Question

Hydroxyl radicals react with and eliminate many atmospheric pollutants. However, the hydroxyl radical does not clean up everything. For example, chlorofluorocarbons, which destroy stratospheric ozone, are not attacked by the hydroxyl radical. Consider the hypothetical reaction by which the hydroxyl radical might react with a chlorofluorocarbon:
OH(g)+CF2Cl2(g)→HOF(g)+CFCl2(g)

Use bond energies to explain why this reaction is improbable. Calculate ΔHrxn of this reaction.

Answer 1

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Answer 2

The enthalpy change ΔHrxn of the given reaction is 295 kJ/mol

Consider the hypothetical reaction whereby hydroxyl radical reacts with Chlorofluorocarbon. The balanced chemical equation can be expressed as:

$\mathbf{OH_{(g)}+ CF_2Cl_{2(g)} \to HOF_{(g)} + CFCl_{2(g)} }$

In the above chemical equation, the chemical species participating in the reaction are:

• O-H, C-F, C-Cl, and O-F

The bond energies of these chemical species are:

• O-H bond = 458.9 kJ/mol
• C-F bond = 485 kJ/mol
• C-Cl bond = 385 kJ/mol
• O-F bond = 190 kJ/mol

The  ΔHrxn is the enthalpy change of the reaction and it can be expressed by using the formula:

$\mathbf{\Delta H_{rxn} = \sum \ B.E (reactants) - \sum \ B.E (products)}$

∴

$\mathbf{\Delta H_{rxn} =\Big ( B.E (O-H) + 2B.E(C-F) +2B.E(C-Cl) \Big) - \Big( B.E (O-H) + B.E(O-F) } \\ \\ \mathbf{\ \ \ \ \ \ \ \ \ \ \ \ \ \ +B.E( C-F )+2B.E(C-Cl) \Big)}$

By replacing the bond energies into the above expression, we have:

$\mathbf{\Delta H_{rxn} =\Big ( 458.9 \ kJ/mol + 2(485 \ kJ/mol) +2(385 \ kJ/mol) \Big)} \\ \\ \mathbf{ - \Big( 458.9 \ kJ/mol + 190 \ kJ/mol +485 \ kJ/mol +2(385 \ kJ/mol)\Big)}$

$\mathbf{\Delta H_{rxn} =\Big ( 2198.9 \ kJ/mol \Big) - \Big( 1903.9\ kJ/mol\Big)}}$

$\mathbf{\Delta H_{rxn} =295 \ kJ/mol}$

Therefore, provided that the enthalpy change is positive, it implies that the reaction is improbable and non-favorable.

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