Given the following entropy values (in J K-1 mol-1) at 298 K and 1 atm: H2(g): 130.6, Cl2(g): 223.0, HCl(g): 186.7 .The entropy change (in J K-1 mol-1) for the reaction H2 (g) + Cl2 (g) arrow 2HCl (g) is

Question

Given the following entropy values (in J K-1 mol-1) at 298 K and 1 atm: H2(g): 130.6, Cl2(g): 223.0, HCl(g): 186.7 .The entropy change (in J K-1 mol-1) for the reaction H2 (g) + Cl2 (g) arrow 2HCl (g) is (A) +540.3 (B) + 727.0 (C) -166.9 (D) +19.8

Tardigrade Admin · Accepted Answer

Entropy change Δ S = Δ S textproduct - Δ S textreactant = 2 (186.7) - (223 + 130.6) = 373.4 - 353.6 = 19.8 JK-1 mol-1

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

+540.3

+ 727.0

-166.9

+19.8

Entropy change
$Δ S = Δ S_{product} - Δ S_{reactant}$
$= 2 (186.7) - (223 + 130.6)$
$= 373.4 - 353.6$
$= 19.8 J K^{- 1} m o l^{- 1}$

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