Question

When $KCl{{O}_{3}}$ is heated, it decomposes into KCl and ${{O}_{2}}$. If some $Mn{{O}_{2}}$ is added, the reaction goes much faster because:
[A] $Mn{{O}_{2}}$ decomposes to give ${{O}_{2}}$
[B] $Mn{{O}_{2}}$ provides heat by reacting.
[C] Better contact is provides by $Mn{{O}_{2}}$
[D] $Mn{{O}_{2}}$ acts as a catalyst.

Answer 1

The formation of potassium chloride and oxygen from potassium perchlorate is a slow reaction itself and requires a high temperature. Addition of manganese dioxide brings down the activation energy and the product is formed at a faster rate and also at a comparatively lower temperature.

Complete step by step answer:
Potassium perchlorate on heating gives us potassium chloride and oxygen. We can carry out this reaction with or without the presence of a catalyst. The products are the same in both the cases but the rate of the reaction and the pathway is altered.
On heating without a catalyst, potassium chlorate turns into potassium perchlorate which on further heating decomposes into oxygen and potassium chloride. We can write the reaction as-

 $$\begin{aligned}  

& 4KCl{{O}{3}}\to 3KCl{{O}{4}}+KCl \\
& KCl{{O}{4}}\to KCl+2{{O}{2}} \\
\end{aligned}$$

However, in the presence of $Mn{{O}_{2}}$, it gives potassium chloride and oxygen directly at a lower temperature. We can write the reaction as-
$2KCl{{O}_{3}}(s)\to 2KCl(s)+3{{O}_{2}}(g)$

In the above reaction, manganese dioxide neither decomposes nor reacts with the given reactant. It does not hinder the formation of the product either. It acts as a positive catalyst here, increasing the speed of the reaction.
So, the correct answer is “Option D”.

Note: We know that the function of a catalyst is to increase the speed of reaction and it does that by bringing down the activation energy so that a larger amount of particles have enough energy to react. A catalyst can bring down the activation energy by orienting the reactant particles in such a way that they collide successfully with each other or by reacting with the reactants and form an intermediate which will require a lower energy compared to the original starting reactant to form the product.