For a first-order reaction A to P,the rate co | vedclass

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

Vedclass · Accepted Answer

$1.0 	imes 10^6 \, s^{-1}$ and $38.3 \, kJ/mol$

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(C) The Arrhenius equation is given by $K = A \cdot e^{-E_a / (RT)}$.Taking the logarithm on both sides: $\log K = \log A - \frac{E_a}{2.303 \cdot R \cdot T}$.Comparing this with the given equation $\log K = -2000 \, (1/T) + 6.0$:$1$. $\log A = 6.0 \implies A = 10^6 \, s^{-1}$.$2$. $\frac{E_a}{2.303 \cdot R} = 2000$.Given $R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$,$E_a = 2000 	imes 2.303 	imes 8.314 \approx 38297 \, J/mol \approx 38.3 \, kJ/mol$.

For a first-order reaction $A \to P$,the rate constant equation is given by $\log K = -2000 \, (1/T) + 6.0$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

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