Answer to Question #177137 in Physical Chemistry for Miriama Tamaivena

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

$Ea = 22.5 kcal mol^{-1}= 22500 cal mol^{-1}$

$T= 40°C = 40 + 273 = 313 K$

$k = 1.8 ×10^{-5}sec ^{-1}$

Substituting the values in the equation

$log A = log k + (\frac{E_a}{2.303RT})$

$log A = log (l.8×10^{-5}) + (\frac{E_a}{2.303×1.987×313})$

$log A = log (l.8) – 5 + (15.7089)$

$log A = (10.9642)$

$A = antilog ( 10.9642)$

$A = 9.208 × 10^{10}$ collisions $s^{-1}$