NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
Calculate Δ H f o for chloride ion from the following data: 1 2 H 2 g + 1 2 Cl 2 g → HCl g ; ΔH f o = − 92.4 kJ HCl g + nH 2 O → H + aq + Cl − aq ; ΔH f o = − 74.8 kJ ΔΗ f o of H + aq = 0.0 kJ
Options
- A- 167.2 k J
- B- 165.2 k J
- C- 157.2 k J
- D- 147.2 k J
Correct answer
A. - 167.2 k J
Step-by-step solution
Given, 1 2 H 2 g + aq → H + aq + e − ; ΔH o = 0 … . i 1 2 H 2 g + 1 2 Cl 2 g → HCl g ; ΔH o = − 92.4 kJ … ii HCl g + nH 2 O l →  H + aq   + Cl − aq   ; ΔH o =   − 74.8  kJ … iii HCl g + nH 2 O l →  H + aq   + Cl − aq   ; ΔH o =   − 74.8  kJ … iii i i + i i i − i ∴ 1 2 Cl 2 g + aq + e − → Cl − aq ; ΔH = − 92.4 − 74.8 kJ − 167.2 kJ … . iv Heat of formation for Cl − aq = − 167.2 k