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)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 6.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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, $Δ H_{1} = Δ H - Δ H_{2} = - 51.1 k J / m o l + 57.1 k J m o l^{- 1} = 6.0 k J / m o l$