Question

$100$ ml $0.25$ $M{H}_{2}S{O}_4$ (strong acid) is neutralised with $200$ ml $0.2$ M $N{H}_4$ OH in a constant pressure calorimeter which results in temperature rise of ${1.4}^{o}C.$ If heat capacity of calorimeter content is $1.5kJ{/}^{o}C.$ Which statement is/are correct?
$HCl+NaOH\to NaCl+H_{2}O+57$ KJ$C{H}_{3}COOH+N{H}_{4}OH\to C{H}_{3}COON{H}_4+H_{2}O+48.1$ KJ

• A. Enthalpy of neutralisation of $HCl$ v/s $N{H}_{4}OH$ is $-52.5$ KJ/mol
• B. Enthalpy of dissociation (ionization) of $N{H}_{4}OH$ is $4.5$ KJ/mol
• C. Enthalpy of dissociationof $C{H}_{3}COOH$ is $4.6$ KJ/mol
• D. $\Delta H$ for $2H_{2}O\left(l\right)\to 2H^{+}(aq.)+2O{H}^{-}(aq.)$ is $114$ KH

Answer 1

Answer: (A), (B), (D)

The correct options are
A Enthalpy of neutralisation of $HCl$ v/s $N{H}_{4}OH$ is $-52.5$ KJ/mol
B Enthalpy of dissociation (ionization) of $N{H}_{4}OH$ is $4.5$ KJ/mol
D $\Delta H$ for $2H_{2}O\left(l\right)\to 2H^{+}(aq.)+2O{H}^{-}(aq.)$ is $114$ KH

As from given data it can be seen that $200$ ml of 0.2M $N{H}_{4}OH$ provides heat of $2.1$ kJ so 1lit of 1M $N{H}_{4}OH$ will provide $2.1\times 5\times 5=52.5kJ/mole$.
Enthalpy of neutralisation of $HCl$ v/s $N{H}_{4}OH$ is $-52.5kJ/mol$.
Enthalpy of dissociation (ionization) of $N{H}_{4}OH$ is $57-52.5=4.5kJ/mol$.
$\Delta H$ for $2H_{2}O\left(l\right)\to 2H^{+}(aq.)+2O{H}^{-}(aq.)$i s $114kJ$ as for one mole of water dissociation energy is $57kJ$.