Question

The activation energy of a certain reaction is 48.4 kJ/mol . At 24 ∘C , the rate constant is 0.0140s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?

Answer 1

Data:

$T \ = \ 297.15K\\E_{A} \ = \ 48400 \frac{J}{mol}\\R \ = \ 8.314 \frac{J}{mol \ K}\\k_{0} = 0.0140s^{-1}$

Explanation:

The rate constant as a function of the temperature can be expressed using the Arrhenius equation, as follows,

$k(T) \ = \ A e^{ \frac{-E_{A}}{RT}}$

Then, find a value for the constant A, according to the data given in the exercise,

$A \ = \frac{ k_{0} }{ e^{ \frac{-E_{A}}{RT}}}$

$A \ = 4512841.2s^{-1}$

As the rate constant has units $[s^{-1}]$ then, the reaction corresponds to one of first order. Then, the reaction corresponds to:

$r \ = \ k(T) \ [A]$

Which tell the amount of mass (mol) is produced by second per unity of volume.

Answer:

If the reaction goes twice fast, the reaction rate corresponds to,

$r \ = \ 2 \ r_{0}$

The expression above can be reduced to:

$k(T) \ = \ 2 \ k_{0}$

Replacing k(T) in order to obtain the temperature T,

$4512841.2 \ e^{\frac{-E_{A}}{RT}}} \ = \ 2k_{0}\\e^{\frac{-E_{A}}{RT}}} \ = \frac{2k_{0}}{4512841.2}\\\frac{-E_{A}}{RT}} \ = \ ln( \frac{2k_{0}}{4512841.2} )\\\frac{-E_{A}}{R \ ln(\frac{2k_{0}}{4512841.2} ) }} \ = \ T$

Replacing all values,

$T \ = \ 308K$