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The rate constant for the decomposition ...

The rate constant for the decomposition of hydrocarbons is `2.418xx10^(-5)s^(-1)` at `546K`. If the energy of activation is `179.9kJ mol^(-1)`, what will be the value of pre`-`exponential factor?

Text Solution

Verified by Experts

The correct Answer is:
`3.9xx10^(12)s^(-10`
`k=2.418xx10^(-5)s^(-1),E_(a)=179.9kJ mol^(-1),T=546K,`
Using Arrhenius equation,
`k=Ae^(-Ea//RT)`
`ln k=lnA-(E_(a))/(RT)`
`or log k=log A-(E_(a))/(2.303RT)`
`or log A= log k +(E_(a))/(2.303RT)`
`=log(2.418xx10^(-5)s^(-1))+(179kJ mol^(-1))/(2.303xx8.314xx10^(-3)kJ mol^(-1)xx546K)`
`or A= Antilog(12.5924)s^)-1)=3.912xx10^(12)s^(-1)`
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The rate constant for the decomposition of hydrocarbons is 2.418xx10^(-5)s^(-1) at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?

The rate constant for the decomposition of a hydrocarbon is 2.418xx10^(-5)s^(-1) at 546 K. If the energy of activation is 179.9" kJ/mol" , what will be the value of pre-exponential factor ?

Knowledge Check

  • The rate constant for the decomposition of N_2O_5 in CCl_4 is 6.2 x 10 ^(-4) s^(-1) at 45°C. Calculate the rate constant at 100°C if the activation energy is 103 kJ mol^(-1) [Ant (2.49) = 309]

    A
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    D
    `1.75 x 10^(-2) s^(-1)`

  • The rate constant of a reaction is 1.5xx10^7 s^(-1) at 50^@C and 4.5xx10^(7)s^(-1) at 100^@C . What is the value of activation energy ?

    A
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