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

For isothermal reversible expansion

w = q = nRT $$ \times $$ 2.303 $$\log {{{v_2}} \over {{v_1}}}$$

= 2RT $$ \times $$ 2.303 $$\log {{{20}} \over {{2}}}$$

= 2 × 2 × T × 2.303 × 1 = 9.2 T

Entropy change, $$\Delta $$S = $${q \over T}$$ = $${{9.2T} \over T}$$ = 9.2 cal/mol K

C_(s) + 2H_2(g) $$ \to $$ CH_4(g), $$\Delta $$H^o = ?

C_(s) + O_2(g) → CO_2(g), $$\Delta $$H = – 94 kcal/mol ....(1)

H_2(g) + $${1 \over 2}$$O_2(g) $$ \to $$ H₂O_(g), $$\Delta $$H = – 68 kcal/mol ....(2)

CH_4(g) + 2O_2(g) $$ \to $$ CO_2(g) + H₂O_(l), $$\Delta $$H = – 213 kcal/mol ....(3)

Performing (1) + 2 × (2) – (3)

$$\Delta $$H^o = [– 94 – 2(68)] – (–213) kcal/mol

= [– 94 – 136] + 213 k cal/mol

= – 230 + 213 = – 17 kcal/mol