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

The decomposition of hydrocarbon follows the equation, `k=(4.5 xx 10^(11) " s"^(-1) )"e"^(-28000K//T)`. Calculate `E_a`.

Text Solution

Verified by Experts

According to Arrhenius equation,
`k = Ae^(-E_(0) //RT)" "...(i)`
According to the available data
`k = (4.5 xx 10^(11)" s"^(-1) )e^(-28000K//7)`
On comparing both equations, we have
`- (E_a)/(RT) = (-28000 K)/(T)`
`E_a = (28000 K ) xx R = (28000 K)xx (8.314 K^(-1) )"J mol"^(-1) )`
`= 232792"J mol"^(-1)`
`E_a = 232.792 "kJ mol"^(-1)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    ARIHANT PUBLICATION|Exercise QUESTIONS FOR PRACTICE (Part I Rate of Chemical Reaction, Differential Rate Law and Rate Constant) Multiple Choice Type questions|2 Videos

  • CHEMICAL KINETICS

    ARIHANT PUBLICATION|Exercise QUESTIONS FOR PRACTICE (Part I Rate of Chemical Reaction, Differential Rate Law and Rate Constant) Very Short Answer Type Questions|5 Videos

  • CARBOXYLIC ACIDS

    ARIHANT PUBLICATION|Exercise CHAPTER PRACTICE (LONG ANSWER TYPE QUESTIONS) |3 Videos

  • CHEMISTRY IN EVERYDAY LIFE

    ARIHANT PUBLICATION|Exercise CHAPTER PRACTICE (LONG ANSWER TYPE QUESTIONS) |2 Videos

Similar Questions

Explore conceptually related problems

For first order decomposition, X to Y +Z rate constant is given by k = 4.6 xx 10^(14)" s"^(-1) e^(-54000// RT) where, the activation energy is in calories. Calculate the temperature where decomposition is 80% complete in 1 hr.

For first order decomposition, X to Y +Z rate constant is given by k = 4.6 xx 10^(14)" s"^(-1) e^(-54000// RT) where, the activation energy is in calories. Calculate the temperature at which 'X' would decompose at the rate of 1 % per second.

The rate constant for the first order decomposition of H_(2) O_(2) is given by the following equation, log k = 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 min?

Mention the factors that affect the rate of chemical reaction. For a reaction, the rate law is k[A]^(1//2)[B] . Can this reaction be an elementary reaction? The rate constant for the first-order decomposition of H_(2)O_(2) is given by the following equation log K =14.2 -(1.0 xx 10^(4)K)/(T) Calculate E_(a) for this reaction and rate constant k if its half life period be 200 min. (R=8.314 JK^(-1)"mol"^(-1))

The rate constant for the decomposition of hydrocarbons is 2.418 xx 10^(-5) "s"^(-1) at 546 K. If the energy of activation is 179.9" kJ mol"^(-1) , what will be the value of pre-exponential factor?

What is the order of the reaction which has rate constant, k=3 xx 10^(-4)" mol"^ L(-1)" s"^(-1) ?

The rate constant of a first order reaction is k =7.39 xx 10^(-5) " s"^(-1) . Find the half-life of the reaction.

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 xx 10^(10)" s"^(-1) . Calculate k at 318 K and E_(a) .

In the equation K=Ae^(-E_a/(RT)) what is A?