Question

The activation energy of a reaction is 22.5 k Cal mol⁻¹ and the value of rate constant at 40°C is 1.8 × 10⁻⁵ s⁻¹. Calculate the frequency factor, A.

Answer 1

Here, we are given that

E_a = 22.5 kcal mol⁻¹ = 22500 cal mol⁻¹

T = 40°C = 40 + 273 = 313 K

k = 1.8 × 10⁻⁵ sec⁻¹

Substituting the values in the equation

log A = `log "K" + (("E"_"a")/(2.303 "RT"))`

log A = `log (1.8 xx 10^-5) + ((22500)/(2.303 xx 1.987 xx 313))`

log A = `log (1.8) - 5 + (15.7089)`

log A = 10.9642

A = antilog ( 10.9642)
A = 9.208 × 10¹⁰ collisions s⁻¹