To solve for the approximate value of $\Delta H$ (enthalpy change) using bond energies for the given reaction:

$\text{CH}_4(g) + 2 \text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 2 \text{H}_2(g)$

Step 1: Identify Bonds Broken and Formed

• Reactants (bonds to be broken):

  • $\text{CH}_4$: Contains 4 C-H bonds
  • $2 \text{Cl}_2$: Contains 2 Cl-Cl bonds

• Products (bonds to be formed):

  • $\text{CCl}_4$: Contains 4 C-Cl bonds
  • $2 \text{H}_2$: Contains 2 H-H bonds

Step 2: Use Bond Energies

Let's denote the bond energies as follows (values are typical and should be verified with the provided bond energy table):

• C-H bond energy: ~413 kJ/mol
• Cl-Cl bond energy: ~243 kJ/mol
• C-Cl bond energy: ~328 kJ/mol
• H-H bond energy: ~436 kJ/mol

Step 3: Calculate Total Energy for Bonds Broken (Reactants)

$\text{Energy for bonds broken} = (4 \times \text{C-H}) + (2 \times \text{Cl-Cl})$ $= (4 \times 413) + (2 \times 243) = 1652 + 486 = 2138 \, \text{kJ}$

Step 4: Calculate Total Energy for Bonds Formed (Products)

$\text{Energy for bonds formed} = (4 \times \text{C-Cl}) + (2 \times \text{H-H})$ $= (4 \times 328) + (2 \times 436) = 1312 + 872 = 2184 \, \text{kJ}$

Step 5: Calculate $\Delta H$

$\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}$ $\Delta H = 2138 - 2184 = -46 \, \text{kJ}$

Answer

The approximate value for $\Delta H$ is $-46$ kJ.

Would you like further details or have any questions?

Related Questions

1. How does bond enthalpy affect the overall enthalpy change in a reaction?
2. Why is it important to consider both bonds broken and bonds formed when calculating $\Delta H$?
3. Can bond enthalpy values vary between different chemical resources?
4. How would the value of $\Delta H$ change if the bond energy of Cl-Cl were slightly different?
5. Why are bond enthalpies considered averages and not exact values for each reaction?

Tip

When calculating enthalpy changes using bond energies, always ensure that you have the most accurate and specific bond energy values from reliable resources, as slight variations can affect the result.