# A certain reaction with an activation energy of 165 kJ/mol was run at 455 K and again at 475 K. What was the ratio of #f# at the higher temperature to #f# at the lower temperature?

##### 1 Answer

#### Explanation:

I presume that you are using

The **Arrhenius equation** gives the relation between temperature and reaction rates:

#color(blue)(|bar(ul(color(white)(a/a) f = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "#

where

If we take the logarithms of both sides, we get

#lnf = lnA - E_"a"/(RT)#

Finally, if we have the rates at two different temperatures, we can derive the expression

#color(blue)(|bar(ul(color(white)(a/a) ln(f_2/f_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#

In your problem,

Now, let's insert the numbers.