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