Question

The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Answer 1

k = 2.418 × 10-5 s-1

T = 546 K

Ea = 179.9 kJ mol-1= 179.9 × 103 J mol-1

According to the Arrhenius equation,

$k = Ae^{-\frac {E_a}{RT}}$

⇒ $ln\ k = ln \ A - \frac {Ea}{RT}$

⇒ $log \ k = log \ A - \frac {Ea}{2.303\ RT}$

⇒ $log \ A = log \ k - \frac {Ea}{2.303\ RT}$

⇒ $log \ A$ = $log \ (2.418 \times 10^{-5} s^{-1})$ + $\frac {179.9 \times 10^3 J\ mol^{-1}}{2.303 \times 8.314J\ k^{-1}mol^{-1} \times 546\ K}$

⇒ $log \ A = = (0.3835 - 5) + 17.2082$

⇒ $log\ A = 12.5917$

Therefore, $A = antilog \ (12.5917)$

$A = 3.9 × 10^{12 }s^{-1 }\ (approximately)$