Question

Estimate ΔH for the reaction: C2H6(g) + Cl2(g)--> C2H5Cl(g) + HCl(g) given the following average bond energies (in kJ/mol): C-C, 348 C-H, 414 CI-CI, 242 C-CI, 327 H-CI, 431 68 kJ/mol 344 kJ/mol +276 kJ/mol -102 kJ/mo 113 kJ/mol

Answer 1

Explanation:

The reaction equation will be as follows.

    $C_{2}H_{6}(g) + Cl_{2}(g) \rightarrow C_{2}H_{5}Cl(g) + HCl(g)$

Using bond energies, expression for calculating the value of $\Delta H$ is as follows.

    $\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

On reactant side, from $C_{2}H_{6}$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 6

From $Cl_{2}$; Cl-Cl bonds = 1

On product side, from $C_{2}H_{5}Cl$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 5

C-Cl bonds = 1

From HCl; H-Cl bonds = 1

Hence, using the bond energies we will calculate the enthalpy of reaction as follows.

  $\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

  =$[(1 \times 348 kJ/mol) + (6 \times 414 kJ/mol) + (1 \times 242 kJ/mol)] - [(1 \times 348 kJ/mol) + (5 \times 414 kJ/mol) + (1 \times 327 kJ/mol) + (1 \times 431 kJ/mol)]$            = -102 kJ/mol

Thus, we can conclude that change in enthalpy for the given reaction is -102 kJ/mol.