$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

Calculate the heat of formation of $PCl_{5(s)}$ from the following data:
$2P_{(s)} + 3Cl_{2(g)} \to 2PCl_{3(l)}; \Delta H_1 = -151.8 \ kcal$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)}; \Delta H_2 = -32.8 \ kcal$
Determine the enthalpy of formation for $PCl_{5(s)}$ in $kcal$.

If the bond energies of $H-H$,$Br-Br$ and $H-Br$ are $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,then $\Delta H^{\circ}$ for the reaction: $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$ is:

For which of the following reactions does the heat of reaction represent the bond energy of $HCl$?

From the following data:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is:

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 $kJ$)