The enthalpy change for the reaction $2CO_{(g)} + O_{2(g)} \rightarrow 2CO_{2(g)}$ is known as:

The correct order of $\Delta_{f} H^{\circ}$ values of Diamond $(I)$, Graphite $(II)$, and Fullerene $(III)$ is:

Given that:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}, \Delta H = -394 \ kJ$
$2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}, \Delta H = -568 \ kJ$
$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}, \Delta H = -892 \ kJ$
Calculate the heat of formation of $CH_{4(g)}$ in $kJ$.

The $\Delta H_f$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. The standard enthalpy change for the reaction is:
$CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$

If for the reaction $CCl_{4(g)} \rightarrow C_{(g)} + 4Cl_{(g)}$ the following data is given:
$\Delta_{vap} H^{\theta} (CCl_{4(l)}) = 30 \ kJ \ mol^{-1}$
$\Delta_{f} H^{\theta} (CCl_{4(l)}) = -136.0 \ kJ \ mol^{-1}$
$\Delta_{a} H^{\theta} (C_{(s)}) = 714.0 \ kJ \ mol^{-1}$
$\Delta_{a} H^{\theta} (Cl_{2(g)}) = 242.0 \ kJ \ mol^{-1}$
Calculate the mean bond enthalpy of $C-Cl$ in $CCl_{4(g)}$.

Which of the following is an endothermic reaction?