Question

Some free radical chemical reactions start with the photodissociation of I2 molecules into Iodine atoms (gas phase) by light with a wavelength shorter than about 792nm. A 100.0mL glass tube, pressure= 55.7mtorr and temperature 25.0C contains I2 molecules What minimum amount of energy must be absorbed by the iodine in the tube to dissociate 15% of the the molecules?

Answer 1

I2 ---> 2I energy required = 792nm

Convert nm to m:

792.0 nm = 792.0 x 10¯⁹ m = 7.920 x 10¯⁷ m

2) Determine the frequency:

λν = c

(7.920 x 10¯⁷ m) (x) = 3.00 x 10⁸ m/s

x = 3.70 x 10¹⁴ s¯¹

3) Determine the energy:

E = hν:

x = (6.626 x 10¯³⁴ J s) (3.7 x 10¹⁴ s¯¹)

x = 2.456 x 10¯¹⁹ J

Important point: this is the energy for one photon.

4) Determine energy for one mole of photons:

(2.456 x 10¯¹⁹ J) (6.022 x 10²³ mol¯¹)

147.8 kJ/mol

PV = nRT n = PV / RT

But we have 100.0mL glass tube, pressure= 55.7mtorr and temperature 25.0C

Or

100.0mL glass tube = 0.1L

, pressure= 55.7mtorr = 55.7* 0.0013 atm = 0.072atm

temperature 25.0C = 298K

R =  0.08206 L atm mol^-1K^-1

contains n = (0.1 L * 0.072) / ( 0.082 * 298) = 2.9*10^-4 Moles I2 will be there

15% of  2.9*10^-4 Moles I2 = 2.9*10^-4 * (15 / 100) = 0.435*10^-4

1 mole I2 need 1 mole photon which means 147.8 kJ/mol

then 0.435*10^-4 Moles I2 need 147.8 *0.435*10^-4= 64.29*10^-4 kJ/mol