Question

Use the data from the table to estimate the enthalpy of formation of CH₃CHO. 
 

Bond Enthalpy	Bond	Enthalpy of Formation
400 kJ mol-1	C–H	C(g)    700 kJ mol-1
350 kJ mol-1	C–C	H(g)    200 kJ mol-1
700 kJ mol-1	C=O	O(g)    250 kJ mol-1

 

• A. –200 kJ/mol
• B. –400 kJ/mol
• C. –350 kJ/mol
• D. –150 kJ/mol

Hint:

Use: \( \Delta H_f = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products} \)

Answer 1

The Correct Option is A

Step 1: Total energy to form atoms in gaseous state:
\( \text{CH}_3\text{CHO} \) has:
- 2 C → \( 2 \times 700 = 1400 \)
- 4 H → \( 4 \times 200 = 800 \)
- 1 O → \( 1 \times 250 = 250 \)
Total = \( 1400 + 800 + 250 = 2450\, \text{kJ/mol} \)
Step 2: Bond formation in product (CH\(_3\)CHO):
Structure:
- 3 C–H → \( 3 \times 400 = 1200 \)
- 1 C–C → 350
- 1 C–H → 400
- 1 C=O → 700
Total = \( 1200 + 350 + 400 + 700 = 2650\, \text{kJ/mol} \) Step 3: Enthalpy of formation: \[ \Delta H_f = \text{Atomization energy} - \text{Bond formation energy} = 2450 - 2650 = -200\, \text{kJ/mol} \]