A reaction has a rate constant of 1.25×10−4 s−1 at 29 ∘C and 0.228 s−1 at 79 ∘C . Part A Determine the activation barrier for the reaction. Express your answer in units of kilojoules per mole. Part B What is the value of the rate constant at 18 ∘C ? Express your answer in units of inverse seconds.
given T1= 29 deg.c= 29+273= 302K, T2= 79deg.c= 79+273= 352K,
K1= rate constant at T1= 1.25*10-4/s and K2= rate constant at T2= 0.228/s
From Arhenius equation, ln (K2/K1)= (Ea/R)*(1/T1-1/T2)
Ea= activation energy , R= 8.314 J/mole.K
ln (0.228/1.25*10-4)= (Ea/8.314)*(1/302-1/352)
6.82= (Ea/R)*0.000437, Ea= 6.82*8.314/0.000437 J/ mole =120552 J/mole= 120.55 Kj/mole
at T3=18 deg.c= 18+273= 291K, Writing Arhenius equations for T3 and T1 and noting that K3= rate constant at T3.
ln (K3/K1)= (Ea/R)*(1/302-1/291)=(120.55*1000/8.314)*(-0.00013)
K3/K1=0.163, K3= 0.163*1.25*10-4/s=2.04*10-5/s
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