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Difference between Internal Energy and Enthalpy

Δf HH2O = -68...

Question

$Δ_{f} H (H_{2} O) = - 68 K c a l m o l^{- 1}$ and $Δ H$ of neutralisation is $- 13.7 K c a l m o l^{- 1}$ , then the heat of formation of $⊖ O H$ is

A

- 68 K c a l m o l^{- 1}

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B

- 54.3 K c a l m o l^{- 1}

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C

54.3 K c a l m o l^{- 1}

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D

- 71.7 K c a l m o l^{- 1}

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Solution

The correct option is B $- 54.3 K c a l m o l^{- 1}$
$H^{\oplus} + ⊖ O H ⇌ H_{2} O$
$Δ H = - 13.7 K c a l$
$Δ H^{⊖} = Δ_{f} H^{⊖} (H_{2} O) - [Δ_{f} H^{⊖} (H^{\oplus}) + Δ_{f} H (⊖ O H)]$
$= - 68 - (0 - 13.7) = - 54.3 K c a l m o l^{- 1}$
$[Δ_{f} H^{⊖} (H^{\oplus}) = 0 (c o n v e n t i o n)]$

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