For the reaction, 2A(g)+B(g)→ 2D(g), Δ U°=-10.5 KJ, Δ S°=-44.1 J K-1 Δ G° for the reaction is

Question

For the reaction, 2A(g)+B(g)→ 2D(g), Δ U°=-10.5 KJ, Δ S°=-44.1 J K-1 Δ G° for the reaction is (A) 0.16 kJ mol-1 (B) 5.5 kJ mol-1 (C) 4.2 kJ mol-1 (D) 8.52 kJ mol-1

Tardigrade Admin · Accepted Answer

Δ H° = Δ U°+Δ ngRT δ U° = 10.5 KJ, R = 8.314 J K-1 mol-1, T=298 K Δ ng = 2-(2+1)=-1 ∴ Δ H°=-10.5 -(1) × 8.314× 10-3 × 298 -10.5-2.48=-12.98 KJ mol-1 Now Δ G° = Δ H°-T Δ S°=-12.98-298 × (-44.1 × 10-3) =-12.98+13.14=0.16 kJ mol-1

Q. For the reaction,
$2 A_{(g)} + B_{(g)} \to 2 D_{(g)}, Δ U^{\circ} = - 10.5 K J, Δ S^{\circ} = - 44.1 J K^{- 1}$ $Δ G^{\circ}$ for the reaction is

$0.16 k J m o l^{- 1}$

$5.5 k J m o l^{- 1}$

$4.2 k J m o l^{- 1}$

$8.52 k J m o l^{- 1}$

Q. For the reaction, 2A(g)​+B(g)​→2D(g)​,ΔU∘=−10.5KJ,ΔS∘=−44.1JK−1 ΔG∘ for the reaction is

0.16kJmol−1

5.5kJmol−1

4.2kJmol−1

8.52kJmol−1

ΔH∘=ΔU∘+Δng​RT δU∘=10.5KJ,R=8.314JK−1mol−1,T=298K Δng​=2−(2+1)=−1 ∴ΔH∘=−10.5−(1)×8.314×10−3×298 −10.5−2.48=−12.98KJmol−1 Now ΔG∘=ΔH∘−TΔS∘=−12.98−298×(−44.1×10−3) =−12.98+13.14=0.16kJmol−1

Q. For the reaction,
$2 A_{(g)} + B_{(g)} \to 2 D_{(g)}, Δ U^{\circ} = - 10.5 K J, Δ S^{\circ} = - 44.1 J K^{- 1}$ $Δ G^{\circ}$ for the reaction is

$0.16 k J m o l^{- 1}$

$5.5 k J m o l^{- 1}$

$4.2 k J m o l^{- 1}$

$8.52 k J m o l^{- 1}$