Question

The heat of neutralization is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt. Heat of neutralization of strong acid with strong base is $-13.7kcalmo{l}^{-1}$.

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(|Enthalpy change in kcal|)}& \text{(Neutralisation)}\\ \text{(I)}<13.7\text{kcal}& \text{(P)}\underset{\text{1 mol}}{HCl}+\underset{\text{1 mol}}{NaOH}\\ \text{(II)}=13.7\text{kcal}& \text{(Q)}\underset{\text{1 mol}}{HF}+\underset{\text{1 mol}}{NaOH}\\ \text{(III)}>13.7\text{kcal}& \text{(R)}\underset{\text{1 mol}}{HCl}+\underset{\text{1 mol}}{N{H}_{4}OH}\\ \text{(IV)}=27.4\text{kcal}& \text{(S)}\underset{\text{2 mol}}{NaOH}+\underset{\text{1 mol}}{H_{2}S{O}_4}\\ & \text{(T)}\underset{\text{1 mol}}{NaOH}+\underset{\text{1 mol}}{C{H}_{3}COOH}\end{array}$

Which of the following options has the correct combination considering List-I and List-II?

• A. $III\to S,T$, $IV\to Q$
• B. $III\to Q,S$, $IV\to S$
• C. $III\to P,S$, $IV\to R$
• D. $III\to R,S$, $IV\to S$

Answer 1

The correct option is B $III\to Q,S$, $IV\to S$

(P) $NaOH\to NaCl+H_{2}O$
Heat of neutralization for strong acid with strong base is $-13.7kcalmo{l}^{-1}$.
(Q) $|\Delta H_{nu}|>13.7kcal$, for $HF$, heat is evolved due to hydration of $F^{-}$.
(R) $N{H}_{4}OH$ is a weak base, so $|\Delta H_{nu}|<13.7kcal$ because some amount of heat will be used up in ionisation of $N{H}_{4}OH$.
(S) For 2 moles of $NaOH$,
$|\Delta H_{nu}|=2\times 13.7=27.4kcal$
(T) $C{H}_{3}COOH$ is a weak acid, so $|\Delta H_{nu}|<13.7kcal$ because some amount of heat will be used up in ionisation of $C{H}_{3}COOH$.