List - I (Reaction) List - II ( ΔG and ΔS of reaction respectively) (P) C (graphite) + O 2 ( g ) → CO 2 ( g ) (1)  = 0 , 0, 0 (R) H

Question

List - I (Reaction) List - II ( ΔG and ΔS of reaction respectively) (P) C (graphite) + O 2 ( g ) → CO 2 ( g ) (1)  = 0 , < 0 (Q) H 2 O ( ℓ ) ⇌ 373 K and 1 atm H 2 O ( g ) (2) > 0, > 0 (R) H 2 O ( s ) ⇌ 270 K and 1 atm H 2 O ( ℓ ) (3) > 0 , ≈ 0 (S) C 6 H 6 ( ℓ ) + 2 15 O 2 ( g ) 2 7 ∘ C 6 CO 2 ( g ) + 3 H 2 O ( ℓ ) (4) = 0 , > 0 (5) < 0 , ≈ 0

TestHub · Accepted Answer

(P) → (5): Combustion of graphite is spontaneous ( Δ G < 0 ), and since Δ n g = 1 − 1 = 0 , the entropy change is minimal ( Δ S ≈ 0 ). (Q) → (4): Phase transition of water at its boiling point ( 373 K , 1 a t m ) is at equilibrium ( Δ G = 0 ), and boiling increases randomness ( Δ S > 0 ). (R) → (2): Melting of ice at 270 K (below freezing point) is non-spontaneous ( Δ G > 0 ), while converting solid to liquid increases randomness ( Δ S > 0 ). (S) → (1): Combustion of benzene is spontaneous ( Δ G < 0 ⇒ Δ G  = 0 ), and since gaseous moles decrease ( Δ n g = 6 − 7.5 = − 1.5 ), entropy decreases ( Δ S < 0 ).

List - I (Reaction)	List - II ( ΔG and ΔS of reaction respectively)
(P) C (graphite) $+ O_{2} (g) \to CO_{2} (g)$	(1) $\neq = 0, < 0$
(Q) $H_{2} O (ℓ) ⇌ 373 K and 1 atm H_{2} O (g)$	(2) > 0, > 0
(R) $H_{2} O (s) ⇌ 270 K and 1 atm H_{2} O (ℓ)$	(3) $> 0, \approx 0$
(S) $C_{6} H_{6} (ℓ) + \frac{15}{2} O_{2} (g) 2 7^{\circ} C 6 CO_{2} (g) + 3 H_{2} O (ℓ)$	(4) $= 0, > 0$
	(5) $< 0, \approx 0$

Chemistry - Thermodynamics - II Question with Solution | TestHub

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