Consider the following reaction:
$C_{(s)} + O_{2(g)} \to CO_{2(g)} + x \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)} + y \ kJ$
The heat of formation of $CO_{(g)}$ is:

Calculate the enthalpy change for the process $CCl_{4(g)} \to C_{(g)} + 4Cl_{(g)}$ and calculate the bond enthalpy of the $C-Cl$ bond in $CCl_{4(g)}$.
$\Delta_{vap} H^{\theta}(CCl_{4}) = 30.5 \, kJ \, mol^{-1}$
$\Delta_{f} H^{\theta}(CCl_{4}) = -135.5 \, kJ \, mol^{-1}$
$\Delta_{a} H^{\theta}(C) = 715.0 \, kJ \, mol^{-1}$ (where $\Delta_{a} H^{\theta}$ is enthalpy of atomisation)
$\Delta_{a} H^{\theta}(Cl_{2}) = 242 \, kJ \, mol^{-1}$

The enthalpies of formation of $N_2O$ and $NO$ are $28$ and $90 \ kJ \ mol^{-1}$ respectively. The enthalpy of the reaction,$2N_2O_{(g)} + O_{2(g)} \to 4NO_{(g)}$ is equal to.....$kJ$

The equation $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 \to HCl$ $(\Delta H_{298} = -22.060 \ kcal)$ means:

Given the following thermochemical equations:
$(i) \ Zn + \frac{1}{2}O_2 \rightarrow ZnO + 84000 \ cal$
$(ii) \ Hg + \frac{1}{2}O_2 \rightarrow HgO + 21700 \ cal$
Calculate the heat of reaction $(\Delta H)$ for the reaction: $Zn + HgO \rightarrow ZnO + Hg$. (in $cal$)

$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = -94 \ kcal$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -67.7 \ kcal$
Then the heat of formation of $CO_{(g)}$ is $..... \ kcal$