The heat of neutralisation of an acid by a base is highest when:

From the following data,the enthalpy of vaporization of liquid water in $KJ \, mol^{-1}$ will be:
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l); \Delta H = -285.77 \, KJ \, mol^{-1}$
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(g); \Delta H = -241.84 \, KJ \, mol^{-1}$

The enthalpy of solution of $BaCl_{2(s)}$ and $BaCl_2 \cdot 2H_2O_{(s)}$ are $-20.6 \, kJ \, mol^{-1}$ and $8.8 \, kJ \, mol^{-1}$ respectively. The enthalpy change for the reaction $BaCl_{2(s)} + 2H_2O_{(l)} \to BaCl_2 \cdot 2H_2O_{(s)}$ is $...... \, kJ \, mol^{-1}$.

Based on Hess's law calculations,what is the average bond energy of $S-O$ in $SO_3$ if $\Delta H_f^o$ of $SO_3$ is $-270 \ kJ \ mol^{-1}$. The bond energy of $O=O$ is $495 \ kJ \ mol^{-1}$ and the heat of sublimation for $S_{(s)}$ is $277 \ kJ \ mol^{-1}$? (Given: Bond energy of $S=O$ is not provided,assume the reaction $S_{(s)} + \frac{3}{2} O_2(g) \rightarrow SO_3(g)$). Note: The provided values in the prompt were inconsistent with standard chemical data; using standard values: $\Delta H_{sub}(S) = 277 \ kJ \ mol^{-1}$,$BE(O=O) = 495 \ kJ \ mol^{-1}$,$BE(S-O) = 330 \ kJ \ mol^{-1}$ is not the goal,we calculate based on the provided logic.

Bond enthalpies of $A_{2}$,$B_{2}$ and $AB$ are in the ratio $2:1:2$. If the enthalpy of formation of $AB$ is $-100 \ kJ \ mol^{-1}$,the bond enthalpy of $B_{2}$ is:

Enthalpy of formation of two compounds $X$ and $Y$ are $-84 \ kJ$ and $-156 \ kJ$ respectively. Which of the following statements is correct?