Compounds with high heat of formation are less stable because

Enthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:

If $\Delta H_f (H_2O) = X$,then the heat of neutralization of $CH_3COOH$ and $NaOH$ will be:

Calculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$

The molar enthalpies of combustion of $C_2H_{2(g)},$ $C$ (graphite) and $H_{2(g)}$ are $-1300,$ $-394$ and $-286 \ kJ \ mol^{-1},$ respectively. The standard enthalpy of formation of $C_2H_{2(g)}$ is.......$kJ \ mol^{-1}$

The $\Delta H_f$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. The standard enthalpy change for the reaction is:
$CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$