Question

Part A The activation energy of a certain reaction is 30.6 kJ/mol . At 22 ∘C , the rate constant is 0.0110s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. Hints T2 = 39 ∘C SubmitMy AnswersGive Up Correct Part B Given that the initial rate constant is 0.0110s−1 at an initial temperature of 22 ∘C , what would the rate constant be at a temperature of 170. ∘C for the same reaction described in Part A? Express your answer with the appropriate units.

Answer 1

A)

T1 = 22.0 oC

=(22.0+273)K

= 295.0 K

K1 = 1.1*10^-2 s-1

K2 = 2*K1 = 2.2*10^-2 s-1

Ea = 30.6 KJ/mol

= 30600.0 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(2.2*10^-2/1.1*10^-2) = (30600.0/8.314)*(1/295.0 - 1/T2)

0.693147 = 3680.53885*(1/295.0 - 1/T2)

T2 = 312

= (312-273) oC

= 39 oC

Answer: 39 oC

B)

T1 = 22.0 oC

=(22.0+273)K

= 295.0 K

T2 = 170.0 oC

=(170.0+273)K

= 443.0 K

K1 = 1.1*10^-2 s-1

Ea = 30.6 KJ/mol

= 30600.0 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(K2/1.1*10^-2) = (30600.0/8.314)*(1/295.0 - 1/443.0)

ln(K2/1.1*10^-2) = 3681*(1.132*10^-3)

K2 = 0.711 s-1

Answer: 0.711 s-1