The value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$. Then,the formation of one mole of water in the gaseous state from $H_{(g)}$ and $O_{(g)}$ atoms is accompanied by:

The $\Delta H_f^o$ 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 (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is

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

Calculate the standard enthalpy of formation of $CH_{3}OH_{(l)}$ from the following data:
$CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$

The reaction $CH_{4(g)} + Cl_{2(g)} \to CH_3Cl_{(g)} + HCl_{(g)}$ has $\Delta H = -25 \, kcal$. Given bond energies $BE(C-H) = 84 \, kcal$,$BE(H-Cl) = 103 \, kcal$,$BE(C-Cl) = x$,and $BE(Cl-Cl) = y$. If $\frac{x}{y} = \frac{9}{5}$,then find the value of $y$. (in $, kcal$)

For the reaction,$H_{2(g)} + I_{2(g)} \to 2HI_{(g)}$,$\Delta H = -12.40 \ kcal$. According to this,the heat of formation of $HI$ will be......$kcal$.