Q120 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Chemistry
Which of the following processes is an example of decrease in entropy?
- ADissolution of solid iodine in water
- Correct answerBFormation of liquid water from hydrogen and oxygen gas
- CDissociation of hydrogen molecules into atoms
- DDecomposition of calcium carbonate on heating
Solution
Entropy is a measure of the randomness or disorder of a system. A decrease in entropy occurs when the randomness of the system decreases, which typically happens when the number of gaseous moles decreases or when a substance changes from a more disordered state (gas) to a less disordered state (liquid or solid).
For the dissolution of solid iodine in water: $I_{2}(s) \rightarrow I_{2}(aq)$. The solid lattice breaks down and molecules disperse in the solvent, leading to an increase in entropy ($\Delta S > 0$).
For the formation of liquid water from hydrogen and oxygen gas: $2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)$. Here, $3$ moles of gaseous reactants form $2$ moles of liquid product. The number of gaseous moles decreases ($\Delta n_{g} = -3$), and the system goes from a highly disordered gaseous state to a more ordered liquid state. Thus, entropy decreases ($\Delta S < 0$).
For the dissociation of hydrogen molecules into atoms: $H_{2}(g) \rightarrow 2H(g)$. One mole of gas forms two moles of gas. The number of particles increases, leading to an increase in entropy ($\Delta S > 0$).
For the decomposition of calcium carbonate on heating: $CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)$. A solid decomposes to form a solid and a gas. The production of gas increases the randomness, leading to an increase in entropy ($\Delta S > 0$).
Therefore, the formation of liquid water from hydrogen and oxygen gas is the only process that results in a decrease in entropy.
Answer: Formation of liquid water from hydrogen and oxygen gas
For the dissolution of solid iodine in water: $I_{2}(s) \rightarrow I_{2}(aq)$. The solid lattice breaks down and molecules disperse in the solvent, leading to an increase in entropy ($\Delta S > 0$).
For the formation of liquid water from hydrogen and oxygen gas: $2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)$. Here, $3$ moles of gaseous reactants form $2$ moles of liquid product. The number of gaseous moles decreases ($\Delta n_{g} = -3$), and the system goes from a highly disordered gaseous state to a more ordered liquid state. Thus, entropy decreases ($\Delta S < 0$).
For the dissociation of hydrogen molecules into atoms: $H_{2}(g) \rightarrow 2H(g)$. One mole of gas forms two moles of gas. The number of particles increases, leading to an increase in entropy ($\Delta S > 0$).
For the decomposition of calcium carbonate on heating: $CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)$. A solid decomposes to form a solid and a gas. The production of gas increases the randomness, leading to an increase in entropy ($\Delta S > 0$).
Therefore, the formation of liquid water from hydrogen and oxygen gas is the only process that results in a decrease in entropy.
Answer: Formation of liquid water from hydrogen and oxygen gas
Hence, the correct answer is option: 'B'.