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On heating, $KCl{O_3}$ 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 oxygen
B. $Mn{O_2}$ provides heat by reacting
C. Better contact is provided by $Mn{O_2}$
D. $Mn{O_2}$ acts as catalyst

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Last updated date: 25th Apr 2024
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Answer
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Hint: Catalyst is a substance which increases or decreases the rate of a chemical reaction. We must remember the properties of a catalyst.

Complete step-by-step answer:
A catalyst is a material that raises reaction rate without getting consumed itself at the end of the reaction and the mechanism is called catalysis. There are certain catalysts that decrease the reaction rate and those catalysts are considered negative catalysts. The catalyst speeding the rate is a positive catalyst. The word positive, however, is seldom used and catalyst itself means positive catalyst. Catalysts are typically foreign substances, but occasionally one of the materials produced will act as a catalyst and this catalyst is called an auto catalyst and the process is known as an auto catalysis.
When $Mn{O_2}$ is added the thermal decomposition of $KCl{O_3}$ increases. Because acts as a $Mn{O_2}$

$2KCl{O_3} + [Mn{O_2}] \to 2KCl + 3{O_2} \uparrow + [Mn{O_2}]$
Here $Mn{O_2}$ which acts as a catalyst reduces the activation energy of the reaction as it provides alternative reaction paths. It also increases the rate of reaction backwards and forward in the same degree. Because of this lower activation energy, equilibrium is reached faster but does not affect the overall reaction enthalpy of the process.

Hence option D is the correct answer and as per this, $KCl{O_3}$ on heating decomposes into $KCl$ and ${O_2}$. If some $Mn{O_2}$ is added the reaction goes much faster because \[Mn{O_2}\] acts as a catalyst.

Note: For answering this type of question we must remember the properties of a catalyst. Here $Mn{O_2}$ which is being used and it increases the rate of reaction in $2KCl{O_3} + [Mn{O_2}] \to 2KCl + 3{O_2} \uparrow + [Mn{O_2}]$. But we can see that at the end of the process it is still present. It does not get consumed. This is an important [property and we must remember this.
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