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

E_a = 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, \(A = antilog \ (12.5917)\)

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