Given the thermochemical reactions: $C(\text{graphite}) + \frac{1}{2}O_2 \rightarrow CO; \Delta H = -110.5 \, kJ$ and $CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H = -283.2 \, kJ$. Calculate the heat of reaction for $C(\text{graphite}) + O_2 \rightarrow CO_2$ in $kJ$.

Observe the following reaction:
$2 A_{2(g)} + B_{2(g)} \xrightarrow{T(K)} 2 A_2 B_{(g)} + 600 \ kJ$
The standard enthalpy of formation $(\Delta_f H^{\circ})$ of $A_2 B_{(g)}$ is:

The heats of combustion of rhombic and monoclinic sulphur are respectively $70960 \ cal$ and $71030 \ cal$. What will be the heat of conversion of rhombic sulphur to monoclinic? $(cal)$

The value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$. Then,the formation of one mole of water in the gaseous state from $H_{(g)}$ and $O_{(g)}$ atoms is accompanied by:

Calculate the enthalpy change for the process $CCl_{4(g)} \to C_{(g)} + 4Cl_{(g)}$ and calculate the bond enthalpy of the $C-Cl$ bond in $CCl_{4(g)}$.
$\Delta_{vap} H^{\theta}(CCl_{4}) = 30.5 \, kJ \, mol^{-1}$
$\Delta_{f} H^{\theta}(CCl_{4}) = -135.5 \, kJ \, mol^{-1}$
$\Delta_{a} H^{\theta}(C) = 715.0 \, kJ \, mol^{-1}$ (where $\Delta_{a} H^{\theta}$ is enthalpy of atomisation)
$\Delta_{a} H^{\theta}(Cl_{2}) = 242 \, kJ \, mol^{-1}$

The enthalpies of combustion of rhombic and monoclinic sulfur are $70,960 \ cal$ and $71,030 \ cal$ respectively. The enthalpy of transition of rhombic sulfur to monoclinic sulfur is ...... $cal$.