Question

Part A

The activation energy of a certain reaction is 36.4 kJ/mol . At 21  ∘C , the rate constant is 0.0160s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Hints

Part B

Given that the initial rate constant is 0.0160s−1 at an initial temperature of 21  ∘C , what would the rate constant be at a temperature of 130.  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

Answer 1

A)

T1 = 21.0 oC

=(21.0+273)K

= 294.0 K

K2/K1 = 2

Ea = 36.4 KJ/mol

= 36400.0 J/mol

use:

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

ln(2) = (36400.0/8.314)*(1/294.0 - 1/T2)

0.693147 = 4378.157325*(1/294.0 - 1/T2)

T2 = 308

= (308-273) oC

= 35 oC

Answer: 35 oC

B)

T1 = 21.0 oC

=(21.0+273)K

= 294.0 K

T2 = 130.0 oC

=(130.0+273)K

= 403.0 K

K1 = 1.6*10^-2 S-1

Ea = 36.4 KJ/mol

= 36400.0 J/mol

use:

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

ln(K2/1.6*10^-2) = (36400.0/8.314)*(1/294.0 - 1/403.0)

ln(K2/1.6*10^-2) = 4378*(9.2*10^-4)

K2 = 0.898 S-1

Answer: 0.898 S-1