Question

Light of wavelengths shorter than 275 nm can be used to photodissociate the hydrogen molecule into hydrogen atoms in the gas phase. A 20.0 mL glass cylinder contains H, (g) at 20.0 mtorr and 25 °C. What minimum amount of light energy must be absorbed by the hydrogen in the tube to dissociate 24.0% of the molecules? energy absorbed: 1.735 X10-19

Answer 1

Given:
P = 20.0 moor
= 0.02 torr
= (0.02/760) atm
= 2.63*10^-5 atm
V = 20.0 mL
= (20.0/1000) L
= 0.02 L
T = 25.0 oC
= (25.0+273) K
= 298 K

find number of moles using:
P * V = n*R*T
2.63*10^-5 atm * 0.02 L = n * 0.08206 atm.L/mol.K * 298 K
n = 2.152*10^-8 mol

Total number of H2 molecule dissociated = 24 % of total H2
= 24 % of number of mol * Avogadro’s number
= 24 % of (2.152*10^-8 * 6.022*10^23)
= 24 % 1.296*10^16
= 0.24 * 1.296*10^16
= 3.11*10^15 molecule

So, it needs 3.11*10^15 photons

Given:
lambda = 2.75*10^-7 m

Find energy of 1 photon first

use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(2.75*10^-7 m)
= 7.228*10^-19 J
This is energy of 1 photon

So,
Energy of 3.11*10^15 photons = 3.11*10^15 * 7.228*10^-19 J
= 2.25*10^-3 J

Answer: 2.25*10^-3 J