The activation energy of a certain reaction is 49.4 kJ/mol. At 20 degrees C, the rate constant is .0130 s^-1. What is the rate constant at 100 degree C?

1 Answer
Aug 8, 2016

Answer:

The rate constant is #"1.00 s"^"-1"#.

Explanation:

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

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

where

#k# = 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

#lnk = 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(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#

In your problem,

#k_1 = "0.0130 s"^"-1"#
#E_"a" = "49.4 kJ/mol" = "49 400 J/mol"#
#R = "8.314 J·K"^"-1""mol"^"-1"#
#T_2 = "100 °C" = "373.15 K"#
#T_1 = "20 °" = "293.15 K"#

Now, let's insert the numbers.

#ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)#

#ln(k_2/k_1) = ("49 400" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))) (1/(293.15 color(red)(cancel(color(black)("K")))) - 1/(373.15 color(red)(cancel(color(black)("K")))))#

#ln(k_2/k_1) = 5942× 7.313 × 10^"-4" = 4.346#

#(k_2/k_1) = e^4.346 = 77.14#

#k_2 = 77.14k_1 = 77.14 × "0.0130 s"^"-1" = "1.00 s"^"-1"#