NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
Given, C H 3 C O O H a q → C H 3 C O O - a q + H + a q ΔH rxn o = 0 .004 kcal gm − 1 Enthalpy change when 1 mole of B a ( O H ) 2 , a strong base, is completely neutralized by C H 3 C O O H ( a q ) is ( Δ H ∘ of neutralization of strong acid with strong base is = − 13 .7 kcal mol − 1 )
Options
- A− 27 .46 kcal/mol
- B27 .46 kcal/mol
- C− 26 .92 kcal/mol
- D− 13 .46 kcal/mol
Correct answer
C. − 26 .92 kcal/mol
Step-by-step solution
ΔH o ionisaton of CH 3 COOH = 0 .04 × 60 = 0 .24 kcal/mol ∴ Enthalpy change = – 13.7 + 0.24 × 2 =   − 26 .92   kcal   mol − 1