In which case change in entropy is negative?

Concept:

• Entropy refers to the randomness present in the system.
• The value of randomness can be positive or negative, depending on the resultant of the system's randomness.
• The order of randomness is (maximum) Gas > liquid > solid (minimum).

• As, when solid converts to its liquid or gaseous state, the entropy increases, as molecules of the solid-state are packed tightly because of more force of attraction between the molecules, but as it is converted to liquid or the gaseous state, the force of attraction reduces and therefore, the entropy increases.

• If the state on both sides of the reaction, i.e., reactant and product, are the same, the entropy value can be determined by calculating the value of Δng, which refers to the change in the moles of gas molecules.

Δng = moles of gas in the product – no. of moles of gas in the reactant.

Explanation:

From the question,

• H₂O(l) \(\rightleftharpoons\) H₂O (v), ΔS > 0
• Expansion of gas at a constant temperature, ΔS > 0
• Sublimation of solid to gas, ΔS > 0
• 2H(g) → H₂(g), ΔS < 0 (∵ Δng < 0)

Hence, the correct option for the given question is, 4) 2H(g) → H2(g).