Question

The iodine molecule can be photodissociated into iodine atoms in the gas phase with light of wavelengths shorter than about 796 nm . A 120.0 mL glass tube contains 48.9 mtorr of gaseous iodine at 19.0 ∘C. What minimum amount of light energy must be absorbed by the iodine in the tube to dissociate 19.0 % of the molecules?

Answer 1

Solution ;-

Pressure = 48.9 mtorr = 48.9 x 10-3 torr

= 48.9 x 10-3/760= 6.434x 10-5 atm

Volume = 120.0 mL = 0.120L

Temperature T = 19°C = 292.15 K

Molar gas constant R = 0.08206 atm.L/mol.K

Moles of I2 = n = PV/RT

= 6.434× 10-5 x 0.120/(0.08206 x 292.15)

= 3.221 x 10-7 mol

Number of I2 molecules = moles x Avogadro's number

= 3.221× 10^-7 x 6.022 x 10^23

= 1.939x 10^17 molecules

Assuming that dissociation of 1 I2 molecule requires 1 photon:

Number of photons required = 19% of the number of iodine molecules

= 19/100 x 1.939 x 10^17

= 3.684x 10^16 photons

Wavelength λ = 796 nm = 796 x 10-9 m

Energy of 1 photon = hc/λ

= 6.626 x 10-34 x 2.998 x 10^8/796 x 10^-9

= 2.495 x 10-19 J

here h is Planck constant and c is speed of light

Light energy required = number of photons required x energy of 1 photon

= 3.684x 10^16 x 2.495 x 10-19

=9.194 x 10-3 J