Consider the following data:
Heat of combustion of $H_{2(g)} = -241.8 \ kJ \ mol^{-1}$
Heat of combustion of $C_{(s)} = -393.5 \ kJ \ mol^{-1}$
Heat of combustion of $C_2H_5OH_{(l)} = -1234.7 \ kJ \ mol^{-1}$
The heat of formation of $C_2H_5OH_{(l)}$ is $(-)$ $...... \ kJ \ mol^{-1}$ (Nearest integer).

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.

Diborane is formed from the elements as shown in equation $(i)$:
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)} \dots (i)$
Given that:
$H_2O_{(l)} \longrightarrow H_2O_{(g)}, \quad \Delta H_1^{\circ} = 44 \, kJ$
$2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$B_2H_{6(g)} + 3 O_{2(g)} \longrightarrow B_2O_{3(s)} + 3 H_2O_{(g)}, \quad \Delta H_3^{\circ} = -2035 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}, \quad \Delta H_4^{\circ} = -286 \, kJ$
The $\Delta H^{\circ}$ for the reaction $(i)$ is $..... \, kJ$.

Which of the following is an exothermic reaction?

Calculate the standard enthalpy of formation of $CH_{3}OH_{(l)}$ from the following data:
$CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$

The average $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following equations correctly represents the bond dissociation of $CH_4$?