Question

Part A: The activation energy of a certain reaction is 39.2 kJ/mol . At 24 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units.

Part B: Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24  ∘C , what would the rate constant be at a temperature of 150.  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

Answer 1

A)

Given:

T1 = 24 oC

=(24+273)K

= 297 K

K1 = 0.0190 s-1 = 1.9*10^-2 s-1

K2 = 2*K1 = 3.8*10^-2 s-1

Ea = 39.2 KJ/mol

= 39200 J/mol

use:

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

ln(3.8*10^-2/1.9*10^-2) = (39200.0/8.314)*(1/297 - 1/T2)

0.6931 = 4714.9387*(1/297 - 1/T2)

T2 = 311 K

= (311-273) oC

= 38 oC

Answer: 38 oC

B)

Given:

T1 = 24 oC

=(24+273)K

= 297 K

T2 = 150 oC

=(150+273)K

= 423 K

K1 = 1.9*10^-2 s-1

Ea = 39.2 KJ/mol

= 39200 J/mol

use:

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

ln(K2/1.9*10^-2) = (39200.0/8.314)*(1/297 - 1/423.0)

ln(K2/1.9*10^-2) = 4715*(1.003*10^-3)

K2 = 2.15 s-1

Answer: 2.15 s-1