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(A) The relationship between the enthalpy change of a reaction $(\Delta H)$,the activation energy of the forward reaction $(E_{a(f)})$,and the activat

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$+300$

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(A) The relationship between the enthalpy change of a reaction $(\Delta H)$,the activation energy of the forward reaction $(E_{a(f)})$,and the activation energy of the reverse reaction $(E_{a(r)})$ is given by the equation: $\Delta H = E_{a(f)} - E_{a(r)}$.Given:$\Delta H = +100 \ kJ/mol$$E_{a(r)} = +200 \ kJ/mol$Substituting these values into the equation:$100 = E_{a(f)} - 200$$E_{a(f)} = 100 + 200 = +300 \ kJ/mol$.Therefore,the activation energy for the forward reaction is $+300 \ kJ/mol$.

For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.

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