In: Chemistry
A reaction has a rate constant of 1.19×10−2 /s at 400. K and 0.693 /s at 450. K.
A) Determine the activation barrier for the reaction
B) What is the value of the rate constant at 425 K
Answer – A) We are given, T1 = 400 K , T2 = 450 K
k1 = 1.19*10-2 s-1 , k2 = 0.693 s-1
Ea = ?
The intergraded Arrhenius equation
ln k1/k2 = Ea / R *(1/T2-1/T1)
ln 1.19*10-2 / 0.693 = Ea/ 8.314 J/mol.K * (1/450 – 1/400)
-4.064 = Ea/ 8.314 J/mol.K * -0.00028
So, Ea = -4.064 * 8.314 J/mol.K / -0.00028
= 1.2165*105 J/mol
= 121.65 kJ/mol
the activation barrier for the reaction is 121.65 kJ/mol
B) We are given, T1 = 400 K , T2 = 425 K
k1 = 1.19*10-2 s-1 , k2 = ?
Ea = 1.22*105 J/mol
The intergraded Arrhenius equation
ln k1/k2 = Ea / R *(1/T2-1/T1)
ln 1.19*10-2 / k2 = 1.22*105 J. mol-1 / 8.314 J/mol.K * (1/425 – 1/400)
ln 1.19*10-2 / k2 = -2.15
so taking anitln from both side
1.19*10-2 / k2 = 0.116
So, k2 = 1.19*10-2 / 0.116
= 0.102 s-1
So, the value of the rate constant at 425 K is 0.102 s-1