Question

Compute the steric factor for the reaction 2?₂?₄ → ?₄?₈ at 300K if the experimentally-measured activation energy is ?? = 146.4 ??/???, effective diameter of ?₂?₄ is 0.49 nm, and experimental rate constant at this temperature is ? = 1.08 x 10⁻¹⁴ ??³ /(???.? )^-1.

Answer 1

As per the collision theory of reaction rates, we know that

k = Zρexp(-E_a/RT)

where

Z = collision frequency

ρ = steric factor

and E_a = activation energy of the reaction.

It is further known that

Z = N_AVO*σ_AB*√8k_BT/πμ_AB

where

N_AVO = Avogadro’s constant = 6.02*10²³ molecules/mol;

σ_AB is the collision radius = ½*(0.49 nm) = 0.245 nm = (0.245 nm)*(1 m)/(1.0*10⁸ nm)

= 2.45*10^-10 m

k_B = Boltzmann’s constant = 1.3806*10^-23 m².kg.s^-2.K^-1

and μ_AB = reduced mass of C₂H₄ = μ_Cμ_H/(μ_C + μ_H)

= (12 amu)*(1 amu)/(12 amu + 1 amu)

= 12/13 amu

= 0.9231 amu = (0.9231 amu)*(1.6605*10^-27 kg)/(1 amu)

= 1.5328*10^-27 amu (1 amu = 1.6605*10^-27 kg)

Plug in values and get

√8k_BT/πμ_AB

= √[8*(1.3806*10^-23 m².kg.s^-2.K^-1)*(300 K)/(3.14)(1.5328*10^-27 kg)]

= √(8*860545.7894 m².s^-2)

= 2623.8076 m/s.

Now determine Z_AB as

Z_AB = (6.02*10²³ molecules/mol)*(2.45*10^-10 m)*(2623.8076 m/s)

= 3.8698*10¹⁷ m².s^-2/mol.

Finally plug in values and determine ρ as

1.08*10^-14 cm³/(mol.s^-1) = (3.8698*10¹⁷ m².s^-2/mol)*ρ*exp [(146.4 kJ/mol)/(8.314 J/mol.K)(300 K)]

=====> [1.08*10^-14 cm³/(mol.s^-1)]*(1 m³)/(1.0*10⁶ cm³)

= (3.8698*10¹⁷ m².s^-2/mol)*ρ*exp [(146.4 kJ/mol)*(1000 J)/(1 kJ)/(2494.2 J/mol)]

=====> 1.08*10^-20 m³/mol.s^-1 = (3.8698*10¹⁷ m².s^-2/mol)*ρ*(3.1004*10²⁵)

=====> ρ = (1.08*10^-20 m³/mol.s^-1)/(1.1998*10⁴³ m².s^-2/mol)

=====> ρ = 9.001*10^-64 m/s

=====> ρ ≈ 9.00*10^-64 m/s (ans).