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Karnataka Board PUCPUC Science 2nd PUC Class 12

The decomposition of hydrocarbon follows the equation k = (4.5 ร— 10^11โข ๐‘ โˆ’1)โข ๐‘’^โˆ’28000โข๐พ/๐‘‡ Calculate Ea.

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Question

The decomposition of hydrocarbon follows the equation

k = `(4.5 xx 10^11 s^-1) e^(-28000 K//T)`

Calculate Ea.

Numerical
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Solution

According to the Arrhenius equation, k = `Ae^((-E_a)/(RT))`

∴ `-E_a/(RT) = -(28000 K)/T`

Ea = 28000 K × R

= 28000 K × 8.314 JK−1 mol−1

= 232.79 kJ mol−1

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Chapter 3: Chemical Kinetics - 'NCERT TEXT-BOOK' Exercises [Page 282]

APPEARS IN

Nootan Chemistry [English] Class 12 ISC
Chapter 3 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.26 | Page 282
NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 3 Chemical Kinetics
Exercises | Q 3.26 | Page 87

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