CH4(g) + Cl(g) -> CH3(g) + HCl(g) dH° = -14 kJ/mol_rxn NH3(g) + Cl(g) -> NH2(g) + HCl(g) dH° = -36 k — Bonding Chemistry Question

STEPS:

1. Understand the relationship between reaction enthalpy and bond energy:
The standard enthalpy change of a reaction ($\Delta H^\circ$) can be estimated by calculating the difference between the energy required to break bonds in the reactants and the energy released when new bonds are formed in the products:
$$\Delta H^\circ = \sum \text{Bond Energies of Broken Bonds} - \sum \text{Bond Energies of Formed Bonds}$$

2. Analyze the bonds broken and formed in each given reaction:
For all three gas-phase reactions, a chlorine atom ($\text{Cl}$) abstracts a hydrogen atom from a gaseous reactant to form a gaseous hydrogen chloride ($\text{HCl}$) molecule.
* In Reaction 1, one $\text{C–H}$ bond in $\text{CH}_4$ is broken, and one $\text{H–Cl}$ bond is formed.
* In Reaction 2, one $\text{N–H}$ bond in $\text{NH}_3$ is broken, and one $\text{H–Cl}$ bond is formed.
* In Reaction 3, one $\text{O–H}$ bond in $\text{H}_2\text{O}$ is broken, and one $\text{H–Cl}$ bond is formed.

3. Express the enthalpy change of each reaction mathematically using bond dissociation energies ($D$):
* Reaction 1: $\Delta H^\circ_1 = D(\text{C–H}) - D(\text{H–Cl}) = -14\text{ kJ/mol}_{\text{rxn}}$
* Reaction 2: $\Delta H^\circ_2 = D(\text{N–H}) - D(\text{H–Cl}) = -36\text{ kJ/mol}_{\text{rxn}}$
* Reaction 3: $\Delta H^\circ_3 = D(\text{O–H}) - D(\text{H–Cl}) = +40\text{ kJ/mol}_{\text{rxn}}$

4. Isolate and compare the relative bond strengths relative to the shared $D(\text{H–Cl})$ term:
By rearranging the equations to solve for the individual bond dissociation energies, we can see how they compare:
* $D(\text{C–H}) = D(\text{H–Cl}) - 14\text{ kJ/mol}$
* $D(\text{N–H}) = D(\text{H–Cl}) - 36\text{ kJ/mol}$
* $D(\text{O–H}) = D(\text{H–Cl}) + 40\text{ kJ/mol}$

5. Conclude which bond is the strongest:
Because $D(\text{H–Cl})$ is a constant value, we can rank the bond energies directly:
$$D(\text{O–H}) > D(\text{C–H}) > D(\text{N–H})$$
* Breaking the $\text{O–H}$ bond requires $40\text{ kJ/mol}$ *more* energy than is released by forming the $\text{H–Cl}$ bond, making this process endothermic.
* Breaking the $\text{C–H}$ and $\text{N–H}$ bonds requires *less* energy than is released by forming the $\text{H–Cl}$ bond, making these processes exothermic.
* Therefore, the $\text{O–H}$ bond is the strongest bond among those shown, which makes Option C the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: The $\text{C–H}$ bond is weaker than the $\text{O–H}$ bond. The reaction involving the cleavage of the $\text{C–H}$ bond is exothermic ($\Delta H^\circ = -14\text{ kJ/mol}_{\text{rxn}}$), indicating that the $\text{C–H}$ bond is easier to break than the $\text{O–H}$ bond (which is endothermic at $\Delta H^\circ = +40\text{ kJ/mol}_{\text{rxn}}$).
• Option B is incorrect: The $\text{N–H}$ bond is the weakest of the three bonds. Its cleavage reaction is the most exothermic ($\Delta H^\circ = -36\text{ kJ/mol}_{\text{rxn}}$), which means it requires the least amount of energy input to break.
• Option D is incorrect: While the absolute values of the bond energies cannot be calculated without knowing the exact value of the $\text{H–Cl}$ bond energy, we can easily determine their relative strengths. Because $D(\text{H–Cl})$ is a constant, comparing the $\Delta H^\circ$ values allows us to establish a definitive comparative ranking.