$$\eqalign{ & 4H\left( g \right) \to 2\,{H_2}\left( g \right),\Delta H = - 869.6\,kJ \cr & 2\,{H_2}\left( g \right) \to 4H\left( g \right),\,\Delta H = 869.6\,kJ \cr & {H_2}\left( g \right) \to 2\,H\left( g \right), \cr & {\text{Dissociation energy of }}H - H\,{\text{bond}} \cr & {\text{ = }}\frac{{869.6}}{2} = 434.8\,kJ \cr} $$

$$\eqalign{ & \Delta H = \sum {B.E{._R}} - \sum {B.E{._P}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 2 \times B.E.\left( {C - H} \right) + 2 \times B.E.\left( {C - Cl} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 2 \times 416 + 2 \times 325 \cr & \,\,\,\,\,\,\,\,\,\,\, = 1482\,kJ \cr} $$

\[\begin{align} & C+2S\to C{{S}_{2}} \\ & \Delta H=\sum{\Delta {{H}_{P}}-\sum{\Delta {{H}_{R}}}} \\ & \,\,\,\,\,\,\,\,\,=-1108.76-\left[ -393.3+2\times \left( -293.7 \right) \right] \\ & \,\,\,\,\,\,\,\,\,=-1108.76+393.3+587.4 \\ & \,\,\,\,\,\,\,\,\,=-128.06\,kJ \\ \end{align}\]

The enthalpy of formation of a compound may be positive or negative as it can be exothermic or endothermic process.

$$\eqalign{ & {C_2}{H_5}OH\left( \ell \right) + 3{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right) \cr & {\text{Bomb calorimeter gives}}\,\,\Delta U\,{\text{of}}\,{\text{the}}\,{\text{reaction}} \cr & {\text{Given, }}\Delta U = {\text{ - }}1364.47\,kJ\,mo{l^{ - 1}} \cr & \Delta {n_g} = - 1 \cr & \Delta H = \Delta U + \Delta {n_g}RT \cr & = - 1364.47 - \frac{{1 \times 8.314 \times 298}}{{1000}} \cr & = - 1366.93\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Given that}} \cr & MgS{O_4}\left( s \right) + n{H_2}O \to MgS{O_4}\,\,n{H_2}O; \cr & {\Delta _r}{H_1} = - 91.2\,kJ/mol\,\,\,...\left( {\text{i}} \right) \cr & MgS{O_4}.7{H_2}O\left( s \right) + \left( {n - 7} \right){H_2}O \to MgS{O_4}\left( {n{H_2}O} \right) \cr & {\Delta _r}{H_2} = 13.8\,kJ/mol\,\,\,\,...\left( {{\text{ii}}} \right) \cr & {\text{or}}\,\,\Delta {{\text{H}}_{hyd}} = {\Delta _r}{H_1} - {\Delta _r}{H_2}\,{\text{equation }}\left( {\text{i}} \right){\text{ - }}\left( {{\text{ii}}} \right) \cr & = - 91.2\,kJ/mol - 13.8\,kJ/mol \cr & = - 105\,kJ/mol \cr} $$

In isolated system, work, energy or heat are not exchanged with the surroundings.

$$\Delta G = \Delta H - T\Delta S$$
$$\Delta G = - ve\,\,{\text{if}}\,\,\Delta H = - ve\,\,$$     $${\text{and}}\,\,\Delta S\,\,{\text{is}}\,\, + ve.$$

Given, $$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right);$$
$${\Delta _f}H = - 393.5\,kJ\,mo{l^{ - 1}}$$
$$\because $$  Heat released on formation of $$44g$$  or $$1\,{\text{mole}}$$
$${\text{C}}{{\text{O}}_2} = - 395.5\,kJ\,mol$$
$$\because $$  Heat released on formation of $$35.2{\text{ }}g$$  of $$C{O_2}$$
$$\eqalign{ & = \frac{{ - 393.5\,kJ\,mo{l^{ - 1}}}}{{44g}} \times 35.2\,g \cr & = - 315\,kJ\,mo{l^{ - 1}} \cr} $$

As we know that,
$$\Delta H = \Delta E + p\Delta V\,\,{\text{or}}\,\,\Delta H = \Delta E + \Delta {n_g}RT$$
where, $$\Delta {n_g} \to $$   number of gaseous moles of product - number of gaseous moles of reactant
If $$\Delta ng = 0$$  ( for reactions in which the total number of moles of gaseous products are equal to total number of moles of gaseous reactants ), therefore $$\Delta H = \Delta E$$  
So, for reaction $$\left( a \right)\Delta n = 2 - 2 = 0$$
Hence, for reaction $$\left( a \right),\Delta H = \Delta E$$