O2* and O*, given the following thermochemical data:

O → O* DHº = 190 kJ/mol and O2 → O2*  DHº = 95 kJ/mol

O3 + hn → O2 + O DHº = +105kJ mol-1.  Thus, O3 + hn → O2* + O* will require light with DHº = 105 + 190 + 95 kJ/mol = 390 kJ/mol.

Converting this to an energy per molecule,

(390 x 103 J/mol)(1 mol/6.02 x 1023 molecules) = 6.48 x 10-19 J/molec.

Now we can convert this to a wavelength: E = hc/l

l = hc/E = (6.63 x 10-34 J·s)(3.00 x 108 m/s)/ 6.48 x 10-19 J/ = 3.07 x 10-7 m = 307 nm

 

Additional Problems: 5

DHfº (Cl) = +121.7 kJ/mol for ½Cl2(g) → Cl(g)

so for 1 mol of Cl2, DHº = 2(121.7 kJ/mol) = 243.4 kJ/mol

on a molecular basis, this is: (243.4 x 103 J/mol)(1 mol/6.02 x 1023 molecules

 = 4.043 x 10-19 J/molec.  This corresponds to l = hc/E =4.92 x10-7 m = 492 nm = visible light