Problem

A sample of ethane ($C_{2}H_{6}$) undergoes dehydrogenation: $C_{2}H_{6}$(g) → $C_{2}H_{4}$(g) + $H_{2}$(g). Given bond enthalpies: C-H = 414 kJ/mol, C-C = 347 kJ/mol, C=C = 611 kJ/mol, H-H = 436 kJ/mol. Calculate: (a) $\Delta H$ for the reaction (b) Whether the reaction is spontaneous at 298 K if $\Delta S = +120$ J/(mol·K) (c) The minimum temperature for spontaneity

Solution

Step 1: Identify bonds broken in reactants ($C_{2}H_{6}$) Structure: $CH_{3}$-$CH_{3}$

• C-C bonds broken: 1 × 347 = 347 kJ
• C-H bonds broken: 6 × 414 = 2484 kJ
• Total energy absorbed = 2831 kJ

Step 2: Identify bonds formed in products ($C_{2}H_{4}$ + $H_{2}$) $C_{2}H_{4}$ structure: C$H_{2}$=C$H_{2}$; $H_{2}$ structure: H-H

• C=C bonds formed: 1 × 611 = 611 kJ
• C-H bonds formed (in $C_{2}H_{4}$): 4 × 414 = 1656 kJ
• H-H bonds formed: 1 × 436 = 436 kJ
• Total energy released = 2703 kJ

Step 3: Calculate $\Delta H$ $\Delta H = \Sigma(BE_{broken}) - \Sigma(BE_{formed}) = 2831 - 2703 = +128\ \text{kJ/mol}$ Units: kJ/mol ✓ (endothermic)

Step 4: Check spontaneity at 298 K $\Delta G = \Delta H - T\Delta S = 128000 - 298 \times 120 = 128000 - 35760 = +92240\ \text{J/mol} = +92.2\ \text{kJ/mol}$ $\Delta G > 0$ → NOT spontaneous at 298 K

Step 5: Find crossover temperature $T_{crossover} = \frac{\Delta H}{\Delta S} = \frac{128000\ \text{J/mol}}{120\ \text{J/(mol·K)}} = 1067\ \text{K} \approx 794°C$ The reaction becomes spontaneous above 1067 K.

Verification: At 1100 K: $\Delta G = 128000 - 1100 \times 120 = 128000 - 132000 = -4000\ \text{J} < 0$ ✓