Question

Photodissociation of CF3CI to form Cl radicals requires 321 kJ/mol. How many photons of wavelength 370 nm are needed to photodissociate 0.425 g of CF3CI? CF3CI+ hv CF3+ C O 245x 1021 photons 7.32 10 photons 4 68x 10 photons 5.33x 10-19 photons O 527 1021 photons

Answer 1

Step 1: Calculate the total energy required

Molar mass of CF3Cl,

MM = 1*MM(C) + 3*MM(F) + 1*MM(Cl)

= 1*12.01 + 3*19.0 + 1*35.45

= 104.46 g/mol

mass(CF3Cl)= 0.425 g

use:

number of mol of CF3Cl,

n = mass of CF3Cl/molar mass of CF3Cl

=(0.425 g)/(1.045*10^2 g/mol)

= 4.069*10^-3 mol

E = number of mol * 321 KJ/mol

= (4.069*10^-3 mol) * 321 KJ/mol

= 1.306 KJ

= 1306 J

Step 2:

Given:

lambda = 3.7*10^-7 m

1st calculate energy of 1 photon

Given:

lambda = 3.7*10^-7 m

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(3.7*10^-7 m)

= 5.372*10^-19 J

number of photon = total energy/energy of 1 photon

n = 1306.0/5.372*10^-19

= 2.431*10^21

Answer: option 1