The enthalpy (H) of an elementary exothermic | vedclass

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Question

(A) For an exothermic reaction,$A \rightleftharpoons B$,the change in enthalpy is negative,i.e.,$\Delta H < 0$. Since $\Delta H = E_{Product} - E_{Rea

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(A) For an exothermic reaction,$A \rightleftharpoons B$,the change in enthalpy is negative,i.e.,$\Delta H Since $\Delta H = E_{Product} - E_{Reactant}$,we have $E_{B} - E_{A} This indicates that the energy level of the product $B$ must be lower than that of the reactant $A$. Furthermore,a catalyst provides an alternative pathway with a lower activation energy but does not change the initial energy of the reactants or the final energy of the products. Therefore,the solid line (uncatalyzed) and dashed line (catalyzed) must start and end at the same energy levels. Plot $A$ correctly shows $E_{B} < E_{A}$ and the same starting and ending points for both catalyzed and uncatalyzed paths.

The enthalpy $(H)$ of an elementary exothermic reaction,$A \rightleftharpoons B$ is schematically plotted against the reaction coordinate. The plots in the presence and absence of a catalyst are shown in dashed and solid lines,respectively. Identify the correct plot for the reaction.

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