If $C + O_2 \to CO_2 + 94.2 \ kcal$,$H_2 + \frac{1}{2} O_2 \to H_2O + 68.3 \ kcal$,and $CH_4 + 2O_2 \to CO_2 + 2H_2O + 210.8 \ kcal$,then the possible heat of formation of methane will be $...... \ kcal$.

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

With the help of the following data,find out the change in heat content for the reaction:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

The following two reactions are known:
$Fe_2O_{3(s)} + 3CO_{(g)} \rightarrow 2Fe_{(s)} + 3CO_{2(g)}; \Delta H = -26.8 \ kJ$
$FeO_{(s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2(g)}; \Delta H = -16.5 \ kJ$
The value of $\Delta H$ for the following reaction:
$Fe_2O_{3(s)} + CO_{(g)} \rightarrow 2FeO_{(s)} + CO_{2(g)}$ is ............. $kJ$.

Calculate the bond enthalpy of the $H-Cl$ bond from the following reaction:
$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$,$\Delta_{r} H^{\circ} = -185 \ kJ \ mol^{-1}$
(Given bond enthalpies of $H-H$ and $Cl-Cl$ bonds are $435.0 \ kJ \ mol^{-1}$ and $244 \ kJ \ mol^{-1}$ respectively.)

Which of the following is an endothermic reaction?