For the reaction A ( g )+ B ( g ) arrow 3 C ( g ), 298 K If Δ U°=-10 kJ and Δ S°=40 JK -1 then Δ G ° for the reaction is nearly

Question

For the reaction A ( g )+ B ( g ) arrow 3 C ( g ), 298 K If Δ U°=-10 kJ and Δ S°=40 JK -1 then Δ G ° for the reaction is nearly (A) 6.1 kJ (B) 12.2 kJ (C) -19.4 kJ (D) -48.8 kJ

Tardigrade Admin · Accepted Answer

Δ H ° =Δ U °+Δ n 9 RT =-10+(3-1-1) × 8.314 × 10-3 × 298 =-7.5 kJ Δ G ° =Δ H °- T Δ S ° =-7.5-298 × 40 × 10-3 =-19.4 kJ

Q. For the reaction
$A (g) + B (g) \to 3 C (g), 298 K$
If $Δ U^{\circ} = - 10 k J$ and $Δ S^{\circ} = 40 J K^{- 1}$ then $Δ G^{\circ}$ for the reaction is nearly

$6.1 k J$

$12.2 k J$

$- 19.4 k J$

$- 48.8 k J$

$Δ H^{\circ} = Δ U^{\circ} + Δ n_{9} RT$
$= - 10 + (3 - 1 - 1) \times 8.314 \times 1 0^{- 3} \times 298$
$= - 7.5 k J$
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
$= - 7.5 - 298 \times 40 \times 1 0^{- 3}$
$= - 19.4 k J$

Q. For the reaction A(g)+B(g)→3C(g),298K If ΔU∘=−10kJ and ΔS∘=40JK−1 then ΔG∘ for the reaction is nearly

6.1kJ

12.2kJ

−19.4kJ

−48.8kJ