The average $S-F$ bond energy in $kJ \ mol^{-1}$ of $SF_{6}$ is $......$ . (Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of $SF_{6(g)}$,$S_{(g)}$ and $F_{(g)}$ are $-1100$,$275$ and $80 \ kJ \ mol^{-1}$ respectively.]

From the following data:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is:

Using the data provided,calculate the bond energy $(kJ \ mol^{-1})$ of a $C \equiv C$ bond in $C_{2}H_{2}$. (Take the bond energy of a $C-H$ bond as $350 \ kJ \ mol^{-1}$)
$2C_{(s)} + H_{2(g)} \longrightarrow C_{2}H_{2(g)} \quad \Delta H = 225 \ kJ \ mol^{-1}$
$2C_{(s)} \longrightarrow 2C_{(g)} \quad \Delta H = 1410 \ kJ \ mol^{-1}$
$H_{2(g)} \longrightarrow 2H_{(g)} \quad \Delta H = 330 \ kJ \ mol^{-1}$

$PbO$ exists in two crystalline forms: yellow and red. The standard enthalpies of formation for these two forms are $-217.3 \ kJ/mol$ and $-219.0 \ kJ/mol$,respectively. Calculate the enthalpy of transition for the process: $PbO \text{ (yellow)} \rightarrow PbO \text{ (red)}$ in $kJ/mol$.

In order to decompose $9 \ g$ of water,$142.5 \ kJ$ of heat is required. Hence,the enthalpy of formation of water is ...... $kJ/mol$.

The enthalpy change at $298 \ K$ for the decomposition of water is given in the following two steps:
Step $I$: $H_2O_{(g)} \to H_{(g)} + OH_{(g)}$; $\Delta H = 498 \ kJ \ mol^{-1}$
Step $II$: $OH_{(g)} \to H_{(g)} + O_{(g)}$; $\Delta H = 428 \ kJ \ mol^{-1}$
The average bond enthalpy of the $O-H$ bond is $.... \ kJ \ mol^{-1}$