Q.Calculate the standard enthalpy of : $\mathrm{N_2H_4(g) + H_2(g) \rightarrow 2\,NH_3(g)}$ if $\Delta H^0$(N-H) = 389 kJ mol$^{-1}$, $\Delta H^0$(H-H) = 435 kJ mol$^{-1}$, $\Delta H^0$ (N-N) = 159 kJ mol$^{-1}$

Bond Enthalpy: The Energy Cost of Breaking a Bond

Think of a chemical bond as a spring holding two atoms together. To pull the atoms apart, you have to do work — you have to put energy in. That energy, per mole of bonds broken, is the bond enthalpy. The stronger the bond, the more energy you need to supply, and the larger the bond enthalpy.

The intuition is simple: breaking bonds costs energy; forming bonds releases energy. A reaction is exothermic if the energy released in forming new bonds is greater than the energy consumed in breaking old ones.

The Precise Definition

Bond enthalpy (symbol: $\Delta H_{\text{bond}}$ or $B.E.$) is defined under a specific set of conditions:

The average enthalpy change when one mole of a particular covalent bond is broken, with all species in the gaseous state.

Three key points are packed into that sentence:

1. One mole of bonds — we measure the energy for Avogadro's number of bonds, not just one.
2. Gaseous state — the atoms or molecules must be in the gas phase. This removes complications from intermolecular forces (like hydrogen bonding in liquid water) and lets us isolate the bond energy itself.
3. Average — in a molecule like water ($H_2O$), the two O–H bonds are not identical in energy. The first O–H bond in $H_2O$ requires about 502 kJ/mol to break, but the second (in the remaining OH radical) requires about 427 kJ/mol. So we report the average bond enthalpy for O–H: roughly 464 kJ/mol.

How It's Used: Estimating Reaction Enthalpies

You can estimate the enthalpy change of a reaction ($\Delta H_{\text{rxn}}$) using bond enthalpies:

The logic: you put energy in to break bonds (positive), and you get energy out when bonds form (negative). So:

Example: Combustion of Methane

Consider: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$

Bonds broken (energy in):

• 4 C–H bonds: $4 \times 413 = 1652$ kJ/mol
• 2 O=O bonds: $2 \times 498 = 996$ kJ/mol
• Total in: $1652 + 996 = 2648$ kJ/mol

Bonds formed (energy out):

• 2 C=O bonds: $2 \times 799 = 1598$ kJ/mol
• 4 O–H bonds: $4 \times 464 = 1856$ kJ/mol
• Total out: $1598 + 1856 = 3454$ kJ/mol

The negative sign tells you the reaction is exothermic — more energy is released in forming bonds than was consumed in breaking them.

Why "Average" Matters

Bond enthalpy is not a universal constant. The C–H bond in methane is not identical to the C–H bond in ethane or benzene. The exact energy depends on the other atoms attached — the bond's environment. But for quick estimates, the average values work well enough.

The table below shows some common average bond enthalpies (in kJ/mol):

Notice the triple bond in $N_2$: 945 kJ/mol. That's enormous — it's why nitrogen gas is so unreactive.

The Bottom Line

Bond enthalpy is the average energy required to break one mole of a specific covalent bond in the gas phase. It lets you estimate reaction enthalpies by comparing the energy cost of breaking bonds with the energy gain from forming new ones. The stronger the bond, the larger the bond enthalpy, and the more stable the molecule.

Bond Enthalpy is a numerical-heavy concept from the Thermodynamics chapter of NCERT/CBSE Class 11 Chemistry, and it shows up often in "Bond Enthalpy numericals", "Bond Enthalpy important questions", and "Bond Enthalpy class 11 chemistry" searches because board exams, JEE Main, and NEET all test it through calculation-based problems.