How Does a 10°C Temperature Increase Double the Rate of a Slow Reaction?

[Sisyphus]

We are currently studying reactions in terms of kinetic energy, reaction rates, collision theory, and so on.

There is a question on my assignment that is kind of boggling me right now:

An increase in temperature of 10 C rarely doubles the kinetic energy of particles, and hence the number of collisions is not doubled. Yet, this temperature increase may be enough to double the rate of a slow reaction. How can this be explained?

I'm not sure if it has to do with the wording of the question, but I'm really not getting this question.

 

[Emieno]

aA+bB -> cC + dD
use abcd in relation to temp to explain the problem

 

[GCT]

Think about the rate of a reaction in relevance to the activation energies of the respective reaction. At a particular temperature and kinetic energy distribution of the molecules, more or less these molecules may have enough energy in surpassing the activation energy.

So if a reaction rate is fast, at a particular starting equilibrium temperature, what does this tell you about the proportion of the molecules which have enough energy surpass the activation energy?

I've given you more than a big hint for this problem,