Question

How to estimate (Delta H) for the bromination of ethylene from bond dissociation energies?

Answer 1

In the above question, we have to find the value of delta H for the bromination of ethylene from bond dissociation energies. This question is based on Hess’s law. We basically have to subtract bond dissociation energies of product from bond dissociation energies of reactants to find the enthalpy change.

Complete step by step solution:
Let us first write a balanced chemical equation of bromination of ethylene:
${{{H}}_{{2}}}{{C = C}}{{{H}}_{{2}}}{{ + Br - Br }} \to {{ Br - C}}{{{H}}_{{2}}}{{ - C}}{{{H}}_{{2}}}{{ - Br}}$
We know that enthalpy change of the reaction is equal to the difference of enthalpy of bonds formation and bond dissociation. Mathematically, it can be written as:
${{\Delta }}{{{H}}_{{{rxn}}}}{{ = }}\sum {{{\Delta }}{{{H}}_{{{bonds broken}}}}{{ - }}\sum {{{\Delta }}{{{H}}_{{{bonds formed}}}}} }$
We know that enthalpy of bond formation and bond dissociation is given by the bond dissociation energy and hence, the above reaction can be written as:
${{\Delta }}{{{H}}_{{{rxn}}}}{{ = }}\sum {{{B}}{{{E}}_{{{reactants}}}}{{ - }}\sum {{{B}}{{{E}}_{{{products}}}}} }$
In the reactant side, one C=C, four C-H and one Br-Br are present and in the product side, 2 C-Br, four C-H bonds and 1 C-C bond is present.
Hence, we can rewrite the equation as:
${{\Delta }}{{{H}}_{{{rxn}}}}{{ = }}\left( {{{4 \times B}}{{{E}}_{{{C - H}}}}{{ + 1 \times B}}{{{E}}_{{{C = C}}}}{{ + 1 \times B}}{{{E}}_{{{Br - Br}}}}} \right){{ - }}\left( {{{4 \times B}}{{{E}}_{{{C - H}}}}{{ + 1 \times B}}{{{E}}_{{{C - C}}}}{{ + 2 \times B}}{{{E}}_{{{C - Br}}}}} \right)$
Since, the 4 C-H bond is present in both the sides, hence, we can cancel it out.
${{\Delta }}{{{H}}_{{{rxn}}}}{{ = }}\left( {{{1 \times B}}{{{E}}_{{{C = C}}}}{{ + 1 \times B}}{{{E}}_{{{Br - Br}}}}} \right){{ - }}\left( {{{1 \times B}}{{{E}}_{{{C - C}}}}{{ + 2 \times B}}{{{E}}_{{{C - Br}}}}} \right)$
We know that bond dissociation energy of C=C, C-C, C-Br and Br-Br is 614 kJ, 384kJ, 276kJ and 193kJ respectively.
Substituting the values, we get:
${{\Delta }}{{{H}}_{{{rxn}}}}{{ = }}\left( {{{614kJ + 193kJ}}} \right){{ - }}\left( {{{348kJ + 2 \times 276kJ}}} \right){{ = - 93}}$ kJ.
Hence, the value of Delta H for the bromination of ethylene is ${{ - 93}}$ kJ.

Note:
From the question, we can infer that bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using Hess’s Law can be used to estimate reaction enthalpies.