Question

Review Constants Periodic Table The lodine molecule can be photodissociated (broken apart with light) into iodine atoms in the gas phase with light of wavelengths shorter than about 792 nm. A 110.0 mL glass tube contains 41.5 mtorr of gaseous iodine at 20.0 °C. Part A What minimum amount of light energy must be absorbed by the iodine in the tube to dissociate 14.0% of the molecules? Express your answer with the appropriate units. " HA O ? Emin • 10-3 Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining

Answer 1

Number of moles of I2 gas present in the glass tube = n

We know that PV = nRT. So, n = PV / RT

Here, P = 41.5 mtorr = 0.0415 torr = 5.46*10-5 atm

1 atm = 760 torr ⇒ 1 torr = 1/760 atm = 0.00132 atm

So, 0.0415 torr = 0.0415*0.00132 atm = 5.46*10-5 atm

V = 110 mL = 0.11 L

T = 20 0C = 20+273 = 293 K

R = 0.0821 L atm / mol K

So, Number of moles of I2 gas present in the glass tube (n) = PV / RT

= (5.46*10-5 * 0.11) / (0.0821*293)

= 2.497*10-7

Number of I2 molecules in the glass tube = n*NA

NA → Avogadro number = 6.022*1023

So, Number of I2 molecules in the glass tube = 2.497*10-7 * 6.022*1023 = 1.5037*1017

14% of the molecule = 1.5037*1017 * (14/100) = 2.105*1016

The highest wavelength of the radiation that can dissociate one I2 molecule (λ) = 792 nm = 792*10-9 m

Minimum energy radiation will have a maximum wavelength.

So, the minimum energy required to dissociate one I2 molecule (E) = hc / λ

h → Plank's constant = 6.63*10-34 J s

c → speed of light in vacuum = 3*108 m

So, E = (6.63*10-34 * 3*108) / 792*10-9 = 2.519*10-19 J

Hence, minim energy required to dissociate 2.105*1016 molecules = 2.519*10-19 * 2.105*1016 = 0.0053 J

= 5.3*10-3 J