In the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}; \Delta H = 80 \ kJ$,$\Delta H$ is known as

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 heats of formation of $CO_2$ and $H_2O$ are $-97 \, kcal$ and $-68 \, kcal$ respectively. The heat of combustion of benzene is $-783 \, kcal$. What is the heat of formation of benzene in $kcal$?

Calculate the enthalpy change of the reaction for $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$, given the bond energies (in $\text{kJ mol}^{-1}$): $H-H = 436$, $Cl-Cl = 242$, $H-Cl = 431$.

The enthalpy of neutralisation of $NH_4OH$ with $HCl$ is $-51.46 \ kJ \ mol^{-1}$ and the enthalpy of neutralisation of $NaOH$ with $HCl$ is $-55.90 \ kJ \ mol^{-1}$. The enthalpy of ionisation of $NH_4OH$ is $...... \ kJ \ mol^{-1}$

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