R1.2 Energy cycles in reactions

The question asks about the energy change specifically associated with bond breaking, using the N≡N bond as an example context.

Bond breaking always requires energy input to overcome the electrostatic attraction between bonded atoms. This makes bond breaking endothermic ($+\Delta H$) by definition — regardless of bond order.

The endothermic nature of bond breaking applies to all bonds — single, double, and triple. It is not limited to triple bonds. The magnitude of energy required is greater for stronger bonds, but the direction (endothermic) is universal.

The correct answer is: Bond breaking is endothermic because energy must be absorbed to separate bonded atoms. This is a fundamental principle of thermochemistry.

The question tests whether students know whether bond breaking is endothermic or exothermic, and whether this applies universally.

'Bond breaking is exothermic because energy is released as bonds are broken' is incorrect. It is bond forming that releases energy. Bond breaking always absorbs energy.

'Bond breaking is endothermic only when the bond broken is a triple bond' is incorrect. All bond breaking is endothermic — the claim is true but the restriction to triple bonds is false.

'Bond breaking has no energy change because atoms are rearranged, not destroyed' is incorrect. Atoms are conserved but bonds are broken, which absolutely requires energy input.

The remaining option — 'Bond breaking is endothermic because energy must be absorbed to separate bonded atoms' — correctly and completely describes bond breaking for all bond types.

The target is the formation of propane from its elements: $3\text{C}(s)+4{\text{H}}_2(g)\to {\text{C}}_3{\text{H}}_8(g)\Delta H_f^{\circ }=?$

Multiply the combustion of C by 3: $3\text{C}(s)+3{\text{O}}_2(g)\to 3{\text{CO}}_2(g),\Delta H=3\times (-394)=-1182{\text{ kJ mol}}^{-1}$. Multiply the combustion of H₂ by 4: $4{\text{H}}_2(g)+2{\text{O}}_2(g)\to 4{\text{H}}_2\text{O}(l),\Delta H=4\times (-286)=-1144{\text{ kJ mol}}^{-1}$. Reverse the combustion of propane: $3{\text{CO}}_2(g)+4{\text{H}}_2\text{O}(l)\to {\text{C}}_3{\text{H}}_8(g)+5{\text{O}}_2(g),\Delta H=+2220{\text{ kJ mol}}^{-1}$.

Adding the three equations, ${\text{CO}}_2$, ${\text{H}}_2\text{O}$, and ${\text{O}}_2$ all cancel, leaving: $3\text{C}(s)+4{\text{H}}_2(g)\to {\text{C}}_3{\text{H}}_8(g)$ ✓

$\Delta H_f^{\circ }=(-1182)+(-1144)+(+2220)=-106\approx -104{\text{ kJ mol}}^{-1}$ The exact calculation gives $-1182-1144+2220=-106$; the closest answer is $-104{\text{ kJ mol}}^{-1}$.

The answer is approximately $-104{\text{ kJ mol}}^{-1}$ (exothermic formation), confirming propane has a negative standard enthalpy of formation.

This question tests the application of Hess's Law using combustion data to find $\Delta H_f^{\circ }$. The result should be a small negative value for a stable hydrocarbon.

$-2900$ and $-3540$ kJ mol${}^{-1}$ are far too large in magnitude. These are closer to combustion enthalpies, not formation enthalpies. Formation of a small organic molecule typically gives values in the range of tens to a few hundred kJ mol${}^{-1}$.

$+104$ kJ mol${}^{-1}$ would indicate an endothermic formation. While possible in principle, applying the Hess's Law calculation gives a negative result, ruling this out.

$-104$ kJ mol${}^{-1}$ is consistent with the Hess's Law calculation: $(-1182)+(-1144)+(+2220)=-106\approx -104$ kJ mol${}^{-1}$. This is the correct answer.