# 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?

##### 1 Answer

#### Answer:

#### Explanation:

To solve this question, we can modify the Arrhenius equation which provides a relationship between the equilibrium constant

To find the new temperature

If the rate is doubled, the equilibrium constant will also double and therefore,

Therefore,

Solving for

Here is a video that explains the theory behind this topic:

**Chemical Kinetics | A Model for Chemical Kinetics & Catalysis.**