H₂O(s) → H₂O(l)

$$\Delta $$H = 6.0 kJ mol^–1

T = 0 + 273 K = 273 K; $$\Delta $$S = ?

$$\Delta $$S = $${{\Delta H} \over T}$$ = $${{6000} \over {273}}$$ = 21.98 JK^–1 mol^–1

For ideal solution,

$$\Delta $$H_soln = $$\Delta $$H₁ + $$\Delta $$H₂ + $$\Delta $$H₃

C_(graphite) $$ \to $$ C_(diamond)

Volume of graphite = $${{12} \over {2.25}} = 5.33$$ cm³ mol^–1

Volume of diamond = $${{12} \over {3.31}} = 3.63$$ cm³ mol^–1

$$\Delta $$V = V_graphite – V_diamond

= 1.70 cm³ mol^–1

= 1.70 × 10^–3 L mol^–1

So, $$\Delta $$G^o = P$$\Delta $$V

$$ \Rightarrow $$ 1895 J mol^–1 = P(1.70 × 10^–3 L mol^–1)

$$ \Rightarrow $$ 1114.70 × 10³ J/L = P

$$ \Rightarrow $$ $${{1114.7 \times {{10}^3}} \over {101.33}}$$ = P

$$ \Rightarrow $$ P = 11000.69 atm × 1.013 × 10⁵ Pa

= 11143.69 × 10⁵ Pa

= 11.14 × 10⁸ Pa

$$\Delta $$H = $$\Delta $$E + $$\Delta $$n_g RT

$$\Delta $$n_g = 3 – (1 + 5) = – 3

$$ \Rightarrow $$ $$\Delta $$H = $$\Delta $$E + (– 3RT)

$$ \Rightarrow $$ $$\Delta $$H – $$\Delta $$E = – 3RT