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(A) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.Taking the logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Rearranging gives $\

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(A) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.Taking the logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Rearranging gives $\ln \frac{k}{A} = -\frac{E_a}{RT}$.Converting to base $10$: $\log_{10} \frac{k}{A} = -\frac{E_a}{2.303 RT}$.Given $\log_{10} \frac{k}{A} = 0.00174$,$T = 300 \ K$,and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$.Substituting the values: $0.00174 = -\frac{E_a}{2.303 	imes 8.314 	imes 300}$.Note: The provided equation $\log_{10} \frac{k}{A} = 0.00174$ implies a positive value,which is physically inconsistent with the Arrhenius equation for activation energy $E_a > 0$. Assuming the magnitude is intended: $E_a = |0.00174 	imes 2.303 	imes 8.314 	imes 300| \approx 10 \ J \ mol^{-1}$.

The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

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The following equation is obtained for a first order reaction at $300 \ K$.$\log_{10} \frac{k}{A} = 0.00174$What is the activation energy (in $J \ mol^{-1}$) of the reaction?$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$