In the preparation of oxygen by heating KClO3 in the presence of MnO2, only moderate heat is needed because the catalyst acts by?

Analysis

The question asks for the mechanism by which a catalyst (in this case, Manganese(IV) oxide, $MnO_2$) facilitates the decomposition of Potassium trioxochlorate(V) ($KClO_3$) at a lower temperature.

Concept Explanation

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed in the process.

In chemical kinetics, for a reaction to occur, reactant particles must collide with a minimum amount of energy known as the Activation Energy ($E_a$). This is the "energy barrier" that must be overcome for the reactants to transform into products.

A catalyst provides an alternative reaction pathway that has a lower activation energy. Because the energy barrier is lower, a larger fraction of reactant molecules possess enough energy to react at a given temperature. Consequently, the reaction proceeds much faster at lower temperatures than it would without the catalyst.

Step-by-Step Solution

1. Identify the role of $MnO_2$: In the laboratory preparation of oxygen, $MnO_2$ is added to $KClO_3$ to speed up the decomposition.
2. Evaluate the options: • A: Catalysts do not change the pressure of a reaction system. • B: While catalysts can be solid surfaces, their primary function is not just "increasing surface area" (that is the role of grinding or powdering reactants). • C: While it is true that a catalyst increases the rate of reaction, the question specifically asks how it allows the reaction to occur at "moderate heat." • D: By lowering the activation energy (the energy barrier), the catalyst ensures that the reaction can proceed efficiently at lower temperatures. This is the fundamental thermodynamic explanation for why "moderate heat" is sufficient.

Conclusion

While option C is a correct statement about what a catalyst does, option D provides the specific mechanism that explains why the heat requirement is reduced. In the context of JAMB chemistry, lowering the activation energy is the standard scientific explanation for this phenomenon.

Exam Tip: Always remember the "Energy Profile Diagram." A catalyst creates a "shortcut" on the graph, lowering the peak (the activation energy) between the reactants and the products. This is a favorite topic for JAMB examiners!