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

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 $125 \, \text{kJ}$.