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(C) The Arrhenius equation is given by $K = A \cdot e^{-E_a / RT}$.Taking $log_{10}$ on both sides: $log_{10} K = log_{10} A - \frac{E_a}{2.303 \cdot

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Pre-exponential factor $A$ is $10^5$

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(C) The Arrhenius equation is given by $K = A \cdot e^{-E_a / RT}$.Taking $log_{10}$ on both sides: $log_{10} K = log_{10} A - \frac{E_a}{2.303 \cdot R \cdot T}$.Comparing this with the given equation $log_{10} K = 5 - \frac{2000}{T}$:$log_{10} A = 5 \implies A = 10^5$.$\frac{E_a}{2.303 \cdot R} = 2000 \implies E_a = 2000 \cdot 2.303 \cdot 8.314 \ J/mol$.$E_a \approx 38294 \ J/mol = 38.294 \ kJ/mol$.Converting to $kcal/mol$: $E_a \approx \frac{38.294}{4.184} \approx 9.15 \ kcal/mol$.Thus,the correct conclusion is that the pre-exponential factor $A$ is $10^5$.

Rate constant varies with temperature by the equation $log_{10} K = 5 - 2000 / T$. We can conclude that $(R = 8.314 \ J \ mol^{-1} K^{-1})$

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