The enthalpy of formation of H2$_{o}$ (l) is - 285.77 kJ mol$^{-1}$. The enthalpy of neutralisation of a strong acid and a strong base is - 56.07 kJ mol$^{-1}$. What is the enthalpy of formation of OH$^{-}$ion? Please help me get the correct answer and explain it.

• A

- 229.70 kJ

• B

+ 229.70 kJ

• C

+ 350.75 kJ

• D

- 350.75 kJ

Posted by: Tanjilur R. 2 months ago

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• From: Gurgaon
• Date: 10.11.2017
Option B is correct H2I+2BOHâ‡’2BI+2H2O=Î”H=âˆ’285.77kJ/mol H+OHâ€“â‡’H2O;Î”H==âˆ’56.07kJ/mol The heat of ionization or the energy of dissociation in this reaction is equal to (-285.77 + 56.07) kJ/mol = -229.7 KJ/mol
• Date: 07.10.2017

H2I+2BOH⇒2BI+2H2O=ΔH=−285.77kJ/mol

H+OH–⇒H2O;ΔH==−56.07kJ/mol

The heat of ionization or the energy of dissociation in this reaction is equal to (–285.77 + 56.07) kJ/mol =-229.7 KJ/mol

Option B

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