Hess’s Law and Bond Energy

HESS'S LAW

This law states that the amount of heat evolved or absorbed in a process, including a chemical change is the same whether the process takes place in one or several steps.

Suppose in a process the system changes from state A to state B in one step and the heat exchanged in this change is q. Now suppose the system changes from state A to state B in three steps involving a change from A to C, C to D and finally from D to B. If q₁, q₂ and q_­3 are the heats exchanged in the first, second and third step, respectively then according to Hess's law

q₁ + q₂ + q₃ = q

Hess's law is simply a corollary of the first law of thermodynamics. It implies that enthalpy change of a reaction depends on the initial and final state and is independent of the manner by which the change is brought about.

Illustration 1. In this case express $\Delta$H in terms of $\Delta$H₁, $\Delta$H₂, $\Delta$H₃ .

Solution: $\Delta$H = $\Delta$H₁ + $\Delta$H₂ + $\Delta$H₃

Illustration 2. H₂O (l)$\to$ H₂(g) + $\frac{1}{2}$O₂(g) $\Delta$H = + 890.36 kJ / mole

What is $\Delta$H for H₂O (l) from its constituent elements

Solution: H₂O(l) $\to$ H₂(g) + $\frac{1}{2}{O}_2$(g) $\Delta$H = + 890.36 kJ / mole

H₂(s) + $\frac{1}{2}$O₂(g) $\to$ H₂O(l) $\Delta$H = – 890.36 kJ / mole

$\therefore$ ${\left((\Delta H_f\right)}_{H_{2}O}$ = –890.36 kJ / mole

APPLICATION OF HESS'S LAW

1. Calculation of enthalpies of formation

There are large number of compounds such as C₆H₆, CO, C₂H₆ etc whose direct synthesis from their constituent element is not possible. Their $\Delta$H⁰_f values can be determined indirectly by Hess's law. e.g. let us consider Hess's law cycle for CO₂ (g) to calculate the $\Delta$H⁰_f of CO(g) which can not determined otherwise.

$C\left(s\right)+O_2\left(g\right)\to CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\Delta H_1$= ?

$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)\Delta H_2=-283KJ/mole$

$C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)\Delta H_3=-0393KJ/mole$

According to Hess's law,

$\Delta H_3=\Delta H_1+\Delta H_2$ or $\Delta H_1=\Delta H_3-\Delta H_2$

= -393 - (-283) $\Rightarrow$ -110 KJ/mole

2. Calculation of standard Enthalpies of reactions

From the knowledge of the standard enthalpies of formation of reactants and products the standard enthalpy of reaction can be calculated using Hess's law.

According to Hess's law

$\sum \Delta H_f^0\left(P\right)=\sum \Delta H_f^0\left(R\right)+\Delta H^0$

$\therefore \Delta H^0=\sum \Delta H_f^0\left(P\right)-\sum \Delta H_f^0\left(R\right)$

$\therefore$$\mathrm{\Delta }{\mathrm{H}}^0=\left[\begin{array}{l}\mathrm{sum}\mathrm{of}\mathrm{standard}\mathrm{enthalpies}\mathrm{of}\\ \mathrm{formation}\mathrm{of}\mathrm{products}\end{array}\right]-\left[\begin{array}{l}\mathrm{sum}\mathrm{of}\mathrm{standard}\mathrm{enthalpies}\\ \mathrm{of}\mathrm{formation}\mathrm{of}\mathrm{reactants}\end{array}\right]$

3. In the calculation of bond energies

BOND ENERGY

Bond energy for any particular type of bond in a compound may be defined as the average amount of energy required to dissociate one mole, viz Avogadro's number of bonds of that type present in the compound. Bond energy is also called the enthalpy of formation of the bond.

Calculation:

For diatomic molecules like H₂, O₂, N₂, HCl, HF etc, the bond energies are equal to their dissociation energies. For polyatomic molecules, the bond energy of a particular bond is found from the values of the enthalpies of formation. Similarly the bond energies of heteronuclear diatomic molecules like HCl, HF etc can be obtained directly from experiments or may be calculated from the bond energies of homonuclear diatomic molecules.

Illustration 3. Calculate the bond energy of HCl. Given that the bond energies of H₂ and Cl₂ are 430 KJmol^-1 and 242 KJ mol^-1 respectively and $\Delta$H⁰_f for HCl is -91 KJ mol^-1.

Solution: $H_2\left(g\right)\to 2H\left(g\right)\Delta H=+430KJmo{l}^{-1}...(i)$

$C{l}_2\left(g\right)\to 2Cl\left(g\right)\Delta H=+242KJmo{l}^{-1}...(ii)$

$HCl\to H\left(g\right)+Cl\left(g\right)\Delta H=?...(iii)$

For the reaction (iii)

$\Delta$H = $\sum \Delta H_f^0\left(product\right)-\sum \Delta H_f^0\left(reac\tan t\right)$

= $\left[\Delta H_f^0\left(H\right)+\Delta H_f^0\left(Cl\right)\right]-\left[\Delta H_f^0\left(HCl\right)\right]$

= $\frac{1}{2}\times 430+\frac{1}{2}\times 242-\left(-91\right)$

$\Delta$H = 427 KJ mol^-1

Illustration 4. Given that

2H₂(g) + O₂(g) $\to$ H₂O(g) , $\Delta$H = –115.4 kcal the bond energy of H–H and O = O bond respectively is 104 kcal and 119 kcal, then the O–H bond energy in water vapour is

(A) 110.6 kcal / mol (B) –110.6 kcal

(C) 105 kcal / mol (D) None

Solution: We know that heat of reaction

$\Delta$H = $\sum$B.E. (reactant) – $\sum$B.E (product)

For the reaction,

2H–H(g) + O = O (g) $\to$ 2H – O–H(g)

$\Delta$H = –115.4 kcal, B.E. of H–H = 104 kcal

B.E. of O=O = 119 kcal

Since one H₂O molecule contains two O–H bonds

–115.4 = (2 x 104) + 119 – 4 (O–H) bond energy

$\therefore$4 (O–H) bond energy = (2x 104) +119+115.4

i.e., O–H bond energy =$\frac{\left(2\times 104\right)+119+115.4}{4}$ = 110.6 kcal mol^–1

Hence, (A) is correct.

Illustration 5. Given the bond energies of N $\equiv$ N, H – H and N – H bonds are 945, 436 and 391 kJ/mol respectively, the enthalpy of the reaction.

N_2(g) + 3H_2(g) $\to$ 2NH_3(g) is

(A) – 93kJ (B) 102kJ

(C) 90kJ (D) 105kJ

Solution: (A)

LATTICE ENERGY OF AN IONIC CRYSTAL (BORN–HABER CYCLE)

The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from its gaseous ions, is known as Lattice energy.

Step 1: Conversion of metal to gaseous atoms

M(s) $\to$ M(g) , $\Delta$H₁ = sublimation

Step 2: Dissociation of X₂ molecules to X atoms

X₂(g) $\to$ 2X (g), $\Delta$H₂ = Dissociation energy

Step 3: Conversion of gaseous metal atom to metal ions by losing electron

M(g) $\to$ M⁺ (g) + e^–, $\Delta$H₃ = (Ionization energy)

Step 4: X(g) atoms gain an electron to form X^– ions

X(g) + e^– $\to$ X^–(g), $\Delta$H₄ = Electron affinity

Step 5: M⁺ (g) and X^– (g) get together and form the crystal lattice

M⁺ (g) + X^– (g) $\to$ MX(s) $\Delta$H₅ = lattice energy

Applying Hess's law we get

$\Delta$H₁ + 1/2 $\Delta$H₂ + $\Delta$H₃ + $\Delta$H₄ + $\Delta$H₅ = $\Delta$H_f (MX)

On putting the various known values, we can calculate the lattice energy.

Illustration 6. What is the expression of lattice energy (U) of CaBr_­2 using BornHaber cycle?

Solution:

= S + IE_­1 + IE₂ + D - 2E.A –$U_{CaB{r}_2}$

Illustration 7. What is the relation between $\Delta$H and $\Delta$E in this reaction?

CH₄(g) + 2O₂(g) $\rightleftharpoons$ CO₂(g) + 2H₂O(l)

Solution: $\Delta$H = $\Delta$E + $\Delta$nRT

$\Delta$n = no. of mole of products - no. of moles of reactants = 1– 3 = –2

$\Delta$H = $\Delta$E – 2RT

Illustration 8. What is the expression of lattice energy (U) of CaBr_­2?

Using Born Haber cycle?

Solution:

$\Delta H_f$= S + IE_­1 + IE₂ + D - 2E.A –$U_{CaB{r}_2}$

Illustration 9. The lattice energy of solid NaCl is 180 kcal/mole. The dissolution of the solid in water in the forms of ions is endothermic to the extent of 1 kcal/mol. If the solution energies of $N{a}^{+}$ and $C{l}^{-}$ are in the ratio 6:5, what is the enthalpy of hydration of $N{a}^{+}$ion?

(A) - 85.6 kcal/mol (B) -97.5 kcal/mol

(C) 82.6 kacl/mol (D) +100 kcal/mol

Solution: (B)

Application of bond energies

(i) Determination of enthalpies of reactions

Suppose we want to determine the enthalpy of the reaction.

If bond energies given for C $-$ C, C = C, C$-$H, and H $-$ H are 347.3, 615.0, 416.2 and

435.1KJ mol^-1 respectively.

$\Delta H=\Delta H_{C=C}+\Delta H_{H-H}+4\Delta H_{C-H}-\left(\Delta H_{C-C}+6\Delta H_{C-H}\right)$

= (615.0 + 435.1) - (347.3 + 832.4) $\Rightarrow$ -129.6 KJ

(ii) Determination of enthalpies of formation of compounds

Consider the formation of acetone.

$\Delta H_f=\left[3\Delta H_{H-H}+\frac{1}{2}\Delta H_{O-O}+3\Delta H_{C_{\left(s\right)}\to C_{\left(g\right)}}\right]-\left[2\Delta H_{C-C}+6\Delta H_{C-H}+\Delta H_{C=O}\right]$

by putting the value of different bond energies you can determine the $\Delta$H_f.

(iii) Determination of resonance energy

If a compound exhibits resonance, there is a considerable difference between the enthalpies of formation as calculated from bond energies and those determined experimentally. As an example we may consider the dissociation of benzene.

$C_6{H}_6\left(g\right)\to 6C\left(g\right)+6H\left(g\right)$

Assuming that benzene ring consists of three single and three double bonds (Kekule's structure) the calculated dissociation energy comes out to be 5384.1 KJ from bond energies data.

$\Delta H_d=3\Delta H_{C-C}+3\Delta H_{C=C}+6\Delta H_{C-H}$

The experimental value is known to be 5535.1 KJ/mol. Evidently, the energy required for the dissociation of benzene is 151 KJ more that the calculated value. The difference of 151 KJ gives the resonance energy of benzene.