In a reaction,the rate constant K_1 is twice | vedclass

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(B) According to the Arrhenius equation,the rate constant is given by $K = A e^{-E_a/RT}$.For two reactions at the same temperature $T$,assuming the f

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$E_1 < E_2$

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(B) According to the Arrhenius equation,the rate constant is given by $K = A e^{-E_a/RT}$.For two reactions at the same temperature $T$,assuming the frequency factor $A$ is the same for both:$\frac{K_1}{K_2} = \frac{e^{-E_1/RT}}{e^{-E_2/RT}} = e^{(E_2 - E_1)/RT}$.Given $K_1 = 2K_2$,we have $\frac{K_1}{K_2} = 2$.Taking the natural logarithm on both sides: $\ln(2) = \frac{E_2 - E_1}{RT}$.Since $\ln(2) > 0$,it follows that $E_2 - E_1 > 0$,which means $E_2 > E_1$ or $E_1 < E_2$.

In a reaction,the rate constant $K_1$ is twice the rate constant $K_2$ of another reaction. Find the relationship between the activation energies $E_1$ and $E_2$ of the two reactions at the same temperature.

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