Calculate the enthalpy for the following reaction using the given bond energies $(kJ/mol)$:
$(C-H = 414; O-H = 463; H-Cl = 431; C-Cl = 326; C-O = 335)$
$CH_3OH_{(g)} + HCl_{(g)} \rightarrow CH_3Cl_{(g)} + H_2O_{(g)}$

Consider the reactions:
$C_{(s)} + 2H_{2(g)} \to CH_{4(g)}, \Delta H = -x \ kcal$
$C_{(g)} + 4H_{(g)} \to CH_{4(g)}, \Delta H = -x_1 \ kcal$
$CH_{4(g)} \to CH_{3(g)} + H_{(g)}, \Delta H = +y \ kcal$
The bond energy of $C-H$ bond is:

$H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}, \Delta H = -44 \ kcal$
$2Na_{(s)} + 2HCl_{(g)} \to 2NaCl_{(s)} + H_{2(g)}, \Delta H = -152 \ kcal$
For the reaction $Na_{(s)} + \frac{1}{2}Cl_{2(g)} \to NaCl_{(s)}, \Delta H = \dots \ kcal$

If heat of neutralization is $-13.7 \, KCal$ at $25 \, ^oC$ and $\Delta H_f^o (H_2O) = -68 \, KCal$,then the standard enthalpy of formation of $OH^{-}$ will be.....$KCal$. (in $.3$)

Combustion of $1$ $mol$ of benzene is expressed as:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of combustion of $2$ $mol$ of benzene is $-x$ $kJ$.
$x = . . . . . . . . . .$
$(1)$ Standard enthalpy of formation of $1$ $mol$ of $C_6H_{6(l)}$ is $48.5$ $kJ \ mol^{-1}$.
$(2)$ Standard enthalpy of formation of $1$ $mol$ of $CO_{2(g)}$ is $-393.5$ $kJ \ mol^{-1}$.
$(3)$ Standard enthalpy of formation of $1$ $mol$ of $H_2O_{(l)}$ is $-286$ $kJ \ mol^{-1}$.

Calculate the heat of formation of $PCl_{5(s)}$ from the following data:
$2P_{(s)} + 3Cl_{2(g)} \to 2PCl_{3(l)}; \Delta H_1 = -151.8 \ kcal$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)}; \Delta H_2 = -32.8 \ kcal$
Determine the enthalpy of formation for $PCl_{5(s)}$ in $kcal$.