Potential Energy Diagram illustrating activation energy and enthalpy change for a forward and reverse reaction.

Here is the step-by-step solution to the problems based on the Potential Energy Diagram provided.

Step-by-Step Reasoning

1. Identify the Energy Levels from the Graph:
First, we read the energy values (in kJ) from the y-axis for each part of the reaction:
* Reactants ($A + B$): The flat line on the left starts at 40 kJ.
* Activated Complex ($ABC$): The highest peak of the curve is at 80 kJ.
* Products ($C + D$): The flat line on the right ends at 20 kJ.

2. Solve Part (a): Activation Energy for the forward reaction
* Definition: This is the energy needed to go from the Reactants up to the top of the hill (Activated Complex).
* Calculation: $\text{Energy of Activated Complex} - \text{Energy of Reactants}$
* $80\text{ kJ} - 40\text{ kJ} = \mathbf{40\text{ kJ}}$

3. Solve Part (b): Activation Energy for the reverse reaction
* Definition: This is the energy needed to go from the Products back up to the top of the hill.
* Calculation: $\text{Energy of Activated Complex} - \text{Energy of Products}$
* $80\text{ kJ} - 20\text{ kJ} = \mathbf{60\text{ kJ}}$

4. Solve Part (c): Enthalpy Change ($\Delta H$) for the forward reaction
* Definition: This is the overall difference in energy between the final products and the starting reactants.
* Calculation: $\text{Energy of Products} - \text{Energy of Reactants}$
* $20\text{ kJ} - 40\text{ kJ} = \mathbf{-20\text{ kJ}}$

5. Solve Part (d): Enthalpy Change ($\Delta H$) for the reverse reaction
* Definition: If you run the reaction backward, the sign of the enthalpy change flips.
* Calculation: The negative of the forward $\Delta H$.
* $-(-20\text{ kJ}) = \mathbf{+20\text{ kJ}}$

6. Solve Part (e): Is the forward reaction exothermic or endothermic?
* Logic: Look at the start and end points. The reaction started at 40 kJ and ended lower at 20 kJ. Because energy was released (the level went down), it is exothermic. Also, a negative $\Delta H$ always means exothermic.

7. Solve Part (f): Is the reverse reaction exothermic or endothermic?
* Logic: The reverse reaction goes from low energy (20 kJ) to higher energy (40 kJ). Because it requires absorbing energy to go "uphill," it is endothermic.

8. Solve Part (g): Which species form the Activated Complex?
* Logic: Look at the label at the very top of the peak on the graph. It is labeled ABC. This represents the temporary combination of atoms A, B, and C before they break apart into products.

9. Solve Part (h): Which bond is stronger, A–B or B–C?
* Logic: In chemistry diagrams like this, the order of letters usually indicates how they are connected. The reactants are $A + B$, meaning A and B are separate. They come together to form the complex $ABC$. Usually, this implies A bonds to B, and B bonds to C ($A-B-C$).
* Reasoning: Bond strength is related to stability. Lower potential energy means a more stable, stronger bond.
* The reactant state involves the A–B interaction (or rather, the formation of the initial link). However, looking at the products $C + D$ isn't helpful here because the labels changed. Let's look at the complex $ABC$.
* Actually, a simpler interpretation often used in these specific textbook problems relies on the position of the "hump". But there is a clearer clue: Stability.
* Let's look at the bonds breaking and forming. We go from $A+B$ to $ABC$ to $C+D$. Wait, the product is $C+D$? No, the product label is $C + D$ but the complex is $ABC$. This implies atom D might be a typo in the product label or part of the background, OR the reaction is actually $AB + C \rightarrow A + BC$. Let's re-read the graph carefully.
* Reactants: $A + B$. Products: $C + D$. Complex: $ABC$. This is chemically inconsistent unless D is a spectator or there's a typo.
* *Correction/Standard Interpretation:* Often in these generic diagrams, if the question asks about A-B vs B-C, it assumes the molecule is arranged $A-B-C$.
* Let's look at energy levels again. The reactants ($A+B$) are at 40. The products ($C+D$) are at 20.
* Let's assume the question implies the bonds within the complex. The bond that forms first or is "stronger" holds the atoms tighter (lower energy).
* However, without explicit bond energy data, we look at the stability of the species. The products are at a lower energy (20 kJ) than the reactants (40 kJ). This suggests the bonds in the products are stronger overall.
* Let's look at the specific question: "Which bond is stronger, A-B or B-C?". This implies the intermediate is $A-B-C$.
* Usually, the side with the lower activation energy has the stronger bonds holding the transition state closer to that side? No.
* Let's use the Hammond Postulate logic simply: The transition state resembles the species it is closest to in energy. The peak (80) is closer to Reactants (40) than Products (20)? No, $80-40=40$ and $80-20=60$. It is closer to reactants.
* *Alternative Logic:* Stronger bonds release more energy when formed. The reaction is exothermic, meaning the new bonds formed (in products) are stronger than the old bonds broken (in reactants).
* If we assume the reaction is effectively $AB \rightarrow A + B$ type logic mapped to $A+B \rightarrow C+D$, it's ambiguous.
* *Most likely intended answer for this specific curriculum:* Look at the "valleys". The product valley is deeper (lower energy) than the reactant valley. Therefore, the bonds in the products are stronger. If the question asks about A-B vs B-C, it likely assumes A-B is the reactant bond and B-C is the product bond (implying a rearrangement $A-B + C \rightarrow A + B-C$). Since the products are lower energy, the B-C bond (product side) is stronger.
* *Reason:* The products have lower potential energy than the reactants. Lower potential energy indicates a more stable system with stronger bonds. Therefore, the bonds in the product phase (assumed to involve B-C) are stronger.

10. Solve Part (i): Which particles are moving faster?
* Logic: Kinetic energy is related to temperature, but in potential energy diagrams, "higher up" on the y-axis means higher Potential Energy.
* Conservation of Energy: Total Energy = Potential Energy + Kinetic Energy.
* When Potential Energy is low, Kinetic Energy is high (assuming constant total energy during the collision process relative to the baseline).
* The products ($C+D$) are at the lowest energy level (20 kJ). The reactants are at 40 kJ.
* Therefore, the particles at the lowest potential energy level have converted that potential energy into kinetic energy.
* Answer: The products ($C + D$) are moving faster because they are at a lower potential energy state, implying higher kinetic energy.

*(Self-Correction on Part i)*: Actually, in many introductory contexts, this question is a trick. Temperature determines average kinetic energy. If the reaction is isolated, the heat released (exothermic) increases the temperature of the surroundings/products. So the products would indeed be hotter/moving faster.
However, a simpler interpretation often accepted is: Lower Potential Energy = Higher Kinetic Energy. Since the products are at the bottom of the well, they have the highest kinetic energy relative to the activated complex. Comparing Reactants vs Products: The drop in PE becomes KE. So Products have more KE.

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Final Answer

a) 40 kJ
b) 60 kJ
c) -20 kJ
d) +20 kJ
e) Exothermic
f) Endothermic
g) ABC
h) B–C (or the product bond).
*Reason:* The products are at a lower potential energy level (20 kJ) than the reactants (40 kJ). Lower potential energy indicates greater stability and stronger bonds. Assuming B-C represents the bond in the lower-energy product state, it is stronger.
i) C + D (The products).
*Reason:* As the reaction proceeds from reactants to products, potential energy decreases (from 40 kJ to 20 kJ). According to the law of conservation of energy, this lost potential energy is converted into kinetic energy. Therefore, the particles with the lowest potential energy (the products) have the highest kinetic energy and are moving the fastest.