The standard heats of formation for $CH_4$, $C_2H_4$, and $C_3H_8$ are $-17.9$, $12.5$, and $-24.8 \ kcal/mol$ respectively. The heat of reaction $(\Delta H)$ for the reaction $CH_4 + C_2H_4 \rightarrow C_3H_8$ in $kcal$ is:

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.)

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 $kJ$.

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction: $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$? Given that the bond energy of $H_2$,$Br_2$,and $HBr$ is $435 \ kJ \ mol^{-1}$,$192 \ kJ \ mol^{-1}$,and $368 \ kJ \ mol^{-1}$ respectively.

Enthalpy of formation of $CO_{(g)}$ and $CO_{2(g)}$ are $-110 \ kJ \ mol^{-1}$ and $-393 \ kJ \ mol^{-1}$ respectively. The enthalpy of combustion of $CO$ (in $kJ \ mol^{-1}$) is:

Enthalpy change for a reaction does not depend upon