Answer
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Hint: 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-
\[\begin{align}
& 4KCl{{O}_{3}}\to 3KCl{{O}_{4}}+KCl \\
& KCl{{O}_{4}}\to KCl+2{{O}_{2}} \\
\end{align}\]
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.
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-
\[\begin{align}
& 4KCl{{O}_{3}}\to 3KCl{{O}_{4}}+KCl \\
& KCl{{O}_{4}}\to KCl+2{{O}_{2}} \\
\end{align}\]
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.
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