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

The standard enthalpy of formation of $CO_{(g)}$,$CO_{2(g)}$,$N_2O_{(g)}$ and $N_2O_{4(g)}$ are respectively $-110$,$-393$,$81$ and $-10 \ kJ \ mol^{-1}$. The enthalpy change (in $kJ$) of the following reaction is $N_2O_{4(g)} + 3CO_{(g)} \longrightarrow N_2O_{(g)} + 3CO_{2(g)}$

Calculate the standard enthalpy of formation of $CH_{3}OH_{(l)}$ from the following data:
$CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$

Given the following thermochemical equations:
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)

When $0.5 \ g$ of sulphur is burnt to $SO_2$,$4.6 \ kJ$ of heat is liberated. What is the enthalpy of formation of sulphur dioxide in $kJ \ mol^{-1}$?

The standard enthalpies of formation of $Al_{2}O_{3}$ and $CaO$ are $-1675 \ kJ \ mol^{-1}$ and $-635 \ kJ \ mol^{-1}$ respectively.
For the reaction $3 CaO + 2 Al \rightarrow 3 Ca + Al_{2}O_{3}$,the standard reaction enthalpy $\Delta_{r}H^{\circ}$ is .......... $kJ$.
(Round off to the Nearest Integer).