For the given reaction, H 2( g )+ Cl 2( g )-2 H +( aq )+2 Cl -( aq ) ; Δ G °=--262 ⋅ 4 kJ The value of free energy of formation (Δ Gf°) for the ion C l-1(a q), therefore will be

Question

For the given reaction, H 2( g )+ Cl 2( g )->2 H +( aq )+2 Cl -( aq ) ; Δ G °=--262 ⋅ 4 kJ The value of free energy of formation (Δ Gf°) for the ion C l-1(a q), therefore will be (A) - 131.2 kJ mol-1 (B)  + 131.2 kJ mol-1 (C) - 262.4 kJ mol-1 (D)  + 262.4 kJ mol-1

Tardigrade Admin · Accepted Answer

(Δ G°) reaction = Δ G0f (Products) - Δ G0f (reactants) ∴ 264.4 = [2 Δ G0f (H+) + 2 Δ G0f (C1-)] or 264.4 = - [Δ Gf0 (H2) + Δ G0f (Cl2)] = [0+ 2Δ G0f (Cl-)] + [0 +0] or, - 262.4 = 2Δ Gf0 (Cl-) or, Δ G0g(Cl-) = -131.2 kJ mol-1

Q. For the given reaction, $H_{2} (g) + C l_{2} (g) - > 2 H^{+} (a q) + 2 C l^{-} (a q); Δ G^{\circ} = - - 262 \cdot 4 k J$ The value of free energy of formation $(Δ G_{f}^{\circ})$ for the ion $C l^{- 1} (a q)$ , therefore will be

$- 131.2 k J m o l^{- 1}$

$+ 131.2 k J m o l^{- 1}$

$- 262.4 k J m o l^{- 1}$

$+ 262.4 k J m o l^{- 1}$

Q. For the given reaction, H2​(g)+Cl2​(g)−>2H+(aq)+2Cl−(aq);ΔG∘=−−262⋅4kJ The value of free energy of formation (ΔGf∘​) for the ion Cl−1(aq), therefore will be

−131.2kJmol−1

+131.2kJmol−1

−262.4kJmol−1

+262.4kJmol−1

(ΔG∘) reaction =ΔGf0​ (Products) −ΔGf0​ (reactants) ∴264.4=[2ΔGf0​(H+)+2ΔGf0​(C1−)] or 264.4=−[ΔGf0​(H2​)+ΔGf0​(Cl2​)] =[0+2ΔGf0​(Cl−)]+[0+0] or, −262.4=2ΔGf0​(Cl−) or, ΔGg0​(Cl−)=−131.2kJmol−1

Q. For the given reaction, $H_{2} (g) + C l_{2} (g) - > 2 H^{+} (a q) + 2 C l^{-} (a q); Δ G^{\circ} = - - 262 \cdot 4 k J$ The value of free energy of formation $(Δ G_{f}^{\circ})$ for the ion $C l^{- 1} (a q)$ , therefore will be

$- 131.2 k J m o l^{- 1}$

$+ 131.2 k J m o l^{- 1}$

$- 262.4 k J m o l^{- 1}$

$+ 262.4 k J m o l^{- 1}$