For a reaction A rightarrow B,the enthalpy of | vedclass

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(B) The enthalpy of reaction $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_a)_f$ and the activation

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(B) The enthalpy of reaction $\Delta H$ is given by the difference between the activation energy of the forward reaction $(E_a)_f$ and the activation energy of the backward reaction $(E_a)_b$:$\Delta H = (E_a)_f - (E_a)_b$Given $\Delta H = -4.2 \ kJ \ mol^{-1}$ and $(E_a)_f = 9.6 \ kJ \ mol^{-1}$:$-4.2 = 9.6 - (E_a)_b$$(E_a)_b = 9.6 + 4.2 = 13.8 \ kJ \ mol^{-1}$Since $\Delta H$ is negative,the reaction is exothermic,meaning the potential energy of the product $B$ must be lower than the potential energy of the reactant $A$. Among the given options,the graph where the product $B$ is at a lower energy level than reactant $A$ is represented by option $B$.

For a reaction $A \rightarrow B$,the enthalpy of reaction is $-4.2 \ kJ \ mol^{-1}$ and the enthalpy of activation is $9.6 \ kJ \ mol^{-1}$. The correct potential energy profile for the reaction is shown in which option?

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