Heats of combustion $(\Delta H^o)$ for $C_{(s)}$,$H_{2(g)}$ and $CH_{4(g)}$ are $-94$,$-68$ and $-213 \ kcal/mol$ respectively. The value of $\Delta H^o$ for the reaction,$C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$ is $..... \ kcal$.

Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ Given that: $\Delta_{f} H^{\circ}(CH_4) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$

Based on the following thermochemical reactions:
$H_2O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = x \ kJ$
The value of $x$ will be:

At $298 \ K$, the enthalpy change (in $kJ$) for the reaction given below is: $CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$
Given:
$1) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H^{\ominus} = -286 \ kJ$
$2) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H^{\ominus} = -394 \ kJ$
$3) \ CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} ; \Delta H^{\ominus} = -890 \ kJ$

If the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:

$S_{(g)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)} + 2x \ kcal$
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)} + y \ kcal$
The heat of formation of $SO_{2(g)}$ is given by :