The standard enthalpy of formation of $CO_{2(g)}$,$CaO_{(s)}$ and $CaCO_{3(s)}$ are $-393, -634, -1210 \ kJ \ mol^{-1}$ respectively. If all the substances are in standard state,the standard enthalpy of decomposition of calcium carbonate to $CaO_{(s)}$ and $CO_{2(g)}$ (in $kJ \ mol^{-1}$) is

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

Given the following thermochemical equations:
$(1) \ H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} ; \Delta H_1 = 100 \ kJ$
$(2) \ CO_{(g)} + \frac{1}{2}O_{2_{(g)}} \to CO_{2_{(g)}} ; \Delta H_2 = -300 \ kJ$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2_{(g)}} \to H_2O_{(g)} ; \Delta H_3 = -250 \ kJ$
Calculate the value of $x$ for the reaction:
$(4) \ C_{(s)} + O_{2_{(g)}} \to CO_{2_{(g)}} ; \Delta H_4 = -x \ kJ$

The enthalpy of atomization of $PH_3(g)$ is $228 \, kcal \, mol^{-1}$ and that of $P_2H_4(g)$ is $355 \, kcal \, mol^{-1}$. The $P-P$ bond energy (in $kcal \, mol^{-1}$) is:

Calculate the enthalpy change for the reaction
$H_2 + F_2 \longrightarrow 2HF$
given that
Bond energy of $H-H$ bond $= 434 \ kJ/mol$
Bond energy of $F-F$ bond $= 158 \ kJ/mol$
Bond energy of $H-F$ bond $= 565 \ kJ/mol$
Result in $kJ$.

The bond dissociation enthalpies of $H_2$, $Cl_2$, and $HCl$ are $434 \text{ kJ mol}^{-1}$, $242 \text{ kJ mol}^{-1}$, and $431 \text{ kJ mol}^{-1}$ respectively. Calculate the enthalpy of formation of $HCl$.