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 _________ \( \text{kJ} \).
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 _________ \( \text{kJ} \).
Correct Answer: 125
Approach Solution - 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 \( 125 \, \text{kJ} \).
Approach Solution -2
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.)
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