Question

1.a)The activation energy of a certain reaction is 46.3 kJ/mol . At 30 ∘C , the rate constant is 0.0180s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

b) Given that the initial rate constant is 0.0180s−1 at an initial temperature of 30 ∘C , what would the rate constant be at a temperature of 140 C for the same reaction described in Part A?

Answer 1

a)
Ea = 46.3 KJ/mol = 46300 J/mol
K2/K1 = 2
T1 = 30 oC = (30 + 273) K = 303 K

use:
ln (K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln (2) = (46300/8.314)*(1/303 - 1/T2)
(1/303 - 1/T2) = 1.2447*10^-4
T2 = 315 K
= (315 - 273 ) oC
= 42 oC

Answer: 42 oC

b)
K1 = 0.0180 s-1
T1 = 30 oC = (30 + 273) K = 303 K
Ea = 46.3 KJ/mol = 46300 J/mol
T2 = 140 oC = (140 + 273) K = 413 K

use:
ln (K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln (K2/0.0180) = (46300/8.314)*(1/303 - 1/413)
ln (K2/0.0180) = 4.895
K2= 2.40 S-1

Answer: 2.40 S-1