Standard enthalpy of formation of water is $-286 \ kJ \ mol^{-1}$. When $1800 \ mg$ of water is formed from its constituent elements in their standard states,the amount of energy liberated is: (in $kJ$)

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

Calculate the enthalpy for the following reaction using the given bond energies $(kJ/mol)$:
$(C-H = 414; O-H = 463; H-Cl = 431; C-Cl = 326; C-O = 335)$
$CH_3OH_{(g)} + HCl_{(g)} \rightarrow CH_3Cl_{(g)} + H_2O_{(g)}$

Hess's law is based on

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 $...... \, kJ \, mol^{-1}$.

Determine the enthalpy of formation for $H_2O_2(\ell)$,using the listed enthalpies of reaction:
$N_2H_{4(\ell)} + 2H_2O_{2(\ell)} \to N_{2(g)} + 4H_2O_{(\ell)}; \Delta _r H_1^o = -818 \, kJ/mol$
$N_2H_{4(\ell)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(\ell)}; \Delta _r H_2^o = -622 \, kJ/mol$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(\ell)}; \Delta _r H_3^o = -285 \, kJ/mol$
Calculate the value in $kJ/mol$.