$$\Delta {G^ \circ } = 0$$   at equilibrium under standard state. Also at equilibrium, $$\Delta G = 0$$
$$\eqalign{ & \therefore \,\,\Delta {H^ \circ } - T\Delta {S^ \circ } = 0 \cr & {\text{Also}},\Delta {G^ \circ } = 2.303\,RT\,\log \,K \cr & \therefore \,\,K = 1\,\,{\text{when}}\,\,\Delta {G^ \circ } = 0 \cr} $$

$$\eqalign{ & {\Delta _f}{H^ \circ }\left( {F{e_{\left( s \right)}}} \right) = 0,{\Delta _f}{H^ \circ }\left( {{H_{2\left( g \right)}}} \right) = 0 \cr & {\Delta _f}{H^ \circ } = 3\left( { - 285.83} \right) - 1\left( { - 824.2} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 857.5 + 824.2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 33.3\,kJ\,mo{l^{ - 1}} \cr & {\text{Reaction is exothermic}}{\text{.}} \cr} $$

When $$\Delta {n_g}RT = 0,$$   then $$\Delta H = \Delta U$$

$$HCl + NaOH \to $$    $$NaCl + {H_2}O + x\,kcal$$
$${H_2}S{O_4} + 2NaOH \to $$     $$N{a_2}S{O_4} + 2{H_2}O + y\,kcal$$
$$1\,molar\,HCl $$    $$ = 1\,g{\text{ - equivalent of}}\,HCl$$

$$1\,molar\,{H_2}S{O_4}$$   $$ = 2\,g\,{\text{ - equivalent of}}\,{H_2}S{O_4}$$
$${\text{so,}}\,\,y = 2x\,\, \Rightarrow \,\,\,\,x = \frac{1}{2}y$$

The work done on the gas can be calculated from $$P{\text{ - }}V$$  graph. Work done is equal to the area $$AB{V_i}{V_f}.$$

$$\eqalign{ & {\text{For the equation}} \cr & {B_2}{H_6}\left( g \right) + 3{O_2}\left( g \right) \to {B_2}{O_3}\left( g \right) + 3{H_2}O\left( g \right) \cr & {\text{Eqs}}{\text{.}}\,\,\left( {\text{i}} \right) + 3\left( {{\text{ii}}} \right) + 3\left( {{\text{iii}}} \right) - \left( {{\text{iv}}} \right) \cr & \Delta H = - 1273 + 3\left( { - 286} \right) + 3\left( {44} \right) - 36 \cr & \,\,\,\,\,\,\,\,\,\,\, = - 1273 - 858 + 132 - 36 \cr & \,\,\,\,\,\,\,\,\,\,\, = - 2035\,kJ/mol \cr} $$

$$\eqalign{ & 2K + L \to 2M \cr & \Delta {n_g} = 2 - 3 = - 1 \cr & \Delta H = \Delta U + \Delta {n_g}RT \cr & \,\,\,\,\,\,\,\,\,\,\, = - 10.5 \times {10^3} + \left( { - 1 \times 8.314 \times 298} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = - 10500 + \left( { - 2477.572} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = - 12977.57\,J \cr & \,\,\,\,\,\,\,\,\,\,\, = - 12.98\,kJ \cr & \Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ } \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 12.98 - 298\left( { - 44.1 \times {{10}^{ - 3}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - 12.98 + 13.14 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 0.16\,kJ \cr} $$
$${\text{Since}}\,\,\Delta {G^ \circ }\,\,{\text{is}}\,\, + ve,$$    $${\text{hence it is non - spontaneous}}{\text{.}}$$

Enthalpy of neutralisation of $$HCl$$  with $$NaOH$$  is $$x.$$  In question $$gev$$  of $$HCl$$
$$\left( {\frac{{500 \times 2}}{{1000}} = 1gev} \right){\text{and}}\,NaOH\left( {\frac{{250 \times 4}}{{1000}} = 1gev} \right)$$
hence the value $$x.$$

From thermodynamics
$$\ln K = \frac{{ - \Delta {H^ \circ }}}{{RT}} + \frac{{\Delta {S^ \circ }}}{R}$$
for exothermic reaction,
$$\eqalign{ & \Delta H = - ve \cr & slope = \frac{{ - \Delta {H^ \circ }}}{R} = + ve \cr} $$
So from graph, line should be A & B.

Enthalpy of neutralisation is $$ - 57.1\,kJ\,mo{l^{ - 1}}$$   for a strong acid and a strong base.