Question

A reaction has a rate constant of 1.26 × 10-4s-1 at 28℃ and 0.226 s-1 at 76℃.

A reaction has a rate constant of 1.26 × 10-4s-1 at 28℃ and 0.226 s-1 at 76℃.

You may want to reference(Pages 606-612) Section 14.6 while completing this problem.


Part B

What is the value of the rate constant at 15℃?

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Answer #1

Step 1: find Ea

Given:

T1 = 28 oC

=(28+273)K

= 301 K

T2 = 76 oC

=(76+273)K

= 349 K

K1 = 1.26*10^-4 s-1

K2 = 0.226 s-1

use:

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

ln(0.226/1.26*10^-4) = ( Ea/8.314)*(1/301 - 1/349)

7.492 = (Ea/8.314)*(4.569*10^-4)

Ea = 136320 J/mol

Step 2:

Given:

T1 = 28 oC

=(28+273)K

= 301 K

T2 = 15 oC

=(15+273)K

= 288 K

K1 = 1.26*10^-4 s-1

Ea = 136.32 KJ/mol

= 136320 J/mol

use:

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

ln(K2/1.26*10^-4) = (136320.0/8.314)*(1/301 - 1/288.0)

ln(K2/1.26*10^-4) = 16396*(-1.5*10^-4)

ln(K2/1.26*10^-4) = -2.459

(K2/1.26*10^-4) = e^(-2.459)

(K2/1.26*10^-4) = 8.553*10^-2

K2 = 1.078*10^-5 s-1

Answer: 1.08*10^-5 s-1

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