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What will be the change in the entropy of the decomposition of $ CaC{O_3} $ ?

Last updated date: 16th Apr 2024
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MVSAT 2024
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Hint: To answer the above question we first need to understand two terms first one is decomposition and the second one is entropy and the change in both of them generally and also with respect to calcium carbonate.

Complete answer:
Decomposition in chemistry is referred to as the breaking down of a compound into two or more entities. It can be done by several methods that are heating a substance etc one common example of decomposition is the reaction of hydrocarbons with oxygen which yields carbon dioxide and water.
Decomposition of calcium carbonate: $ CaC{O_3}(s) \to CaO(s) + C{O_2}(g) $
entropy can be understood as the randomness of a system it cannot be calculated easily but on increasing the randomness of the product entropy of the reaction also increases now in the above reaction e can observe that the reactant is in solid-state but the product which is yield is in the solid-state and in the gaseous state hence the randomness of the molecules of a gas is greater than the randomness of the solid molecules. Therefore there is an increase in randomness of the system hence entropy also increases.
The entropy of the system increases in the decomposition of calcium carbonate to calcium oxide and carbon dioxide.

The reason for the increase in randomness of gases molecules than that of solid molecules is because of the arrangement of the atoms in solid-state and gases state. In solid the atoms closely attached to each other while in gaseous state atoms move randomly in the space.
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