Question

Part A

The activation energy of a certain reaction is 30.5 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?

Express your answer with the appropriate units.

T2 =

Part 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 170.  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

k2 =

s-1

Answer 1

A)

T1 = 30.0 oC

=(30.0+273)K

= 303.0 K

K1 = 1.8*10^-2 s-1

K2 = 1.8*10^-2 * 2 = 3.6*10^-2 s-1

Ea = 30.5 KJ/mol

= 30500 J/mol

use:

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

ln(3.6*10^-2/1.8*10^-2) = (30500/8.314)*(1/303.0 - 1/T2)

0.6931 = 3668.5109*(1/303.0 - 1/T2)

T2 = 321

= (321-273) oC

= 48 oC

Answer: 48 oC

B)

T1 = 30.0 oC

=(30.0+273)K

= 303.0 K

T2 = 170.0 oC

=(170.0+273)K

= 443.0 K

K1 = 1.8*10^-2 s-1

Ea = 30.5 KJ/mol

= 30500.0 J/mol

use:

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

ln(K2/1.8*10^-2) = (30500.0/8.314)*(1/303.0 - 1/443.0)

ln(K2/1.8*10^-2) = 3669*(1.043*10^-3)

K2 = 0.826 s-1

Answer: 0.826 s-1