The bond energies of $H-H$ and $Cl-Cl$ are $430 \, kJ \, mol^{-1}$ and $242 \, kJ \, mol^{-1}$ respectively. $\Delta H_f$ for $HCl$ is $-91 \, kJ \, mol^{-1}$. The bond energy of $HCl$ will be ............. $kJ \, mol^{-1}$.

$C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$; $\Delta H = -3267.7 \ kJ \ mol^{-1}$. Given that the standard enthalpies of formation of $CO_{2(g)}$ and $H_2O_{(g)}$ are $-393.5 \ kJ \ mol^{-1}$ and $-285.85 \ kJ \ mol^{-1}$ respectively,calculate the standard enthalpy of formation of benzene $(C_6H_{6(l)})$.

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 $.3$)

If bond energy for $H_{2(g)}$,$Br_{2(g)}$ and $HBr_{(g)}$ is $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,then $\Delta H^o$ for the reaction,$H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$ is......$kJ$.

On the basis of the following thermochemical data: $(\Delta_fG^o H^{+}_{(aq)} = 0)$
$H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \,; \, \Delta H = 57.32 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \,; \, \Delta H = -286.20 \, kJ$
The value of enthalpy of formation of $OH^{-}$ ion at $25 \, ^oC$ is : .............. $kJ$

The compounds with negative heat of formation are known as: