Question

Heat liberated in the neutralization of $500mL$ of $1N$ $HCl$ and $500mL$ of $1N$ $N{H_4}OH$ is $1.36Kcal$ . The heat of ionization of $N{H_4}OH$ is
$1){\text{ }}10.95K.Cals$
$2){\text{ }} - 12.34K.Cals$
$3){\text{ }} - 10.98K.Cals$
$4){\text{ }}12.34{\text{ }}K.Cals$

Answer 1

The heat of neutralization is the enthalpy change during the reaction of one equivalent of both acid and base to produce salt and water.

Complete step by step answer:
The given reaction is a neutralization reaction of an acid and a base. The acid present here is hydrochloric acid and the base used is ammonium hydroxide. Equal volume and equal strength of both acid and base are used for the neutralization reaction.
The reaction of strong acid $HCl$ and strong base $NaOH$ produces a salt sodium chloride $NaCl$ and water. The corresponding reaction is.
$NaOH + HCl \to NaCl + {H_2}O$ ,
The enthalpy change in this neutralization reaction is$\Delta H = - 13.7kcal/eq$. The heat of neutralization is also referred as the heat of formation of water.
Thus, for the reaction, ${H^ + } + O{H^ - } \to {H_{2}}O$ , $\Delta H = - 13.7kcal$.
Or ${H_{2}}O \to {H^ + } + O{H^ - }$, $\Delta H = 13.7kcal$.....eq(A)
For the given reaction the heat of neutralization of $500mL$ of $1N$ $HCl$ and $500mL$ of $1N$ $N{H_4}OH$ is$- 1.36Kcal$. The normality of a solution is the ratio of the gram equivalent of the solute and the volume of the solution in liters.
Thus normality, $N = \dfrac{{gram{\text{ }}equivalent{\text{ }}of{\text{ }}solute}}{{volume{\text{ }}of{\text{ }}the{\text{ }}solvent{\text{ }}in{\text{ }}litres}}$
So, g equiv. of $HCl$ = g equiv. of $N{H_4}OH$​= $1 \times \dfrac{{500}}{{1000}} = 0.5g$
The amount of heat liberated using $0.5\;g{\text{ }}equiv.$of $HCl$ and $N{H_4}OH$=$- 1.36kcal$ . So the heat liberated using $1\;g{\text{ }}equiv.$ of $HCl$ and $N{H_4}OH$= $- \dfrac{{1.36}}{{0.5}} = - 2.72kcal$. The corresponding equation is written as ${H^ + } + N{H_4}OH \to N{H_4}{^ + } + {H_2}O$……eq(B)
The heat of formation of water for the reaction is $\Delta H = - 2.72kcal.$
Adding eq(A)&(B),
${H_2}O + {\text{ }}{H^ + } + N{H_4}OH \to {H^ + } + O{H^ - } + N{H_4}^{ - } + {H_2}O$
Cancelling ${H^ + }$ and ${H_2}O$ from both sides,
$N{H_4}OH \to N{H_{4}}^ + + O{H^ - }$
Therefore, $\Delta H = 13.7 + ( - 2.72) = 10.98kcal$
Thus the heat of ionization of $N{H_4}OH$ is $10.98K.Cals$ , i.e. option A is the correct answer.

Note: The heat of neutralization is always exothermic and carries a negative sign. The heat of neutralization is a maximum for neutralization of strong acid and strong base.