For the forward reaction X rightarrow Y,the a | vedclass

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(A) The relationship between the activation energy of the forward reaction $(E_{a,f})$,the activation energy of the reverse reaction $(E_{a,r})$,and t

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$80$

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(A) The relationship between the activation energy of the forward reaction $(E_{a,f})$,the activation energy of the reverse reaction $(E_{a,r})$,and the enthalpy change $(\Delta H)$ is given by the equation: $\Delta H = E_{a,f} - E_{a,r}$.Given: $E_{a,f} = 60 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$.Substituting these values into the equation: $-20 = 60 - E_{a,r}$.Rearranging to solve for $E_{a,r}$: $E_{a,r} = 60 + 20 = 80 \ kJ \ mol^{-1}$.

For the forward reaction $X \rightarrow Y$,the activation energy is $60 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$. What is the activation energy for the reverse reaction $Y \rightarrow X$ in $kJ \ mol^{-1}$?

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