calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl, molecule

/ Exams & Certification / By

Given that AH; CI(g)| = 121.3 kJ-mol- AH; C(g)] = 716.7 kJ-mol- AH; (CC,(g) = -95.7 kJ-mol-! calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl, molecule. None O 16 6 Given that Y H N AH; [CI(g)] = 121.3 kJ.mol-¹ AH; [C(g)] = 716.7 kJ·mol¯¹ AH; [CC14(g)] = −95.7 kJ·mol−¹ calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl4 molecule. AHC-Cl = 17 & 7 J f8 3 e 8 U UT about us K fg careers privacy policy terms of use W 9 alt hp O L f10 V O F11 P ctri ? contact us help { f12 ✈ [ 44 11 ins Question Source: MRG – General Chemistry | Publisher: Universit prt sc ] pause delete backspace kJ.mol-1 8 home enter num lock T shift 10: 7/21 end 7 home 4

The Correct Answer and Explanation is :

To calculate the average molar bond enthalpy of the carbon-chlorine (C-Cl) bond in a CCl₄ molecule, we need to use the following given information:

  • Enthalpy of formation of chlorine gas (Cl₂(g)): ( \Delta H_f^\circ[\text{Cl}(g)] = 121.3 \, \text{kJ/mol} )
  • Enthalpy of formation of carbon (C(g)): ( \Delta H_f^\circ[\text{C}(g)] = 716.7 \, \text{kJ/mol} )
  • Enthalpy of formation of carbon tetrachloride (CCl₄(g)): ( \Delta H_f^\circ[\text{CCl}_4(g)] = -95.7 \, \text{kJ/mol} )

Step 1: Use Hess’s Law

We can apply Hess’s Law, which states that the enthalpy change for a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.

The formation of CCl₄ from its elements in their standard states can be written as:

[ \text{C}(s) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) ]

We can break this into several steps:

  1. Sublimation of carbon:
    [ \text{C}(s) \rightarrow \text{C}(g) \quad \Delta H = 716.7 \, \text{kJ/mol} ]
  2. Dissociation of chlorine gas:
    [ \text{Cl}_2(g) \rightarrow 2\text{Cl}(g) \quad \Delta H = 2 \times 121.3 = 242.6 \, \text{kJ/mol} ]
  3. Formation of CCl₄:
    [ \text{C}(g) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \quad \Delta H = -95.7 \, \text{kJ/mol} ]

Step 2: Calculate the enthalpy change for breaking the C-Cl bonds

Now, we can sum these steps:

[
\Delta H = 716.7 + 242.6 – 95.7 = 863.6 \, \text{kJ/mol}
]

This value represents the energy required to form CCl₄ in its gaseous form from the elements. Since CCl₄ has 4 C-Cl bonds, we divide the total enthalpy change by 4 to find the bond enthalpy for one C-Cl bond:

[
\text{Average bond enthalpy of C-Cl} = \frac{863.6}{4} = 215.9 \, \text{kJ/mol}
]

Final Answer:

The average molar bond enthalpy of the carbon-chlorine bond in a CCl₄ molecule is approximately 215.9 kJ/mol.

Related Posts

Scroll to Top