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Match column -I to column-II standard en...

Match column -I to column-II standard entropy in J/K-molar at 298 K
`{:("column"-1,"Column"-2),((A)"C(s, diamond)" , (p)5.7),((B)"C(s, graphite)",(q)2.37),((C)H_(2)(g),(r)117.6),((D)H(g),(s)130.6):}`

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To solve the problem of matching the standard entropy values from Column I to Column II, we will analyze the given substances and their respective entropies at 298 K. ### Step-by-Step Solution: 1. **Identify the substances in Column I:** - A: C(s, diamond) - B: C(s, graphite) - C: H₂(g) - D: H(g) 2. **Understand the concept of entropy:** - Entropy is a measure of disorder or randomness in a system. Generally, gases have higher entropy than solids due to the greater freedom of movement of gas molecules. Among solids, the structure and bonding can affect entropy. 3. **Compare the entropy of diamond and graphite:** - Diamond (C(s, diamond)) has a more ordered structure due to sp³ hybridization, resulting in lower entropy. - Graphite (C(s, graphite)) has a layered structure with sp² hybridization, allowing for more disorder and higher entropy. - From the video transcript, we know: - Entropy of diamond = 2.37 J/K·mol - Entropy of graphite = 5.7 J/K·mol Therefore, we can match: - A (C(s, diamond)) → q (2.37) - B (C(s, graphite)) → p (5.7) 4. **Compare the entropy of H₂ and H:** - H₂(g) is a diatomic gas, while H(g) is a monatomic gas. - The diatomic molecule (H₂) has more degrees of freedom and thus higher entropy than the monatomic hydrogen atom (H). - From the video transcript, we know: - Entropy of H₂(g) = 130.6 J/K·mol - Entropy of H(g) = 117.6 J/K·mol Therefore, we can match: - C (H₂(g)) → s (130.6) - D (H(g)) → r (117.6) 5. **Final Matching:** - A → q (C(s, diamond) → 2.37) - B → p (C(s, graphite) → 5.7) - C → s (H₂(g) → 130.6) - D → r (H(g) → 117.6) ### Final Answer: - A → q (2.37) - B → p (5.7) - C → s (130.6) - D → r (117.6)
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