H₂O(l) $$\buildrel {100^\circ \,C} \over \longrightarrow $$ H₂O(g)

∆H^o = 40.66kJ mol^–1

∆H^o = ∆u^o + $$\Delta $$n_g RT

$$\Delta $$n_g = 1, R = 8.314 × 10^–3 kJ mol^–1 k^–1

T = 100 + 273 = 373 K

$$ \Rightarrow $$ 40.66 = ∆u^o + (1) (8.314 × 10^–3) × 373

$$ \Rightarrow $$ ∆u^o = 37.56 kJ mol^–1

$$\Delta $$S^o positive means entropy of products is more than that of reactants. Among the given reactions only in option (a) the $$\Delta $$S^o is positive cause products has 1 mole of gaseous products CO while reaction have half mole of gaseous O₂ and Carbon (C) in solid state. Thus, reactions have less randomness than products. Also, the reaction is a combustion reaction which has a negative value of $$\Delta $$H. According to Gibbs energy change ($$\Delta $$S^o).

$$\Delta $$G^o = $$\Delta $$H^o - T$$\Delta $$S^o

$$ \Rightarrow $$ $$\Delta $$G^o = -ve - T(+ve)

Thus, on increasing temperature value of $$\Delta $$G^o decreases sharply.

	H (kJ/mol)
1/2A $$ \to $$ B	+150
3B  $$ \to $$ 2C + D	-125
E + A  $$ \to $$ 2D	+350

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   B + D $$ \to $$ E + 2C;

$$\Delta $$H = (300 - 125 - 350) = - 175 kJ/mol

For adiabatic condition, change in heat does not take place thus, q = 0.

If q = 0 then change in temperature does not take place thus, $$\Delta $$T = 0.

Also for a free expansion of ideal gas work done, W = 0 as there is no external pressure on it.