Home
Class 12
CHEMISTRY
The decomposition of hydrocarbon follow...

The decomposition of hydrocarbon follows the equation `k=(4.5xx10^(11)s^(-1))e^(-28000K//T)`
Calculate `E_(a)`.

Text Solution

Verified by Experts

`K=(4.5xx10^11 s^(-1)) e^(-28000k//T)`
Comparing the equation with Arrhenius equation,
`K=Ae^(-Ea//RT)`
`-E_a/R=-28000` K
`E_a=28000xx8.314`
`=232192 J "mol"^(-1)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CHEMICAL KINETICS

    CBSE COMPLEMENTARY MATERIAL|Exercise SHORT ANSWER TYPE-II QUESTIONS|23 Videos

  • CHEMICAL KINETICS

    CBSE COMPLEMENTARY MATERIAL|Exercise LONG ANSWER TYPE QUESTIONS|12 Videos

  • CHEMICAL KINETICS

    CBSE COMPLEMENTARY MATERIAL|Exercise VERY SHORT ANSWER TYPE QUESTIONS|24 Videos

  • BIOMOLECULES

    CBSE COMPLEMENTARY MATERIAL|Exercise SHORT ANSWER-II TYPE QUESTION|13 Videos

  • CHEMISTRY IN EVERYDAY LIFE

    CBSE COMPLEMENTARY MATERIAL|Exercise LONG ANSWER TYPE QUESTIONS|2 Videos

Similar Questions

Explore conceptually related problems

The decomposition of a hydrocarbon follows the equation k=(4.5xx10^(11)s^(-1))e^(-28000" K"//"T") Calculate E_(a) .

Show that time required for 99% completionis twice the time required for the completion of 90% of reactions for a first order reaction. The decomposition of hydrocarbon follow the equation K=(4.5xx10^(11)5^(-1))e^(-28000K//T)" Calculate Ea" .

Knowledge Check

  • 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

    A
    `2.9xx10^(4) kJ "mol"^(-1)`
    B
    `2.41xx10^(2) kJ "mol"^(-1)`
    C
    `1.45 xx10^(4) kJ "mol"^(-1)`
    D
    None of these

  • 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

    A
    `5.0xx10^(11)` k cal
    B
    `29000` k cal
    C
    58 k cal
    D
    `-29000` cal

  • 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:

    A
    `275 kJ//mol e`
    B
    `250 kJ// mol e`
    C
    `241 kJ//mol e`
    D
    `225KJ//mol e`

    • Similar Questions

    Explore conceptually related problems

    The rate constant for the first order decomposition of N_(2) O_(5) is given by the following equation: k=(2.5 xx 10^(14) s^(-1) ) e^((-25000K)//T) Calculate E_a for this reaction and rate constant if its half-life period be 300 minutes.

    The decomoposition of N_(2)O into N_(2) and O in the presence of argon follows second kinetics with k = (5.0 xx 10^(11)) e^(-29000K//T) Calculate the energy of activation .

    The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.34 -1.25 xx 10^(4)K//T Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes?

    The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : logk=14.34-1.25xx10^(4)"K"//"T" Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes ?

    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.

    Home
    Profile