Question

If the mean bond dissociation energies of C–C and C–H are 80 and 90 kJ mol\(^{-1}\) respectively, the standard enthalpy of atomization of n-butane(g) (in kJ mol\(^{-1}\)) is

• A. –980
• B. 1040
• C. 1140
• D. –1140

Hint:

Atomization enthalpy = Sum of all bond energies required to break all bonds in the molecule.

Answer 1

The Correct Option is C

n-butane (\(C_4H_{10}\)) structure:
- 3 C–C bonds
- 10 C–H bonds
Given:
- C–C bond energy = 80 kJ/mol
- C–H bond energy = 90 kJ/mol
\[ \text{Total atomization enthalpy} = (3 \times 80) + (10 \times 90) = 240 + 900 = \boxed{1140~\text{kJ/mol}} \]

Answer 2

If the mean bond dissociation energies of C–C and C–H are 80 and 90 kJ mol⁻¹ respectively, the standard enthalpy of atomization of n-butane (g) is:

Step 1: Understand what atomization means:
The enthalpy of atomization is the energy required to break all the bonds in a molecule to obtain all its atoms in the gaseous state.
So we need to calculate the energy required to break all C–C and C–H bonds in a molecule of n-butane (C₄H₁₀).

Step 2: Structure of n-butane:
n-butane has the following structure:
CH₃–CH₂–CH₂–CH₃

In this structure:
- Number of C–C bonds = 3
- Number of C–H bonds = 10

Step 3: Apply bond dissociation energies:
- Energy to break 3 C–C bonds = 3 × 80 = 240 kJ/mol
- Energy to break 10 C–H bonds = 10 × 90 = 900 kJ/mol

Step 4: Total enthalpy of atomization:
\[ \Delta H_{\text{atomization}} = 240 + 900 = 1140\ \text{kJ/mol} \]

Final Answer:
\[ \boxed{1140\ \text{kJ/mol}} \]