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?
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)|)))" "#
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.