Question

0. The following questions use the following value for Bond Dissociation Energy for a triple bond between two nitrogen atoms: 945kJ/mol a. Given that purple light has a wavelength of 430nm; how many purple light photons will need to completely break this triple bond? b. What is the minimum wavelength that will allow me to break that bond if I had 3.5 moles of photon? In which region of the electromagnetic spectrum will I find this light? Is this light more or less energetic than the light in part A?

Answer 1

I have taken the bond dissociation energy for 1 molecule for solving the question as in question is not specified. If your answer does not match please follow the same procedures.

a Date: Boud Page: dissociation Energy = 945 kJ/mol 23 of NSN molecule A of purple light = 430nm i Boud dissociation energy = 945 kJ per molecule NA - " Ny = Avogadoro No = 6:023x153) total energy of light = nxhe where na no of photos - d = wavelength of light h=Planks const. I c = speed of light nx hc - 945x13 - 6.023x123 n . 945 × 430 x 10 9 410 6-023 6163441034 x 3 x 18 4123 - 343 89 x 10 - 3 To = 39 389 x 10 n = 3.389 photon Can not n ñ 4 s' no of be fractional

6 n = 3.5X NA given no of photons =n=3.5 X6-023X183 nxhc 945 X10 NA 3 -34 8.. 3.5XN AX 6.634 Nox 340 945 X6 -26, A = 3.5 X6.634 X 3 X 10 x (6.023x105) 945x 10 A = 20673 x -26 10 x 46 10 x = 2 673 x 10 37 m (radio waves) A Wavelength in part Ŕ is more than wavelength in part A so light in part À will be more energetic than part B because Energy & wardlength