Skip to content

Chemistry · Ch 4 — Chemical Thermodynamics

Bond enthalpy

4.10.8

Bond enthalpy

Consider the reaction

H2(g)⟶H(g)+H(g),ΔrH0=436.4 kJ\mathrm{H_2(g)} \longrightarrow \mathrm{H(g) + H(g)}, \quad \Delta_r H^0 = 436.4\ \mathrm{kJ}

It shows that the H-H bond in one mole of H2_2(g) is decomposed, producing gaseous H atoms. The enthalpy change of the reaction, 436.4 kJ, is the bond enthalpy of the H-H bond. The enthalpy change required to break a particular covalent bond in one mole of gaseous molecule to produce gaseous atoms and/or radicals, is called bond enthalpy.

Remember

Remember...

For diatomic molecules, the bond enthalpy is the same as the enthalpy of atomization.

The HCl molecule dissociates as

HCl(g)⟶H(g)+Cl(g),ΔrH0=431.9 kJ\mathrm{HCl(g)} \longrightarrow \mathrm{H(g) + Cl(g)}, \quad \Delta_r H^0 = 431.9\ \mathrm{kJ}

ΔH0 (H-Cl bond)=431.9 kJ mol−1\Delta H^0\,(\text{H-Cl bond}) = 431.9\ \mathrm{kJ\,mol^{-1}}

Average bond enthalpy in polyatomic molecules : Each covalent bond in polyatomic molecules is associated with its own specific bond enthalpy. The thermochemical equation for the dissociation of H2_2O molecules is

H2O(g)⟶2 H(g)+O(g),ΔrH0=927 kJ\mathrm{H_2O(g)} \longrightarrow 2\,\mathrm{H(g)} + \mathrm{O(g)}, \quad \Delta_r H^0 = 927\ \mathrm{kJ}

The above equation implies that the enthalpy change for breaking the two O-H bonds in one mole of gaseous H2_2O molecules is 927 kJ. Though the two O-H bonds in H2_2O are identical, the energies needed to break the individual O-H bonds are different. The bonds in H2_2O are broken in successive steps, as shown:

i.  H2O(g)⟶OH(g)+H(g),ΔrH0=499 kJ\mathrm{i.\ \ H_2O(g)} \longrightarrow \mathrm{OH(g) + H(g)}, \quad \Delta_r H^0 = 499\ \mathrm{kJ}

ii.  OH(g)⟶O(g)+H(g),ΔrH0=428 kJ\mathrm{ii.\ \ OH(g)} \longrightarrow \mathrm{O(g) + H(g)}, \quad \Delta_r H^0 = 428\ \mathrm{kJ}

Adding (the book sums the two steps above a horizontal rule):

H2O(g)⟶2 H(g)+O(g),ΔrH0=927 kJ\mathrm{H_2O(g)} \longrightarrow 2\,\mathrm{H(g) + O(g)}, \quad \Delta_r H^0 = 927\ \mathrm{kJ}

The total enthalpy change is 927 kJ — not twice as large as either single O-H bond enthalpy. What is the enthalpy of the O-H bond in the H2_2O molecule? For polyatomic molecules, the average bond enthalpy of a particular bond is considered. Thus, the average bond enthalpy of the O-H bond is

9272=463.5 kJorΔH0(O-H)=463.5 kJ mol−1\frac{927}{2} = 463.5\ \mathrm{kJ} \quad \text{or} \quad \Delta H^0(\text{O-H}) = 463.5\ \mathrm{kJ\,mol^{-1}}

Note

Do you know?

In the CH4_4 molecule there are four identical C-H bonds, yet the bond enthalpies of all four C-H bonds are different. The breaking of the C-H bonds in CH4_4 occurs in four steps:

CH4(g)→CH3(g)+H(g),ΔrH0=427 kJ\mathrm{CH_4(g)} \rightarrow \mathrm{CH_3(g) + H(g)}, \quad \Delta_r H^0 = 427\ \mathrm{kJ}

CH3(g)→CH2(g)+H(g),ΔrH0=439 kJ\mathrm{CH_3(g)} \rightarrow \mathrm{CH_2(g) + H(g)}, \quad \Delta_r H^0 = 439\ \mathrm{kJ}

CH2(g)→CH(g)+H(g),ΔrH0=452 kJ\mathrm{CH_2(g)} \rightarrow \mathrm{CH(g) + H(g)}, \quad \Delta_r H^0 = 452\ \mathrm{kJ}

CH(g)→C(g) +H(g),ΔrH0=347 kJ\mathrm{CH(g)} \rightarrow \mathrm{C(g)\ + H(g)}, \quad \Delta_r H^0 = 347\ \mathrm{kJ}

Adding the four steps: CH4(g)→C(g)+4 H(g)\mathrm{CH_4(g)} \rightarrow \mathrm{C(g)} + 4\,\mathrm{H(g)}, ΔrH0=1665\Delta_r H^0 = 1665 kJ. The average C-H bond enthalpy =1665 kJ/4=416 kJ= 1665\ \mathrm{kJ}/4 = 416\ \mathrm{kJ}. Hence, ΔrH0(C-H)=416 kJ mol−1\Delta_r H^0(\text{C-H}) = 416\ \mathrm{kJ\,mol^{-1}}.

Reaction and bond enthalpies : In a chemical reaction, bonds are broken and formed. The enthalpies of reactions involving substances having covalent bonds are calculated by knowing the bond enthalpies of reactants and those in products. The calculations assume that all the bonds of a given type are identical.

Enthalpy change of a reaction

ΔrH0=∑ΔH0 (reactant)−∑ΔH0 (product)...(4.31)\Delta_r H^0 = \sum \Delta H^0\,(\text{reactant}) - \sum \Delta H^0\,(\text{product}) \qquad \text{...(4.31)} …

Start your 14-day free trial to unlock the full solution →
← Previous · 4.10.7Problem 4.11 — Estimate the standard enthalpy of combustion of CH₄(g) if Δf H⁰(CH₄) =…Next · 4.10.8 →Q1 — Write equation for bond enthalpy of Cl-Cl bond in mathrmCl₂ molecule. …