Given are the following entropy values at 298 K and 1 atm: H 2 (g) = 130.6 J K –1 mol –1…

Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics

Given are the following entropy values at 298 K and 1 atm:
H₂ (g) = 130.6 J K^–1 mol^–1; Cl₂ (g) = 223.0 J K^–1 mol^–1;
HCl (g) = 186.7 J K^–1 mol^–1
The entropy change for the reaction
H₂ (g) + Cl₂ (g) 2HCl (g) would be

1. +540.3 J K^–1 mol^–1
2. +727 J K^–1 mol^–1
3. –166.9 J K^–1 mol^–1
4. +19.8 J K^–1 mol^–1

Answer

(D) +19.8 J K –1 mol –1

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