Calculate the standard enthalpy of formation of $ICl_{(g)}$ based on the following reactions. The standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$ respectively.
$(i)$ $Cl_{2(g)} = 2Cl_{(g)}$,$\Delta H = 242.3 \text{ kJ mol}^{-1}$
$(ii)$ $I_{2(g)} = 2I_{(g)}$,$\Delta H = 151.0 \text{ kJ mol}^{-1}$
$(iii)$ $ICl_{(g)} = I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \text{ kJ mol}^{-1}$
$(iv)$ $I_{2(s)} = I_{2(g)}$,$\Delta H = 62.76 \text{ kJ mol}^{-1}$
Result in $\text{kJ mol}^{-1}$:

The $\Delta H^o_f$ for $CO_{2(g)}$,$CO_{(g)}$,and $H_2O_{(g)}$ are $-393.5$,$-110.5$,and $-241.8 \ kJ/mol$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is:

When $2 \ mol$ of $C_2H_{6(g)}$ is completely combusted,it releases $3129 \ kJ$ of heat. What is the enthalpy of formation of $C_2H_{6(g)}$? The $\Delta H_f$ values for $CO_{2(g)}$ and $H_2O_{(l)}$ are $-395 \ kJ \ mol^{-1}$ and $-286 \ kJ \ mol^{-1}$ respectively.

The average $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following equations correctly represents the bond dissociation of $CH_4$?

Based on Hess's law calculations,what is the average bond energy of $S-O$ in $SO_3$ if $\Delta H_f^o$ of $SO_3$ is $-270 \ kJ \ mol^{-1}$. The bond energy of $O=O$ is $495 \ kJ \ mol^{-1}$ and the heat of sublimation for $S_{(s)}$ is $277 \ kJ \ mol^{-1}$? (Given: Bond energy of $S=O$ is not provided,assume the reaction $S_{(s)} + \frac{3}{2} O_2(g) \rightarrow SO_3(g)$). Note: The provided values in the prompt were inconsistent with standard chemical data; using standard values: $\Delta H_{sub}(S) = 277 \ kJ \ mol^{-1}$,$BE(O=O) = 495 \ kJ \ mol^{-1}$,$BE(S-O) = 330 \ kJ \ mol^{-1}$ is not the goal,we calculate based on the provided logic.

If $8.84 \ kJ$ heat is liberated for the formation of $3 \ g$ of ethane,calculate its $\Delta_{f} H^{\circ}$.