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
Jul 12, 2016

Answer:

#(f_2/f_1) = 6.27#

Explanation:

I presume that you are using #f# as a symbol for the rate constant.

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

#f# = the rate constant
#A# = the pre-exponential factor
#E_"a"# = the activation energy
#R# = the Universal Gas Constant
#T# = the temperature

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,

#E_"a" = "165 kJ/mol" = "165 000 J/mol"#
#T_2 = "475 K"#
#T_1 = "455 K"#

Now, let's insert the numbers.

#ln(f_2/f_1) = E_"a"/R(1/T_1 -1/T_2)#

#ln(f_2/f_1) = ("165 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))) (1/(455 color(red)(cancel(color(black)("K")))) - 1/(475 color(red)(cancel(color(black)("K")))))#

#ln(f_2/f_1) = "19 846" × 9.254 × 10^"-5" = 1.837#

#(f_2/f_1) = e^1.837 = 6.27#