Question

The enthalpy of formation of ethane (\( \text{C}_2\text{H}_6 \)) from ethylene by addition of hydrogen,
where the bond energies of \( \text{C} - \text{H} \), \( \text{C} - \text{C} \), \( \text{H} - \text{H} \) are \( 414 \, \text{kJ} \), \( 347 \, \text{kJ} \), \( 615 \, \text{kJ} \), and \( 435 \, \text{kJ} \) respectively, is _________ \( \text{kJ} \).

Answer 1

Correct Answer: 125

Given Information:

Bond energies:

• \( \text{C–H} = 414 \, \text{kJ/mol} \)
• \( \text{C–C} = 347 \, \text{kJ/mol} \)
• \( \text{H–H} = 435 \, \text{kJ/mol} \)
• \( \text{C=C (double bond)} = 615 \, \text{kJ/mol} \)

Reaction for Formation of Ethane from Ethylene:

The reaction can be represented as:

\[ \text{C}_2\text{H}_4(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) \]

Bond Energy Calculations:

Breaking Bonds:

• One \( \text{C=C} \) bond in ethylene: \( 615 \, \text{kJ} \)
• One \( \text{H–H} \) bond: \( 435 \, \text{kJ} \)
• Total energy required to break bonds = \( 615 + 435 = 1050 \, \text{kJ} \)

Forming Bonds:

• One \( \text{C–C} \) bond in ethane: \( 347 \, \text{kJ} \)
• Two \( \text{C–H} \) bonds: \( 2 \times 414 = 828 \, \text{kJ} \)
• Total energy released in forming bonds = \( 347 + 828 = 1175 \, \text{kJ} \)

Enthalpy Change (\( \Delta H \)):

\[ \Delta H = \text{Energy required to break bonds} - \text{Energy released in forming bonds} \]

\[ \Delta H = 1175 - 1050 = 125 \, \text{kJ} \]

Conclusion:

The enthalpy of formation of ethane from ethylene by addition of hydrogen is \( 125 \, \text{kJ} \).