Question

the enthalpy of adsorption of co on palladium is -146kJ/mol estimate the residence time of a co molecule on a palladium surface at 300k and 500k(assume that t°=1.0*10^-12s

Answer 1

The residence time of an adsorbate molecule on an adsorbent surface is given as,

t = t^oexp(-E_a/RT) .....................(1)

where t = residence time

t^o = average time between two successive attempts to escape from the surface (pre-exponential factor)

E_a = enthalpy of adsorption

R = gas constant

T = temperature of the system

Now, t^o = 1.0x10^-12 s

E_a = -146 kJ/mol = -146000 J/mol

R = 8.314 JK^-1mol^-1

putting in equation (1), at T = 300 K

t₃₀₀ = 1.0x10^-12 s x exp[-(-146000 Jmol^-1)/ (8.314 JK^-1mol^-1)(300 K)]

= 1.0x10^-12 s x 2.64x10²⁵

= 2.64x10¹³ s

at T = 500 K,

t₅₀₀ = 1.0x10^-12 s x exp[-(-146000 Jmol^-1)/ (8.314 JK^-1mol^-1)(500 K)]

= 1.0x10^-12 s x 1.79 x 10¹⁵

=1.79x10³ s