Question

Find the longest wavelengths of light that can cleave the bonds in elemental nitrogen, oxygen, and fluorine. the average bond energy of n―n bond is 945 kj/mol, oxygen is 498 kj/mol, and f―f is 159 kj/mol. give your answers in scientific notation.

Answer 1

Step-by-step explanation:

Longest wavelengths of light that can cleave the bonds in elemental nitrogen

Energy to cleave 1 mol N-N bond = 945 kJ/mol = 945000 J/mol

1 mol=
`6.022* 10^(23)`

Energy to break 1 N-N bond =
`(945000 J/mol)/(6.022* 10^(23) mol^(-1))=1.56* 10^(-18) J`

`E=(hc)/(\lambda )`

`\lambda =(6.626* 10^(-34)J s* 3* 10^(8) m/s)/(1.56* 10^(-18) J)=12.74* 10^(-8) m=127.4 nm`

Longest wavelengths of light that can cleave the bonds in elemental nitrogen is 127.4 nm.

Similarly

For oxygen:

Energy to cleave 1 mol O-O bond = 498 kJ/mol = 498000 J/mol

Energy to break 1 O-O bond =
`(498000 J/mol)/(6.022* 10^(23) mol^(-1))=8.26* 10^(-19) J`

`E=(hc)/(\lambda )`

`\lambda =(6.626* 10^(-34)J s* 3* 10^(8) m/s)/(8.26* 10^(-19) J)=2.406* 10^(-7)m=240.6 nm`

Longest wavelengths of light that can cleave the bonds in elemental oxygen is 240.6 nm.

For fluorine

Energy to cleave 1 mol F-Fbond = 159 kJ/mol = 159000 J/mol

Energy to break 1 F-F bond =
`(159000 J/mol)/(6.022* 10^(23) mol^(-1))=2.64* 10^(-19) J`

`E=(hc)/(\lambda )`

`\lambda =(6.626* 10^(-34)J s* 3* 10^(8) m/s)/(2.64* 10^(-19) J)=7.529* 10^(-7)m=752.9 nm`

Longest wavelengths of light that can cleave the bonds in elemental fluorine is 752.9 nm.

Answer 2

For Nitrogen:
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (945 x 10³)
= 1.27 x 10⁻⁷ m
For Oxygen::
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (498 x 10³)
= 2.40 x 10⁻⁷ m
For Fluorine:
wavelength = NA x h C / bond energy
= 6.022 x 10²³ x 6.626 x 10⁻³⁴ x 3 x 10⁸ / (159 x 10³)
= 7.52 x 10⁻⁷ m
(NA is Avogadro's number, h is Planck's constant and C is speed of light )