Enthalpy of neutralization of HCl by NaOH is $- 57.1 k J / m o l$ and by $N H_{4} O H$ is $- 51.1 k J / m o l$ . Calculate the enthalpy of dissociation of $N H_{4} O H$ (in kJ)

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answer is 6.

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Detailed Solution

Given that $H^{+} (a q) + N H_{4} O H (a q) \to N H_{4} (a q) + H_{2} O (l) Δ H = - 51.1 k J / m o l e$
We may consider neutralization in two steps.
i) $\begin{matrix} I o n i z a t i o n & N H_{4} O H (a q) \to N {H_{4}}^{+} (a q) + O H^{-} (a q) & Δ H_{1} = \end{matrix} ?$
ii) $\begin{matrix} N e u t r a l i z a t i o n & H^{+} (a q) + O H^{-} (a q) \Rightarrow H_{2} O (l) & Δ H_{2} = - 57.1 k J / m o l e \end{matrix}$
Thus, $Δ H = Δ H_{1} + Δ H_{2}$
Therefore, $Δ H_{1} = Δ H - Δ H_{2} = - 51.1 k J / m o l + 57.1 k J m o l^{- 1} = 6.0 k J / m o l$