The minimum energy needed to dissociate iodine molecules, I2, is 151 kJ/mol. What is the wavelength of photons (in nm) that supplies this energy, assuming each bond dissociated by absorbing one photon?

1 Answer
Mar 24, 2018

792792 nanometers (or scientifically 7.92*10^2*mum7.92102μm.)

Explanation:

Equations this solution applied:

  1. N=n*N_AN=nNA where NN is the quantity of nn moles of particles and N_A=6.02*10^23*"mol"^(-1)NA=6.021023mol1 is the Avagordoro's number

  2. The Planck's Law E=h*fE=hf where EE is the energy of a single photon of frequency ff and hh is the Planck's Constant, h=6.63 × 10^(-34)* "m"^2* "kg" *"s"^(-1)= 6.63*10^(-34) color(blue)("J")*"s"h=6.63×1034m2kgs1=6.631034Js [1]

  3. lambda=v/fλ=vf where lambdaλ is the wavelength of a wave or an electromagnetic (EM) radiation of frequency ff.

From the question, breaking N_ANA (one mole of) iodine molecules consumes
E("one mole")=E*N_A=151*color(red)("kJ")*"mol"^(-1)E(one mole)=ENA=151kJmol1
=1.51*10^5*color(blue)("J") *"mol"^(-1)=1.51105Jmol1
where EE is the energy input required to break a single molecule.

Thus it takes
E=(E("one mole"))/(N_A)=(1.51*10^5)/(6.02*10^23)=2.51* 10^(-19) * color(blue)("J")E=E(one mole)NA=1.511056.021023=2.511019J
to break a single iodine molecule.

Apply the Planck's Law to find the maximum frequency of the EM radiation capable of breaking one such molecule:
f=E/h= (2.51* 10^(-19) * color(blue)("J"))/(6.63*10^(-34) color(blue)("J")*"s")f=Eh=2.511019J6.631034Js
=3.79*10^14 "s"^(-1)=3.791014s1

*Make sure that you get the unit that corresponds to the quantity after canceling out corresponding pairs. Here we are expecting "s"^-1s1 for frequency, which does appear to be the case. *

Assuming 3.00*10^8 * color(magenta)("m")* "s"^-1=3.00\cdot 10^(17)* color(green)(mu "m")\cdot "s"^-13.00108ms1=3.001017μms1 to be the speed of light. A photon of such EM radiation is expected to have the wavelength
lambda=v/f=(3.00*10^17* color(green)(mu"m")*"s"^(-1))/(3.79*10^14* "s"^(-1))=7.92*10^2* color(green)(mu "m)λ=vf=3.001017μms13.791014s1=7.92102μm

Sources:
1. Units ("dimensions") of the Planck's Constant: https://www.askiitians.com/forums/General-Physics/find-the-dimension-of-planck-constant-h-from-the-e_74309.htm