In the preparation of oxygen by heating KCIO$_3$ , in the presence of MnO$_2$ only moderate heat is needed because the catalyst acts by $_2$

Analysis of the Question

The question asks why the addition of Manganese(IV) oxide ($MnO_2$) allows the decomposition of Potassium trioxochlorate(V) ($KClO_3$) to occur at a lower temperature than it would otherwise. This is a classic question about the role of catalysts in chemical reactions.

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 ($E_a$).

• Without a catalyst, the reaction requires a high amount of heat to provide enough particles with the energy to overcome this high activation energy barrier.
• When a catalyst like $MnO_2$ is added, it provides an alternative reaction pathway that has a lower activation energy.
• Because the energy barrier is lowered, more reactant molecules have sufficient energy to react at a lower temperature, thereby increasing the overall rate of the reaction.

Step-by-Step Solution

1. Identify the reaction: The decomposition of $KClO_3$ is: $2KClO_3(s) \xrightarrow{MnO_2, \Delta} 2KCl(s) + 3O_2(g)$
2. Evaluate the options: • A (Lowering pressure): Catalysts do not affect the pressure of the system. • B (Increasing surface area): While surface area affects reaction rates in heterogeneous systems, it is not the mechanism by which a catalyst functions. • C (Increasing the rate of reaction): This is the definition of what a catalyst does. By providing an alternative pathway with a lower activation energy, the reaction proceeds much faster at a given temperature (or at a lower temperature). • D (Increasing the energy barrier): This is incorrect. A catalyst decreases the energy barrier (activation energy), it never increases it.

Conclusion

The presence of $MnO_2$ acts as a catalyst, which speeds up the decomposition of $KClO_3$ by providing an alternative pathway with a lower activation energy. Therefore, the reaction proceeds more rapidly (and at lower temperatures) than it would without the catalyst.

Exam Tip: Always remember that a catalyst lowers the activation energy but does not change the enthalpy change ($\Delta H$) of the reaction. If you see an energy profile diagram in your exam, the "hump" (activation energy) is always smaller when a catalyst is present!