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 (J)/(mol)\\R \ = \ 8.314 (J)/(mol \ K)\\k_(0) = 0.0140s^(-1)`

Step-by-step explanation:

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

`k(T) \ = \ A e^{ (-E_(A))/(RT)}`

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

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

Replacing all values,

`T \ = \ 308K`