Homework SourseThe activation energy for the gas phase isomerization of cyc
The activation energy for the gas phase isomerization of cyclobutene is 136 kJ.
(CH2)2(CH)2------->CH2=CHCH=CH2
The rate constant at 439 K is 7.72×10-4 /s. The rate constant will be_____ /s at 471 K.
Solution
ln(k1/k2) = Ea/R(1/T2 - 1/T1)
where,
k1 and k2 are rate constants, k1=7.72×10-4s-1
Ea = activation energy, 136kJ/mol = 136000J/mol
T1 = 439K
T2 = 471K
substituting the values
ln(7.72×10-4s-1/k2) = 136000J/mol/8.314J/K mol (1/471K-1/439K)
ln(7.72×10-4s-1/k2) = 16358K (0.002123K-1 - 0.002278K-1)
ln(7.72×10-4s-1/k2) = - 2.535
log(7.72×10-4s-1/k2) =- 1.1007
7.72×10-4s-1/k2 = 0.0793
k2 = 9.74×10-3 s-1
so, the rate constant at 471K is 9.74×10-3s-1
