“At absolute zero temperature, entropy of a perfectly crystalline substance is taken to be zero.” It is called third law of thermodynamics.

The given reaction is combustion reaction, so it takes place by evolution of heat and hence, the sign of $$\Delta H = $$  negative and there is a increase in the number of moles of gaseous products, so entropy also increases and hence, $$\Delta S = $$  positive.
$$\eqalign{ & {\text{Thus,}}\,\Delta G = \Delta H - T\Delta S \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - ve - T\left( { + ve} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - ve\,\,{\text{at any temperature}} \cr} $$

$$\eqalign{ & Q = \frac{1}{2}438 \times \frac{{22.3}}{{223}} \cr & = 21.9\,kJ \cr} $$

$${\Delta _r}H = \frac{5}{8} \times 40 + \frac{3}{8} \times 50 = 43.75\,kJ/mol$$

We know that, $$\,\Delta H = \Delta E + \Delta {n_g}RT$$
where, $$\,\Delta {n_g} = $$  total number of moles of gaseous product $$-$$ total number of moles of gaseous reactant
$$\eqalign{ & = 2 - 3 = - 1 \cr & {\text{So,}}\,\Delta H = \Delta E - RT \cr} $$

$$\eqalign{ & {C_{{\text{(diarmond)}}}} + {O_{2\left( g \right)}} \to C{O_{2\left( g \right)}} \cr & \Delta H = - 393.7 + \left( { - 2.1} \right) = - 395.8\,kJ \cr & 12\,g\,\,{\text{diamond gives}}\,\,395.8\,kJ. \cr & 800\,kJ\,\,{\text{will be given by}}\,\,\frac{{12}}{{395.8}} \times 800 = 24.25\,g \cr} $$

$$\eqalign{ & {\text{Applying Hess's}}\,\,{\text{Law}} \cr & {\Delta _f}{H^ \circ } = {\Delta _{sub}}H + \frac{1}{2}{\Delta _{diss}}H + I.E. + E.A + {\Delta _{lattice}}H \cr & - 617 = 161 + 520 + 77 + E.A. + \left( { - 1047} \right) \cr & E.A. = - 617 + 289 = - 328\,kJ\,mo{l^{ - 1}} \cr & \therefore \,\,{\text{electron affinity of fluorine}} \cr & = - 328\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {C_{v,m}}\left( {{H_2}O,g} \right) = 33.314 - 8.314 \cr & = 25\,J/K\,mol \cr & \Delta {U_2} = \Delta {H_2} - \Delta {n_g}RT = 37.6 \cr & \Delta {U_{total}} = \Delta {u_1} + \Delta {u_2} + \Delta {u_3} \cr & = {C_{v,m}}\left( l \right)\Delta T + \Delta {V_{vap.}} + {C_{v,m}}\left( g \right)\Delta T \cr & = \frac{{75 \times 50}}{{1000}} + 37.6 + \frac{{25 \times 50}}{{1000}} \cr & = 42.6\,kJ/mol \cr} $$

Enthalpy of reaction $$\left( {\Delta H} \right) = {E_{{a_{\left( f \right)}}}} - {E_{{a_{\left( b \right)}}}}$$
for an endothermic reaction $$\Delta H = + ve$$   hence for $$\Delta H$$ to be negative
$${E_{{a_{\left( b \right)}}}} < {E_{{a_{\left( f \right)}}}}$$

$$\Delta G = \Delta H - T\Delta S,\,\Delta H\,\,{\text{is}}\,\, + ve,$$       $$\Delta S\,\,{\text{is}}\,\, + ve;\,T\Delta S > \Delta H$$
$$\eqalign{ & {\text{for spontaneous process}}{\text{.}} \cr & {\text{It will make}}\,\Delta G, - ve \cr} $$