IIT JEE| NEET | NCERT CHEMICAL KINETICS: Question 4.23 - Videos

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The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5s-1 at 546 K. If the energy of activation is 179.9 kJ/mole, what will be the value of pre-exponential factor?
Solution:
Ea = 179.9 kJ/mol = 1.799×10^5Jmole-1

Arrhenius equation
K = Ae^-Ea/RT (on taking logarithm 10, equation becomes as given below)
log K = log A – Ea/2.303RT, Find out A value.

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