The stratospheric ozone (O3) layer helps to protect us from harmful ultraviolet radiation. It does so by absorbing ultraviolet light and falling apart into an O2 molecule and an oxygen atom, a process known as photodissociation. O3(g)→O2(g)+O(g)?
The stratospheric ozone (O3) layer helps to protect us from harmful ultraviolet radiation. It does so by absorbing ultraviolet light and falling apart into an O2 molecule and an oxygen atom, a process known as photodissociation.
O3(g)→O2(g)+O(g)
a) Calculate the enthalpy change for this reaction.
b) What is the maximum wavelength a photon can have if it is to possess sufficient energy to cause this dissociation?
c) In what portion of the spectrum does this wavelength occur?
The stratospheric ozone (O3) layer helps to protect us from harmful ultraviolet radiation. It does so by absorbing ultraviolet light and falling apart into an O2 molecule and an oxygen atom, a process known as photodissociation.
O3(g)→O2(g)+O(g)
a) Calculate the enthalpy change for this reaction.
b) What is the maximum wavelength a photon can have if it is to possess sufficient energy to cause this dissociation?
c) In what portion of the spectrum does this wavelength occur?
1 Answer
You should be looking up these data yourself. Your textbook has an appendix which gives...
#DeltaH_f^@("O"_3(g)) = "142.67 kJ/mol"#
http://webbook.nist.gov/cgi/cbook.cgi?ID=C10028156&Mask=1#Thermo-Gas
#DeltaH_f^@("O"_2(g)) = "0 kJ/mol"#
(Why don't I need a reference for this value?)
#DeltaH_f^@("O"(g)) = "249.18 kJ/mol"#
http://webbook.nist.gov/cgi/cbook.cgi?ID=C17778802&Units=SI&Mask=1#Thermo-Gas
And so, we utilize the state function property of enthalpy to add the following reactions via Hess's Law (how else can you get the same result?):
#- (cancel(3/2"O"_2(g)) -> "O"_3(g))# ,#" "# #DeltaH_f^@ = -"142.67 kJ/mol"#
#" "cancel("O"_2(g)) -> "O"_2(g)# ,#" "" "" "# #DeltaH_f^@ = "0 kJ/mol"#
#" "cancel(1/2"O"_2(g)) -> "O"(g)# ,#" "" "# #DeltaH_f^@ = "249.18 kJ/mol"#
#"----------------------------"#
#"O"_3(g) -> "O"_2(g) + "O"(g)# ,#" "" "color(blue)(DeltaH_(rxn)^@ = "106.51 kJ/mol")#
And this holds at
This dissociation requires input of energy, so if the sun is to do that...
#"106.51 kJ/mol" = E_"photon" = (hc)/lambda# where
#h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant, and the speed of light is known to be#2.998 xx 10^8 "m/s"# .
The maximum wavelength required is simply the minimum energy or frequency required.
#color(blue)(lambda) = (hc)/E_"photon"#
#= (6.626 xx 10^(-37) cancel"kJ"cdotcancel"s" cdot 2.998 xx 10^8 "m/"cancel"s")/(106.51 cancel"kJ""/"cancel"mol" xx (cancel"1 mol photons")/(6.0221413 xx 10^(23) "photons"))#
#= 1.123 xx 10^(-6) "m/photon"#
#=# #color(blue)(1.123 mu"m/photon")#
And this lies in the near infrared region, i.e. in the
Again, since this is the maximum wavelength, it provides the minimum energy required. Ultraviolet light is at a shorter wavelength (and hence higher energy) than infrared, so this is reasonable.
