The decomposition of hydrocarbon follows the equation `k=(4.5xx10^(11)s^(-1))e^(-28000K//T)` Calculate `E_(a)`.
Text Solution
Verified by Experts
The given equation is `k = (4.5 xx 10^(11) s^(-1)) e^(-28000 K//T) (i)` Arrhenius equation is given by, `k = Ae^(-E_(a) //RT) (ii)` From equation (i) and (ii) , we obtain `(E_(a))/(RT) = (28000 K)/(T)` `implies E_(a) = R xx 28000 K` = `232792 J mol^(-1)` = `232.792 kJ mol^(-1)`
Topper's Solved these Questions
CHEMICAL KINETICS
NCERT|Exercise Exercise|1 Videos
CHEMICAL KINETICS
NCERT|Exercise SOLVED EXAMPLE|1 Videos
BIOMOLECULES
NCERT|Exercise Exercise|33 Videos
CHEMISTRY IN EVERYDAY LIFE
NCERT|Exercise Exercise|32 Videos
Similar Questions
Explore conceptually related problems
The decomposition of N_(2)O into N_(2) and O in the presence of gaseous argon follows second kinetics with k=(5.0xx10^(11) L "mol"^(-1) s^(-1))e^(-29000 K//T) The energy of activation is
The decomposition of N_(2)O into N_(2) and O in the presence of gaseous argon follows second order kinetics with k=(5.0xx10^(11) L " mol"^(-1)s^(-1))e^(-29000 K//T) The activation energy for the reaction E_(a) is
The rate constant of decomposition of methyl nitrite (K_(1)) and ethyl nitrite (K_(2)) follow the equations. K_(1)(s^(-1))=10^(13)e^([(-152.3xx10^(3))/(RT)Jmol^(-1)]) and K_(2)(s^(-1))=10^(14)e^([(-152.3xx10^(3))/(RT)Jmol^(-1)]) Calculate the temperature at which both have same rate of decomposition if 0.1M of each is taken and both show I order kinetics.
The rate constant for a second order reaction is given by k=(5 times 10^11)e^(-29000k//T) . The value of E_a will be:
