Enthalpy change for the reaction,$4H_{(g)} \rightarrow 2H_{2_{(g)}}$ is $-869.6 \ kJ$. The dissociation energy of $H-H$ bond is $............ \ kJ$.

Given
$(i) \, 2Fe_2O_{3(s)} \to 4Fe_{(s)} + 3O_{2(g)}$
$\Delta _rG^o = + 1487.0 \, kJ \, mol^{-1}$
$(ii) \, 2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}$
$\Delta _rG^o = - 514.4 \, kJ \, mol^{-1}$
Free energy change,$\Delta _rG^o$ for the reaction
$2Fe_2O_{3(s)} + 6CO_{(g)} \to 4Fe_{(s)} + 6CO_{2(g)}$ will be ..... $kJ \, mol^{-1}$

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

If the enthalpy of formation of $N_2O$ is $82 \, kJ \, mol^{-1}$,calculate the resonance energy of $N_2O$ in $kJ \, mol^{-1}$.
$N \equiv N \, (946 \, kJ \, mol^{-1}); \, N = N \, (418 \, kJ \, mol^{-1})$
$O = O \, (498 \, kJ \, mol^{-1}); \, N = O \, (607 \, kJ \, mol^{-1})$

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 $-286$ $kJ \ mol^{-1}$.

If standard enthalpy of formation $(\Delta_{f} H^{\circ})$ of $CO_2, H_2 O$ and $CH_4$ are $-393, -286$ and $-74.0 \ kJ \ mol^{-1}$ respectively,the standard enthalpy of combustion of methane in $kJ \ mol^{-1}$ is