While performing a thermodynamics experiment,a student made the following observations:
$HCl + NaOH \rightarrow NaCl + H_{2}O$ $\Delta H = -57.3 \ kJ \ mol^{-1}$
$CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$ $\Delta H = -55.3 \ kJ \ mol^{-1}$
The enthalpy of ionization of $CH_{3}COOH$ as calculated by the student is $kJ \ mol^{-1}$. (nearest integer)

One mole of ethanol $(l)$ was completely burnt in oxygen to form $CO_{2(g)}$ and $H_2O_{(l)}$. What is the $\Delta_r H^{\ominus}$ (in $kJ \ mol^{-1}$) for this reaction?
(The standard enthalpy of formation $(\Delta_f H^{\ominus})$ of $C_2H_5OH_{(l)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ is respectively $-277, -393$ and $-286 \ kJ \ mol^{-1}$.)

Consider the following cases of standard enthalpy of reaction $\Delta H_{r}^{\circ}$ in $kJ \ mol^{-1}$:
$C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \rightarrow 2 CO_{2(g)} + 3 H_{2}O(\ell)$,$\Delta H_{1}^{\circ} = -1550$
$C(\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_{2}^{\circ} = -393.5$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O(\ell)$,$\Delta H_{3}^{\circ} = -286$
The magnitude of $\Delta H_{f, C_{2}H_{6(g)}}^{\circ}$ is $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

If $S + O_2 \to SO_2; (\Delta H = -298.2 \ kJ)$,$SO_2 + \frac{1}{2} O_2 \to SO_3; (\Delta H = -98.2 \ kJ)$,$SO_3 + H_2O \to H_2SO_4; (\Delta H = -130.2 \ kJ)$,$H_2 + \frac{1}{2} O_2 \to H_2O; (\Delta H = -287.3 \ kJ)$,then the enthalpy of formation of $H_2SO_4$ at $298 \ K$ will be......$kJ$.

Enthalpy of formation of $CO_{(g)}$,$CO_{2(g)}$,$N_2O_{(g)}$ and $N_2O_{4(g)}$ are $-110$,$-393$,$81$,and $9.7 \ kJ \ mol^{-1}$ respectively. Calculate $\Delta_r H$ for the following reaction: $N_2O_{4(g)} + 3 CO_{(g)} \longrightarrow N_2O_{(g)} + 3 CO_{2(g)}$

The value of enthalpy of neutralization is highest for the reaction between: