Define: (a) Reversible process (b) Standard enthalpy of combustion. Calculate the enthalpy change for the reac

Question 95 of 115

Q. Define:

(a) Reversible process

(b) Standard enthalpy of combustion. Calculate the enthalpy change for the reaction:

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

. The bond enthalpies are given in a table: Bond

N\equiv N

\Delta H^\circ = 946\ kJ\ mol^{-1}

; Bond

H-H

\Delta H^\circ = 435\ kJ\ mol^{-1}

; Bond

N-H

\Delta H^\circ = 389\ kJ\ mol^{-1}

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 4mImportance★★★★★

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Concept understanding — Bond Enthalpy

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:

One mole of bonds — we measure the energy for Avogadro's number of bonds, not just one.
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.
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.

Important

Bond enthalpy is always positive — it is the energy absorbed to break a bond. Bond formation releases the same amount of energy (negative enthalpy change).

How It's Used: Estimating Reaction Enthalpies

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

\Delta H_{\text{rxn}} = \sum \text{(bond enthalpies of bonds broken)} - \sum \text{(bond enthalpies of bonds formed)}

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

\Delta H_{\text{rxn}} = \text{Energy in} - \text{Energy out}

Watch out

This method gives an estimate, not an exact value. Bond enthalpies are averages taken from many different molecules, so they don't account for the exact molecular environment. For precise work, use standard enthalpies of formation.

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

\Delta H_{\text{rxn}} = 2648 - 3454 = -806 \text{ kJ/mol}

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

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A reversible process proceeds through equilibrium states; combustion enthalpy is the heat of complete burning; bond-enthalpy bookkeeping gives ΔH = −83 kJ/mol for ammonia synthesis.

(a) Reversible process: A thermodynamic process that is carried out in an infinite number of infinitesimally small steps, so slowly that the system remains in virtual equilibrium with its surroundings at every stage. Such a process can be exactly reversed, restoring both system and surroundings to their original states, with no net change left behind.

(b) Standard enthalpy of combustion ( $\Delta_cH^\circ$ ): The enthalpy change accompanying the complete combustion (in excess oxygen) of exactly one mole of a substance, with all reactants and products in their standard states, at a specified temperature (usually 298 K) and 1 bar pressure.

Bond enthalpy calculation for $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ :

$\Delta H = \Sigma BE_{reactants} - \Sigma BE_{products}$ …

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