Question

The activation energy of a reaction is 94.14KJ/mole and rate constant at 298K is 18x10^-5

Answer 1

Answer:

This is not a bio's question.

Answer 2

Answer:

By using Arrhenius equation:

$log \: k = - \frac{ - Ea}{2.303RT} + log \: A \\ we \: get \\ log \: A \: = log(1.8 \times {10}^{ - 5} ) + \frac{94140}{2.303 \times 8.314 \times 298} \\ (log \: 1.8) - 5 +15.7081 ) \\ 0.2553 - 5 +15.7081 = 10.964 \\ log \: A = (10.9634) = 9.914 \times {10}^{10}$