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

If $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = r$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = s$,then the heat of formation of $CO$ is

At $298 \ K$ and constant pressure, the heat of formation of $C_2H_{2(g)}$ and $C_6H_{6(g)}$ is $230 \ kJ \ mol^{-1}$ and $85 \ kJ \ mol^{-1}$ respectively. What will be the change in heat for the reaction?
$3 \ C_2H_{2(g)} \rightarrow C_6H_{6(g)}$

From the following bond energies:

$H-H$ bond energy	$431.37 \text{ kJ mol}^{-1}$
$C=C$ bond energy	$606.10 \text{ kJ mol}^{-1}$
$C-C$ bond energy	$336.49 \text{ kJ mol}^{-1}$
$C-H$ bond energy	$410.50 \text{ kJ mol}^{-1}$

Enthalpy for the reaction $CH_2=CH_2 + H-H \to CH_3-CH_3$ will be .............. $\text{kJ mol}^{-1}$

Enthalpies of formation of $CCl_{4(g)}$,$H_2O_{(g)}$,$CO_{2(g)}$ and $HCl_{(g)}$ are $-105$,$-242$,$-394$ and $-92 \ kJ \ mol^{-1}$ respectively. The magnitude of enthalpy of the reaction given below is $...... \ kJ \ mol^{-1}$ (nearest integer): $CCl_{4(g)} + 2H_2O_{(g)} \rightarrow CO_{2(g)} + 4HCl_{(g)}$

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is: