Chemistry Enthalpies for Different Types of Reactions

Topics Covered :

● Enthalpies for Different Types of Reaction
● Standard Enthalpy of Combustion
● Enthalpy of Atomization
● Bond Enthalpy
● Enthalpy of Solution
● Lattice Enthalpy

Enthalpies For Different Types of Reactions :

It is convenient to give name to enthalpies specifying the types of reactions.

Standard enthalpy of combustion (symbol : `Δ_cH^⊖`) :

`=>` Combustion reactions are exothermic in nature.

`=>` These are important in industry, rocketry, and other walks of life.

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
`color{green}("Definition ")` Standard enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) of a substance, when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature.

`=>` Cooking gas in cylinders contains mostly butane `color{purple}(C_4H_(10))`.

● During complete combustion of one mole of butane, `2658` `kJ` of heat is released. We can write the thermochemical reactions for this as :

`color{purple}(C_4H_(10) (g) +(13)/2 O_2 (g) → 4CO_2 (g) +5H_2 O (l) ; Delta_cH^(⊖) = -2658.0kJ mol^(-1))`

`=>` Similarly, combustion of glucose gives out `color{purple}(2802.0 kJ//mol)` of heat, for which the overall equation is :

`color{purple}(C_6H_(12)O_6 (g) +6O_2(g) → 6CO_2(g) +6H_2O (l) ; Delta_c H^⊖ = -2802.0 kJ mol^(-1))`

● Our body also generates energy from food by the same overall process as combustion, although the final products are produced after a series of complex bio-chemical reactions involving enzymes.
Q 3007123988

The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, `CO_2(g)` and `H_2O (1)` are produced and `3267.0 kJ` of heat is liberated. Calculate the standard enthalpy of formation, `Delta_f H^(⊖)` of benzene. Standard enthalpies of formation of `CO_2(g)` and ` H_2O( l )` are `–393.5 kJ mol^(–1 )`and `– 285.83 kJ mol^(–1)` respectively.


The formation reaction of benezene is given by :

`6C text{(graphite)} +3H_2 (g) →C_6H_6 (l) ; Delta_fH^(⊖) = ? ` ......(1)

The enthalpy of combustion of 1 mol of benzene is :

`C_6H_6 (l) +15/2 O_2 → 6CO_2 (g) +3H_2O(l) ; Delta_c H^(⊖) = -3267 kJ mol^(-1)` ........(2)

The enthalpy of formation of 1 mol of `CO_2(g)` :

`C text{(graphite)} + O_2 (g) → CO_2 (g) ; Delta_f H^(⊖) = - 393.5 kJ mol^(-1)` .........(3)

The enthalpy of formation of 1 mol of `H_2O(l)` is :

`H_2 (g) +1/2 O_2 (g) → H_2 O (l) ; Delta_f H^(⊖) = -285.83kJ mol^(-1)` .......(4)

multiplying eqn. (3) by 6 and eqn. (4) by 3 we get:

`6 C text{(graphite)} +6O_2 (g) → 6CO_2 (g) ; Delta_fH^(⊖) = -2361 kJ mol^(-1)`

`3H_2(g) +3/2O_2 (g) → 3H_2O (l) ; Delta_fH^(⊖) = -857.49 kJ mol^(-1)`

Summing up the above two equations :

`6C text{(graphite)} +3H_2 (g) +15/2 O_2 (g) → 6CO_2 (g) +3H_2O(l) ; Delta_fH^(⊖) = -3218.49 kJ mol^(-1)` ....(5)

Reversing equation (2);

`6CO_2 (g) +3H_2O (l) → C_6H_6 (l) +15/2 O_2 ; Delta_fH^(⊖) = 3267.0 kJ mol^(-1)` ......(6)

Adding equations (5) and (6), we get

`6C text{(graphite)} +3H_2 (g) → C_6H_6 (l) ; Delta_fH^(⊖) = 48.51 kJ mol^(-1)`

Enthalpy of atomization (symbol : `D_a H^⊖`) :

`=>` Consider the following example of atomization of dihydrogen

`color{purple}(H_2 (g) → 2 H (g) ; Delta_aH^⊖ = 435.0kJ mol^(-1))`

● You can see that `H` atoms are formed by breaking `H–H` bonds in dihydrogen. The enthalpy change in this process is known as enthalpy of atomization, `color{purple}(Delta_a H^(⊖))`.

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
Enthalpy of atomization: It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.

`color{red}("Note ")` In case of diatomic molecules, like dihydrogen (given above), the enthalpy of atomization is also the bond dissociation enthalpy.

`=>` The other examples of enthalpy of atomization can be

`color{purple}(CH_4 (g) → C (g) +4H (g) ; Delta_aH^⊖ = 1665 kJ mol^(-1))`

`color{red}("Note ")` The products are only atoms of `C` and `H` in gaseous phase.

`=>` Now see the following reaction :

`color{purple}(Na (s) → Na(g) ; Delta_a H^(⊖) = 108.4 kJ mol^(-1))`

● In this case, the enthalpy of atomization is same as the enthalpy of sublimation.

Bond Enthalpy (symbol : `Delta_text(bond) H^(⊖)`) :

`=>` Chemical reactions involve the breaking and making of chemical bonds.

● Energy is required to break a bond and energy is released when a bond is formed.

● It is possible to relate heat of reaction to changes in energy associated with breaking and making of chemical bonds.

`=>` With reference to the enthalpy changes associated with chemical bonds, two different terms are used in thermodynamics.

(i) Bond dissociation enthalpy

(ii) Mean bond enthalpy

`=>` Now discuss these terms with reference to diatomic and polyatomic molecules.

`color{green}("Diatomic Molecules ")` Consider the following process in which the bonds in one mole of dihydrogen gas `(H_2)` are broken :

`color{purple}(H_2 (g) → 2H (g) ; H_(H-H) H^(⊖) = 435.0 kJ mol^(-1))`

● The enthalpy change involved in this process is the bond dissociation enthalpy of `H–H` bond.

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
● The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase.

`color{red}("Note ")` (i) It is the same as the enthalpy of atomization of dihydrogen.

(ii) This is true for all diatomic molecules.

`color{red}("Example ")`

`color{purple}(Cl_2 (g) → 2Cl (g) ; Delta_(Cl -Cl)H^(⊖) = kJ mol^(-1))`

`color{purple}(O_2 (g) → 2O(g) ; Delta_(O = O)H^(⊖) = 428 kJ mol^(-1))`

`=>` In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule.

`color{green}("Polyatomic Molecules ")` Let us now consider a polyatomic molecule like methane, `color{purple}(CH_4)`.

● The overall thermochemical equation for its atomization reaction is given below :

`CH_4 (g) → C (g) +4 H (g) ; Delta_a H^(⊖) = 1665 kJ mol^(-1)`

● In methane, all the four `C – H` bonds are identical in bond length and energy.

● However, the energies required to break the individual `C – H` bonds in each successive step differ :

`color{purple}(CH_4 (g) → CH_3 (g) +H(g) ; Delta_text(bond)H^(⊖) = +427 kJ mol^(-1))`

`color{purple}(CH_3 (g) → CH_2 (g) +H(g) ; Delta_text(bond)H^(⊖) = +439kJ mol^(-1))`

`color{purple}(CH_2 (g) → CH (g) +H (g) ; Delta_text(bond)H^(⊖) = +452kJ mol^(-1))`

`color{purple}(CH(g) → C(g) +H (g) ; Delta_text(bond)H^(⊖) = +347 kJ mol^(-1))`

Therefore `color{purple}(CH_4(g) → C(g) +4H (g) ; Delta_a H^(⊖) = 1665 kJ mol^(-1))`

● In such cases we use mean bond enthalpy of `C – H` bond.

● For example in `color{purple}(CH_4 , Delta_(C-H) H^(⊖)) ` is calculated as :

`color{purple}(Delta_(C-H) H^(⊖) = 1/4 ( Delta_a H^(⊖)) = 1/4 ( 1665 kJ mol^(-1) ))`

` color{purple}(= 416 kJ mol^(-1))`

● We find that mean `C–H` bond enthalpy in methane is `416 kJ//mol`.

● It has been found that mean `C–H` bond enthalpies differ slightly from compound to compound, as in `CH_3 CH_2 Cl`,`CH_3 NO_2`, etc, but it does not differ in a great deal.

`color{red}("Note ")` Symbol used for bond dissociation enthalpy and mean bond enthalpy is the same.

`=>` Using Hess’s law, bond enthalpies can be calculated.

`=>` Bond enthalpy values of some single and multiple bonds are given in Table 6.3.

`=>` The reaction enthalpies are very important quantities as these arise from the changes that accompany the breaking of old bonds and formation of the new bonds.

`=>` We can predict enthalpy of a reaction in gas phase, if we know different bond enthalpies.

`=>` The standard enthalpy of reaction, `color{purple}(D_rH^(⊖))` is related to bond enthalpies of the reactants and products in gas phase reactions as :

`color{purple}(Delta_r H^(⊖) = Sigma)`bond enthalpies `color{purple}(text()_text(reactants) - Sigma text(bond enthalpies)_text(products))`

● This relationship is particularly more useful when the required values of `color{purple}(Delta_f H^(⊖))` are not available.

● The net enthalpy change of a reaction is the amount of energy required to break all the bonds in the reactant molecules minus the amount of energy required to break all the bonds in the product molecules.

● Remember that this relationship is approximate and is valid when all substances (reactants and products) in the reaction are in gaseous state.

Enthalpy of Solution (symbol : `Delta_text(sol)H^(⊖)`) :

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
`color{green}("Enthalpy of Solution ")` Enthalpy of solution of a substance is the enthalpy change when one mole of it dissolves in a specified amount of solvent.

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
● The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions (or solute molecules) are negligible.

`=>` When an ionic compound dissolves in a solvent, the ions leave their ordered positions on the crystal lattice.

● These are now more free in solution. But solvation of these ions (hydration in case solvent is water) also occurs at the same time.

● This is shown diagrammatically, for an ionic compound, `AB (s)`.

`=>` The enthalpy of solution of `color{purple}(AB(s)`, `Delta_text(sol)H^(⊖))`, in water is, therefore, determined by the selective values of the lattice enthalpy, `color{purple}Delta_text(lattice) H^(⊖))` and enthalpy of hydration of ions, `color{purple}(Delta_text(hyd)H^(⊖))` as

`color{purple}(Delta_text(sol) H^(⊖) = Delta_text(lattice) H^(⊖) + Delta_text(hyd)H^(⊖))`

`=>` For most of the ionic compounds, `color{purple}(D_text(sol) H^(⊖))` is positive and the dissociation process is endothermic.

● Therefore the solubility of most salts in water increases with rise of temperature.

● If the lattice enthalpy is very high, the dissolution of the compound may not take place at all.

`=>` Why do many fluorides tend to be less soluble than the corresponding chlorides?

● Estimates of the magnitudes of enthalpy changes may be made by using tables of bond energies (enthalpies) and lattice energies (enthalpies).

Lattice Enthalpy :

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
`color{green}("Definition ")` The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

`color{purple}(Na^(+) Cl^(-) (s) → Na^(+) (g) +Cl^(-) ; Delta_text(lattice) H^(⊖) = +788 kJ mol^(-1))`

`=>` Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a `color{red}("Born-Haber Cycle")` (Fig. 6.9).

`=>` Let us now calculate the lattice enthalpy of `color{purple}(Na^(+)` `Cl^(-) (s))` by following steps given below :

(i) `color{purple}(Na(s)→ Na(g))` , sublimation of sodium metal , `color{purple}(Delta_text(sub)H^(⊖) = 108.4 kJ mol^(-1))`

(ii) `color{purple}(Na(g) → Na^(+) (g) e^(-1) (g) )` , the ionization of sodium atoms, ionization enthalpy `color{purple}(Delta_t H^(⊖) = 496 kJ mol^(-1))`

(iii) `color{purple}(1/2 Cl_2 (g) → Cl (g)) `, the dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy.

`color{purple}(1/2 Delta_text(bond) H^(⊖) = 121 kJ mol^(-1))`

(iv) `color{purple}(Cl(g)+ e^(-1) (g) → C barl(g))` electron gained by chlorine atoms. The electron gain enthalpy, `color{purple}(Delta_(eg) H^(⊖)=
–348.6 kJ mol^(–1))` .

● Ionization enthalpy and electron gain enthalpy have been taken from thermodynamics.

● Earlier terms, ionization energy and electron affinity were in practice in place of the above terms.

(v) `color{purple}(Na^(+) (g) Cl^(-) (g) → Na^(+) Cl^(-) (s))`

● The sequence of steps is shown in Fig. 6.9, and is known as a Born-Haber cycle.

`text(Importance :)` The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.

● Applying Hess’s law, we get,

`color{purple}(Delta_text(lattice) H^(⊖) = 411.2+108.4+121+496-348.6)`

`color{purple}(Delta_text(lattice) H^(⊖) = +788 kJ)`

for `color{purple}(NaCl (s) → Na^(+) + Cl^(-) (g))`

● Internal energy is smaller by `2RT` (because `color{purple}(Deltan_g =2)`) and is equal to `+ 783 kJ mol^(–1)`.

`=>` Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression :

`color{purple}(Delta_text(sol) H^(⊖) = Delta_text(lattice) H^(⊖) + Delta_text(hyd) H^(⊖))`

For one mole of `color{purple}(NaCl(s)),` lattice enthalpy `= + 788 kJ mol^(–1)` and `color{purple}(Delta_text(hyd)^(⊖)) = -784 kJ mol^(-1) ` ( from the literature)

`color{purple}(Delta_text(sol)H^(⊖) = +784 kJ mol^(-1) -784 kJ mol^(-1) = +4 kJ mol^(-1))`

● The dissolution of `color{purple}(NaCl(s))` is accompanied by very little heat change.