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प्रश्न
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?
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उत्तर
Given: k = 2.418 × 10−5 s−1
T = 546 K
Ea = 179.9 kJ mol−1
According to the Arrhenius equation,
log A = `log k + E_a/(2.303 RT)`
= `log 2.418 × 10^-5 + 179.9/(2.303 xx 8.314 xx 10^-3 xx 546)`
= (−5 + 0.3834) + 17.2081
= −4.6166 + 17.2081
log A = 12.5915
⇒ A = Antilog (12.5915)
= 3.902× 1012 s−1
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