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The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor. - CBSE (Science) Class 12 - Chemistry

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Question

The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Solution

k = 2.418 × 10−5 s−1

T = 546 K

Ea = 179.9 kJ mol−1 = 179.9 × 103 J mol−1

According to the Arrhenius equation,

`k = Ae^((-E_a)/"RT")`

`=>In k = In A - E_a/"RT"`

`=>log k = log A - E_a/(2.303 RT)`

`=> logA = log k + E_a/(2.303 RT)`

`=log(2.418 xx10^(-5) s^(-1))+ (179.9xx10^3 "J mol"^-1)/(2.303xx8.314 " JK"^(-1) mol^(-1)xx546K)`

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 1012 s−1 (approximately)

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Chemistry Textbook for Class 12 (2018 to Current)
Chapter 4: Chemical Kinetics
Q: 23 | Page no. 120

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Solution The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor. Concept: Temperature Dependence of the Rate of a Reaction.
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