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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?
उत्तर
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
= 12.5924 s−1
or, A = Antilog (12.5924) s−1 = 3.902 × 1012 s−1
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