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: $2Zn + O_2 \rightarrow 2ZnO, \Delta G^o = -616 \, J$; $2Zn + S_2 \rightarrow 2ZnS, \Delta G^o = -293 \, J$; $S_2 + 2O_2 \rightarrow 2SO_2, \Delta G^o = -408 \, J$. The value of $\Delta G^o$ for the reaction $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$ is ....... $J$.

On combustion,carbon forms two oxides $CO$ and $CO_2$. The heat of formation of $CO_2$ gas is $-94.3 \, kcal$ and that of $CO$ is $-26.0 \, kcal$. The heat of combustion of carbon is.......$kcal$.

The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$. The $C-H$ bond enthalpy of each successive step in $CH_4$ $\rightarrow CH_3$ $\rightarrow CH_2$ $\rightarrow CH$ are $+15, +30$ and $+45 \ kJ \ mol^{-1}$ higher than the mean bond enthalpy of $C-H$ bonds,respectively. The bond enthalpy of the last $C-H$ unit is

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)

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