Question

Evaluate the collision frequency of H2 molecules in a gas at 1.00atm and 25 degrees celsius

Answer 1

According to the kinetic Molecular Theory,

=

Where, = Collision frequency

= Root-mean-square velocity of the molecules

= Mean free path

Root-mean-square velocity :

The formula relating the rms velocity to the temperature and molar mass is:

v_rms =

Where, R = The universal Gas constant

T = The temperature

M =Tthe molar mass

For H₂ at 25 °C ,

T = ( 25 + 273.15 ) K = 298.15 K

M = 2.016 g.mol^-1 = 2.016 10^-3 kg.mol^-1

v_rms = =

v_rms = m.s^-1 = 1920 m.s^-1

The mean free path :

For a hydrogen molecule, Kinetic diameter σ = 289 pm.

The formula for the mean free path is

λ =

Where = Mean free path

R = 8.314 10^-5 bar.m³.K^-1 .mol^-1

T = 298.15 K

N_A = Avogadro number = 6.022 10²³ mol^-1

P = Pressure = 1.00 atm = 1.01325 bar

λ =

=

= m

= 1.094 10^-7 m = 10.94 nm

Now Collision frequency,

=

=

= 1.755 10^-10 s^-1

There are 1.755 10^-10 collisions per second.