For the following reaction,
$C (diamond) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -97.6 \ kcal$
$C (graphite) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -94.3 \ kcal$
The heat change for the conversion of $1 \ g$ of $C (diamond) \rightarrow C (graphite)$ is (in $kcal$)

Energy required to dissociate $16 \ g$ of $O_{2(g)}$ into free atoms is $x \ kJ$. The value of bond enthalpy of $O=O$ bond is

Calculate the heat of formation of $Ca(OH)_{2(s)}$ at $1.8\,^{\circ}C$ from the following data:
$CaO_{(s)} + H_2O_{(l)} \to Ca(OH)_{2(s)}$; $\Delta H_{1.8\,^{\circ}C} = -15.26\,K\,cal$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H_{1.8\,^{\circ}C} = -68.37\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} \to CaO_{(s)}$; $\Delta H_{1.8\,^{\circ}C} = -151.80\,K\,cal$

Given $:$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$

The enthalpy changes at $298 \ K$ in successive breaking of $O-H$ bonds of $H_2O$ are:
$H_2O_{(g)} \to H_{(g)} + OH_{(g)}, \Delta H = 498 \ kJ \ mol^{-1}$
$OH_{(g)} \to H_{(g)} + O_{(g)}, \Delta H = 428 \ kJ \ mol^{-1}$
The bond enthalpy of the $O-H$ bond is ..... $kJ \ mol^{-1}$.

For the reaction $Cu_{(g)}^{+} + I_{(g)}^{-} \to CuI_{(s)}$,the value of $\Delta H^o$ is $-446 \ kJ \ mol^{-1}$. If the ionization energy of $Cu_{(g)}$ is $745 \ kJ \ mol^{-1}$ and the electron affinity of $I_{(g)}$ is $-295 \ kJ \ mol^{-1}$,calculate the value of $\Delta H^o$ for the formation of $CuI_{(s)}$ from $Cu_{(g)}$ and $I_{(g)}$ in $kJ \ mol^{-1}$.