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

Find the value $\Delta H_f^o[NH_{3(g)}]$ if $\Delta H_r$ for the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is $-\,183.6 \ kJ/mol$ of $N_{2(g)}$.

For the dissolution of $BaCl_2(s)$ and $BaCl_2 \cdot 2H_2O(s)$,the values of $\Delta H_{sol}$ are $-a \ kJ$ and $b \ kJ$ respectively. What is the value of $\Delta H_{hydration}$ for $BaCl_2(s)$?

Standard molar enthalpy of formation,$\Delta _{f}H^{o}$ is just a special case of enthalpy of reaction,$\Delta _{r}H^{o}$. Is the $\Delta _{r}H^{o}$ for the following reaction same as $\Delta _{f}H^{o}$? Give reason for your answer. $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$; $\Delta _{r}H^{o} = -178.3 \ kJ \ mol^{-1}$

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

The $\Delta H_f^{\circ}$ values (in $kJ \ mol^{-1}$) for graphite,diamond,and $C_{60}$ are respectively: