Question

He is impressed with your work but he wants you to calculate the value of A as well. Calculate the value of A at the same conditions. (The rate constants of the reaction at 500 K and 700 K are $0.02s^{-1}$ and $0.07s^{-1}$ respectively.)

• A. $1.585s^{-1}$
• B. $1s^{-1}$
• C. $0.5s^{-1}$
• D. $0.02s^{-1}$

Answer 1

The correct option is A $1.585s^{-1}$

We know the activation energy, substitute the values in the Arrhenius equation at the given temperature.
Since, $k=A{e}^{-Ea/RT}logk=logA-\frac{E_a}{2.303RT}logA=logK+\frac{E_a}{2.303RT}logA=log\left(0.02\right)+\frac{18230.8Jmo{l}^{-1}}{2.303\times 8.314J{k}^{-1}\times 500K}=-1.699+1.90=0.0201\approx 0.2A=Antilog\left(0.2\right)\approx 1.585$