Comparing strength of the C-H, N-H, and O-H bonds

Question

$$ \begin{align} \ce{CH4(g) + Cl(g) &-> CH3(g) + HCl(g)} &\quad ΔH^\circ &= \pu{-14 kJ mol-1}_\mathrm{rxn}\\ \ce{NH4(g) + Cl(g) &-> NH2(g) + HCl(g)} &\quad ΔH^\circ &= \pu{-36 kJ mol-1}_\mathrm{rxn}\\ \ce{H2O(g) + Cl(g) &-> OH(g) + HCl(g)} &\quad ΔH^\circ &= \pu{+40 kJ mol-1}_\mathrm{rxn} \end{align} $$

41. Based on the data above, what can be concluded regarding the strength of the $\ce{C-H}$, $\ce{N-H}$, and $\ce{O-H}$ bonds in the molecules shown?

(A) The $\ce{C-H}$ bond is the strongest.
(B) The $\ce{N-H}$ bond is the strongest.
(C) The $\ce{O-H}$ bond is the strongest.
(D) Nothing can be concluded without knowing the strength of the $\ce{H-Cl}$ bond.

I am not sure how to approach the problem 41. In exothermic reactions, more energy is released when the bonds are formed in the products than is used to break the bonds in the reactants. So would the answer be B because the $ΔH$ is exothermic and is the greater than $ΔH$ of -14 for A?

Answer 1

The answer should be C. One way would be to write the thermochemical equations, however since we just want to compare, we can do it without going into that. The formation of the H-Cl bond, (like all bonds) is exothermic. And the breaking of the C-H, N-H and O-H bonds is endothermic (since their formation is exothermic too). Since the contribution of the H-Cl bond to the enthalpy is the same for all three, we can focus on the C-H, N-H and O-H bonds alone.

Now, the stronger the bond, the more energy it would need, and hence make a positive contribution to the enthalpy change. Hence we can see that since the total enthalpy of the reaction is highest(most positive) for the reaction involving O-H bond, the O-H bond is the strongest. (Infact, it is stronger than the H-Cl bond too as the enthalpy change is positive).

Note that this question also demonstrates the order of electronegativity, which is O>Cl >~ N> C>H. Since the bond with most electronegativity difference would be the strongest, it would be the O-H bond.