The enthalpies of dissolution of $BaCl_2(s)$ and $BaCl_2 \cdot 2H_2O(s)$ are $-20.6 \ kJ \ mol^{-1}$ and $8.8 \ kJ \ mol^{-1}$ respectively. Calculate the enthalpy of hydration for the given reaction: $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$

The enthalpy changes for the following reactions are given:
$Cl_{2(g)} = 2Cl_{(g)}, 242.3 \, kJ \, mol^{-1}$; $I_{2(g)} = 2I_{(g)}, 151.0 \, kJ \, mol^{-1}$
$ICl_{(g)} = I_{(g)} + Cl_{(g)}, 211.3 \, kJ \, mol^{-1}$; $I_{2(s)} = I_{2(g)}, 62.76 \, kJ \, mol^{-1}$
Given that the standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is $...... \, kJ \, mol^{-1}$.

From the following,the heat of neutralization is maximum in:

The enthalpy of formation of ammonia is $-46.0 \ kJ \ mol^{-1}$. The enthalpy change for the reaction $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$ is ............... $kJ \ mol^{-1}$.

The bond dissociation energies of $X_2$,$Y_2$ and $XY$ are in the ratio of $1 : 0.5 : 1$. $\Delta H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$. The bond dissociation energy of $X_2$ will be $...... \ kJ \ mol^{-1}$.

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)