The entropy values (in .JK -1 mol -1) of H 2(g)=130.6, Cl 2(g)=223.0 and HCl (g)=186.7 at 298 K and 1 atm pressure. Then entropy change for the reaction: H 2(g)+ Cl 2(g) longrightarrow 2 HCl (g) is :

Question

The entropy values (in .JK -1 mol -1) of H 2(g)=130.6, Cl 2(g)=223.0 and HCl (g)=186.7 at 298 K and 1 atm pressure. Then entropy change for the reaction: H 2(g)+ Cl 2(g) longrightarrow 2 HCl (g) is : (A) + 540.3 (B) + 727.3 (C) - 166.9 (D) + 19.8

Tardigrade Admin · Accepted Answer

Δ S° =2 S° HCl -(S H 2°+S Cl 2°) =2 × 186.7-(130.6+223.0) =19.8 JK -1 mol -1

Q. The entropy values (in $J K^{- 1} m o l^{- 1})$ of $H_{2} (g) = 130.6, C l_{2} (g) = 223.0$ and $H Cl (g) = 186.7$ at $298 K$ and 1 atm pressure. Then entropy change for the reaction : $H_{2} (g) + C l_{2} (g) ⟶ 2 H Cl (g)$ is :

+ 540.3

+ 727.3

- 166.9

+ 19.8

$Δ S^{\circ} = 2 S^{\circ}_{H Cl} - (S_{H_{2}}^{\circ} + S_{C l_{2}}^{\circ})$
$= 2 \times 186.7 - (130.6 + 223.0)$
$= 19.8 J K^{- 1} m o l^{- 1}$

Q. The entropy values (in JK−1mol−1) of H2​(g)=130.6,Cl2​(g)=223.0 and HCl(g)=186.7 at 298K and 1 atm pressure. Then entropy change for the reaction : H2​(g)+Cl2​(g)⟶2HCl(g) is :