For the reaction $$\mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{R}+\mathrm{Z} \quad \Delta H=+295 \mathrm{~kJ},$$ draw a reaction-energy diagram for the reaction if its activation energy is \(378 \mathrm{~kJ} .\)

Question: Draw and label a reaction-energy diagram for the chemical reaction $$\mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{R}+\mathrm{Z}$$ with its enthalpy change (\(\Delta H\)) equal to \(+295 \mathrm{~kJ}\) and activation energy (\(E_a\)) equal to \(378 \mathrm{~kJ}\). Answer: To draw a reaction-energy diagram for the given reaction, follow these steps: 1. Draw the axes for the reaction-energy diagram, labeling the x-axis as "Reaction Progress" and the y-axis as "Energy." 2. Indicate the initial energy state of the reactants (point A) and the final energy state of the products (point B), with B being higher on the y-axis than A due to the positive enthalpy change. 3. Mark the point representing the activation energy (point C), which is \(378 \mathrm{~kJ}\) higher on the y-axis than point A. 4. Indicate the enthalpy change (295 kJ) between A and B. 5. Connect points A, C, and B with a smooth curve to create the reaction-energy diagram, showing the energy changes during the reaction.