Match the columns
Column I
Column II
P) Fe(s)
i) Δ_f H° = -ve, Δ_f S° = -ve
Q) Br₂(g)
ii) Δ_f H° = 0, Δ_f S° = 0
R) NO₂(g)
iii) Δ_f H° = +ve, Δ_f S° = +ve
S) NH₃(g)
iv) Δ_f H° = +ve, Δ_f S° = -ve
(Column I = Substance; Column II = Signs of Δ_f H° and Δ_f S°)

Question

Match the columns

Column I	Column II
P) Fe(s)	i) Δ_f H° = -ve, Δ_f S° = -ve
Q) Br₂(g)	ii) Δ_f H° = 0, Δ_f S° = 0
R) NO₂(g)	iii) Δ_f H° = +ve, Δ_f S° = +ve
S) NH₃(g)	iv) Δ_f H° = +ve, Δ_f S° = -ve

(Column I = Substance; Column II = Signs of Δ_f H° and Δ_f S°)

TestHub · Accepted Answer

P: Fe(s) is the standard state of iron, so both Δ f H° and Δ f S° are zero, P-ii. Q: Br 2 (l) → Br 2 (g) is vaporization, endothermic with a large entropy increase (liquid to gas), Q-iii. R: 1/2N 2 (g) + O 2 (g) → NO 2 (g) has Δ f H° = +33 kJ/mol and 1.5 mol of gas becomes 1 mol, so Δ f S° < 0, R-iv. S: 1/2N 2 (g) + 3/2H 2 (g) → NH 3 (g) has Δ f H° = -46 kJ/mol and 2 mol of gas become 1 mol, so Δ f S° < 0, S-i. Ans: P-ii; Q-iii; R-iv; S-i.

Chemistry - Thermochemistry Question with Solution | TestHub

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