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 \( 125 \, \text{kJ} \).

Answer 2

Step 1: Write the given reaction.
The reaction given is the hydrogenation of ethylene to form ethane:
\[ \text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6 \] We are asked to find the enthalpy of formation (ΔH) using bond energies.

Step 2: Write all given bond energies.
C–H = 414 kJ/mol
C–C = 347 kJ/mol
C=C = 615 kJ/mol
H–H = 435 kJ/mol

Step 3: Identify bonds broken and bonds formed.
In the reactants:
Ethylene (C₂H₄): 1 × C=C bond, 4 × C–H bonds
Hydrogen (H₂): 1 × H–H bond

In the products:
Ethane (C₂H₆): 1 × C–C bond, 6 × C–H bonds

Step 4: Calculate total bond energy.
Bonds broken (energy absorbed):
\[ E_{\text{broken}} = (1 \times 615) + (4 \times 414) + (1 \times 435) \] \[ E_{\text{broken}} = 615 + 1656 + 435 = 2706 \, \text{kJ/mol} \]
Bonds formed (energy released):
\[ E_{\text{formed}} = (1 \times 347) + (6 \times 414) \] \[ E_{\text{formed}} = 347 + 2484 = 2831 \, \text{kJ/mol} \]
Step 5: Find enthalpy change (ΔH).
\[ \Delta H = E_{\text{broken}} - E_{\text{formed}} \] \[ \Delta H = 2706 - 2831 = -125 \, \text{kJ/mol} \] The negative sign indicates the reaction is exothermic.

Step 6: Final Answer.
\[ \boxed{125 \, \text{kJ/mol}} \] (The enthalpy of formation of ethane from ethylene is 125 kJ released per mole.)