Rate constant k varies with temperature by equation log⁡kmin−1 =log⁡5−2000kcalRT×2.303. We can conclude that

Compare Arrhenius equation with the given equation log⁡k=log⁡A−Ea2.3RT2−T1T1T2Given equation: log⁡k=log⁡5−20002.3RTlog⁡A=log⁡5⇒A=105Ea=2000kcal

Rate constant k varies with temperature by equation log⁡kmin−1 =log⁡5−2000kcalRT×2.303. We can conclude that

Compare Arrhenius equation with the given equation log⁡k=log⁡A−Ea2.3RT2−T1T1T2Given equation: log⁡k=log⁡5−20002.3RTlog⁡A=log⁡5⇒A=105Ea=2000kcal

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Rate constant k varies with temperature by equation $\log k (min^{- 1}) = \log 5 - \frac{2000 kcal}{RT \times 2 .303}$ . We can conclude that

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The pre-exponential factor A is 10⁵

E_a is 2000 kcal

E_a is 9.12 kcal

The pre-exponential factor A is 5

answer is A, B.

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Detailed Solution

Compare Arrhenius equation with the given equation $\log k = \log A - \frac{E_{a}}{2.3 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
Given equation: $\log k = \log 5 - \frac{2000}{2.3 RT}$
$\begin{array}{l} \log A = \log 5 \Rightarrow A = 10^{5} \\ E_{a} = 2000 kcal \end{array}$