Question

The activation energy of a certain reaction is 35.3 kJ/mol. At 20 degrees C, the rate constant is `0.0130 s^-1`. At what temperature would this reaction go twice as fast? please help?

Answer 1

`T_2=308K=35^@C`

Explanation:

To solve this question, we can modify the Arrhenius equation which provides a relationship between the equilibrium constant `k` and temperature `T`:

`lnk=-(E_a)/R(1/T)+lnA`

To find the new temperature `T_2` of the reaction in this question, we can use the following expression:

`ln((k_2)/(k_1))=(E_a)/R(1/(T_1)-1/(T_2))`

If the rate is doubled, the equilibrium constant will also double and therefore, `k_2=2xxk_1=2xx0.0130s^(-1)=0.0260s^(-1)`

Therefore, `ln((0.0260cancel(s^(-1)))/(0.0130cancel(s^(-1))))=(35300(cancel(J))/(cancel(mol)*cancel(K)))/(8.31(cancel(J))/(cancel(mol)*cancel(K)))(1/(293K)-1/(T_2))`

Solving for `T_2`: `=>T_2=308K=35^@C`

Here is a video that explains the theory behind this topic:
Chemical Kinetics | A Model for Chemical Kinetics & Catalysis.