Question

If for the reaction \[ \text{CCL}_3(g) \rightarrow \text{C}(g) + 3\text{Cl}(g), \] the following data is given: \[ \Delta H^\circ_{\text{CCL}_3(l)} = 30 \, \text{kJ mol}^{-1}, \, \text{vap} = \text{vaporization} \] \[ \Delta H^\circ_{\text{CCl}_3} = -136.0 \, \text{kJ mol}^{-1}, \, f = \text{formation} \] \[ \Delta H^\circ_{\text{C}} = 714.0 \, \text{kJ mol}^{-1}, \, a = \text{atomization} \] \[ \Delta H^\circ_{\text{Cl}} = 242.0 \, \text{kJ mol}^{-1}, \, a = \text{atomization} \] The bond mean enthalpy of C-Cl in CCl\(_4\)(l) is: \[ \text{CCL}_4(g) \rightarrow \text{C}(g) + 3\text{Cl}(g) \text{ gives the bond mean enthalpy.} \]

• A. -319
• B. 326
• C. -326
• D. 292

Hint:

When calculating bond enthalpies, remember to account for all enthalpy changes during the formation and vaporization of the molecule.

Answer 1

The Correct Option is A

To find the bond mean enthalpy of C-Cl in CCl\(_3\), we use the following equation: \[ \Delta H^\circ_{\text{bond}} = \Delta H^\circ_{\text{formation}} + \Delta H^\circ_{\text{atomization}} - \Delta H^\circ_{\text{vaporization}} \] Substituting the given values: \[ \Delta H^\circ_{\text{bond}} = 30 + 242 - 136 = -319 \, \text{kJ/mol} \]