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

2KBr + I₂ $$ \to $$ 2KI + Br₂

reaction is not possible because Br^– ion is not oxidised in Br₂ with I₂ due to higher electrode (oxidation) potential of I₂ than bromine.

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

When in a closed insulated container, a liquid is stirred, it increases the temperature of liquid thus it started behaving as adiabatic process resulting no change in heat and only temperature increases. Therefore, q = 0 for this process.

Also, $$\Delta $$E = q + W

$$\Delta $$E = W (As, q = 0)

As, there is rise in temperature thus, there must be change in internal energy, i.e., $$\Delta $$E = W ≠ 0.