Question

100 ml 0.5 N $H_{2}S{O}_4$ (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 of the following statements is/are correct?
Given: $HCl+NaOH\to NaCl+H_{2}O+57kJ$
$C{H}_{3}COOH+N{H}_{4}OH\to C{H}_{3}COON{H}_4+H_{2}O+48.1kJ$

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

Answer 1

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

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

The reaction of $H_{2}S{O}_4$ and $N{H}_{4}OH$ is
$H_{2}S{O}_4+2N{H}_{4}OH\to (N{H}_4{)}_{2}S{O}_4+2H_{2}O$
Again,
$100mlof0.5N{H}_{2}S{O}_4=50meq=25mmol=50mmolof{H}^{+}ion$
$200mlof0.2MN{H}_{4}OH=40meq=40mmol=40mmolofO{H}^{-}ion$
Heat evolved during the reaction = $C\Delta T=1.5\times 1.4=2.10kJ$
a) So enthalpy of neutralisation of $HCl$ v/s $N{H}_{4}OH$ is
$=\frac{2.1}{40}\times 1000=-52.5kJ/mol$
b) Enthalpy of dissociation (ionization) of $N{H}_{4}OH$ is $(57-52.5)=4.5kJ/mol$
c) Enthalpy of dissociation of $C{H}_{3}COOH$ is $57-48.1-4.5=4.4kJ/mol$
d) For the reaction $2H_{2}O\left(l\right)\to 2H^{+}\left(aq\right)+2O{H}^{-}\left(aq\right)$, enthalpy change will be $\Delta H=-57\times 2=-114kJ$