Rate constant ‘K’ varies with temperature by equation. loglog Kmin-1=5-2000T. We can conclude that R=8.314J Mol-1K-1

log Kmin-1=5-2000T It is in the form oflog K=logA-Ea2.303RT

log Kmin-1=5-2000T It is in the form oflog K=logA-Ea2.303RT

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Rate constant ‘K’ varies with temperature by equation. log $\log K (m i n^{- 1}) = 5 - \frac{(2000)}{T}$ . We can conclude that $(R = 8.314 J {Mol}^{- 1} K^{- 1})$

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$Pre - exponential factor, ‘ A ’ is 10^{5}$

$E_{a} is 9.212 kcal / mol$

Pre-exponential factor, ‘A’ is ‘5’.

$E_{a} is 4 k cal / mol$

answer is C, D.

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$\log K (m i n^{- 1}) = 5 - \frac{2000}{T}$ It is in the form of

$\log K = \log A - \frac{E_{a}}{2.303 R T}$

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