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Heat of Formation

The enthalpy ...

Question

The enthalpy of neutralisation of a strong acid by a strong base is $- 57.32 k J {m o l}^{- 1}$ . The enthalpy of formation of water is $- 285.84 k J {m o l}^{- 1}$ . The enthalpy of formation of hydroxy ion is:

+ 228.52 k J {m o l}^{- 1}

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- 114.26 k J {m o l}^{- 1}

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- 228.52 k J {m o l}^{- 1}

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+ 114.2 k J {m o l}^{- 1}

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Solution

The correct option is C $- 228.52 k J {m o l}^{- 1}$
$Δ H_{r e a c t i o n} = \sum H e a t o f f o r m a t i o n o f p r o d u c t s - \sum H e a t o f f o r m a t i o n o f r e a c t a n t s$
$- 57.32 = - 285.84 - (0 + x)$
$x = - 228.52 k J {m o l}^{- 1}$

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