Posted by lef on Friday, November 16, 2012 at 8:23am.
I came up with:
EI-F = 0.5[EI-I + EF-F] + (χF – χI)2
EI-F = 0.5[1.5546+ 1.6583] + (3.98 – 2.66)2
Solving this gives
EI-F = 3.34885 eV = 323 kJ/mol
Similarly ECl-F = 265 kJ/mol and
EI-Cl = 219 kJ/mol
So to break the I-Cl bond,
Wavelength of radiation= 6.625x10^-34 Js x 3x10^8 m⁄s/2.19x10^3 J x6.023 x 10^23 = 5.47 x 10^-5 m
Not ... not sure if this is 100% correct ... the result will change depending on decimals. What was your result?
Correct answer is:
1)
E I-Cl = sqrt(150X240) + 96.3 (2.66-3.16)^2
2)
E in kJ/mole -> E in J/bond:
E I-Cl/ 6.02*10^23
3)
Wavelength = (6.626*10^(-34) X 3*10^8) / (E in J/bond)
:))
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