Rate constant k varies with temperature by equation log10k(min-1) = 5 2000/T. We can concludea)Pre exponential factor A is 105b)Eais 2000 k calc)Eais 9.212 k cald)Pre exponential factor A is 5Correct answer is option 'A,C'. Can you explain this answer?

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Rate constant k varies with temperature by equation log₁₀k(min^-1) = 5 – ²⁰⁰⁰/_T. We can conclude

a)
Pre exponential factor A is 10⁵
b)
E_a is 2000 k cal
c)
E_a is 9.212 k cal
d)
Pre exponential factor A is 5

Correct answer is option 'A,C'. Can you explain this answer?

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Chemical Kinetics MCQ - 1 (Advanced)

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Verified Answer

Rate constant k varies with temperature by equation log10k(min-1) = 5 ...

log A = 5 ⇒ A = 10⁵

E_a = 2000 × 2.303 × 8.314 cal
E_a = 9.212 Kcal

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Rate constant k varies with temperature by equation log10k(min-1) = 5 2000/T. We can concludea)Pre exponential factor A is 105b)Eais 2000 k calc)Eais 9.212 k cald)Pre exponential factor A is 5Correct answer is option 'A,C'. Can you explain this answer?

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