Molar entropy change for the melting of ice,
$$\eqalign{ & \Delta {S_{melt}} = \frac{{\Delta {H_{fusion}}}}{T} \cr & = \frac{{1.435\,kcal/mol}}{{\left( {0 + 273} \right)\,K}} \cr & = 5.26 \times {10^{ - 3}}\,kcal/mol\,K \cr & = 5.26\,cal/mol\,K \cr} $$

$$\Delta S$$  [ change in entropy ] and $$\Delta H$$ [ change in enthalpy ] are related by the equation $$\Delta G = \Delta H - T\Delta S$$
[ Here, $$\Delta G = $$  change in Gibbs free energy ]
For adsorption of a gas, $$\Delta S$$  is negative because randomness decreases. Thus, in order to make $$\Delta G$$ negative [ for spontaneous reaction ], $$\Delta H$$ must be highly negative. Hence for the adsorption of a gas, if $$\Delta S$$  is negative, therefore, $$\Delta H$$ should be highly negative.

$$\eqalign{ & {\text{For reaction,}} \cr & {H_2}\left( g \right) + B{r_2}\left( g \right) \to 2HBr\left( g \right)\Delta {H^ \circ } = ? \cr} $$
$$\Delta {H^ \circ } = - \left[ {\left( {2 \times {\text{bond energy of }}HBr} \right)} \right.$$       $$\left. { - \left( {{\text{bond energy}}\,{\text{of}}\,{H_2} + {\text{bond energy of}}\,C{l_2}} \right)} \right]$$
$$\eqalign{ & \Delta {H^ \circ } = - \left[ {2 \times \left( {364} \right) - \left( {433 + 192} \right)} \right]kJ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - \left[ {728 - \left( {625} \right)} \right]kJ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - 103\,kJ \cr} $$

For combustion reaction, $$\Delta H$$  is negative, $$\Delta n = \left( {16 + 18} \right) - \left( {25 + 2} \right) = + 7,$$       so $$\Delta S$$  is $$ + ve,$$  reaction is spontaneous, hence $$\Delta G$$  is $$ - ve.$$

$$\Delta G = \Delta H - T\Delta S,\Delta S = 0\,\,{\text{or}}\,\,T\Delta S = 0,\Delta G = \Delta H$$

Work is not a state function because it depends upon the path followed.

$$\eqalign{ & \Delta {U_{AB}} = q + w = + 5 + \left( { - 8} \right) = - 3 \cr & q = - 3,\,\Delta {U_{BA}} = + 3 \cr & \,\Delta {U_{BA}} = q + w \cr & \Rightarrow 3 = - 3 + w \Rightarrow w = + 6\,J\,{\text{(work done on the system)}}{\text{.}} \cr} $$

$$\eqalign{ & {X_2} + {Y_2} \to 2XY,\,\Delta H = 2\left( { - 200} \right) \cr & {\text{Let}}\,x\,{\text{be}}\,{\text{the}}\,{\text{bond}}\,{\text{dissociation}}\,{\text{energy}}\,{\text{of}}\,{X_2}.\,{\text{Then}} \cr & \Delta H = - 400 = {\xi _{x - x}} + {\xi _{y - y}} - 2{\xi _{x - y}} = x + 0.5x - 2x = - 0.5x \cr & or\,x = \frac{{400}}{{0.5}} = 800\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & \Delta {G^ \circ } = - RT\,\,\ln \,\,{K_p}\,\,{\text{or}}\,\,\ln {K_p} = - \frac{{\Delta {G^ \circ }}}{{RT}} \cr & {\text{or}}\,\,\,{K_p} = {e^{ - \,\frac{{\Delta {G^ \circ }}}{{RT}}}} \cr} $$

$$\eqalign{ & {\text{For}}\,\,\left( {\text{i}} \right),\Delta {n_g} = 1 - \frac{3}{2} = - \frac{1}{2} \cr & {\text{For}}\left( {{\text{ii}}} \right),\Delta {n_g} = \frac{1}{2} - 0 = \frac{1}{2} \cr} $$