Heat of neutralization, heat of solution, heat of combustion, heat of vapourization, heat of formation and bond energy

The calorimeter is kept open to the atmosphere in the determination of change in enthalpy of a reaction. The calorimeter is immersed in an insulated water bath fitted with stirrer and thermometer. The temperature of the bath is recorded in the beginning and after the end of the reaction and the change in temperature is calculated. Knowing the heat capacity of water bath and calorimeter and also the change in temperature the heat absorbed or evolved in the reaction is calculated. This gives the enthalpy change (DH) of the reaction.

Enthalpy of Combustion

The enthalpy of combustion of a compound is the enthalpy change at normal pressure and at constant temperature accompanying complete combustion of one mole of the compound. It is denoted by DH_c. Combustion here means the burning of the given compound to the highest oxides of the constituent elements in the presence of excess of oxygen.

For example, the enthalpy of combustion of benzene at 298 K is the enthalpy change of the reaction.

Thus, DH_comb C₆H_6(l) = -3268kJmol^-1_.

Enthalpy of combustion is generally obtained experimentally. For cases, where it is not possible to measure it experimentally, it is estimated from the enthalpies of formation of the various compounds involved in the process.

Problems

11. The heats of combustion of CH₄ and C₂H₆ are -890.3 and -1559.7kJ mol^-1 respectively. Which of the two has greater efficiency of fuel per gram?

Solution

The fuel efficiency can be predicted from the amount of heat evolved for every gram of fuel consumed.

(i) The combustion of methane is as follows:

DHc = 890.3 kJmol-1

Molar mass of CH₄ = 16

(ii) The combustion of ethane is as follows:

DH_c = 1559.7kJmol^-1

Molar mass of C₂H₆ = 30

Thus, methane has greater fuel efficiency than ethane.

12. (a) A cylinder of gas supplied by a company is assumed to contain 14 kg of Butane. If a normal family requires 20,000 kJ of energy per day for cooking, how long will the cylinder last?

(b) If the air supplied to the burner is insufficient, a portion of gas escapes without combustion. Assuming that 25% of the gas is wasted due to this inefficiency, how long will the cylinder last? (Heat of combustion of butane = 2658 kJ/mol).

Solution

(a) Molecular formula of butane = C₄H₁₀

Molecular mass of butane = 4 x 12 + 10 x 1 = 58

Heat of combustion of butane = 2658kJmol^-1

1 mole of 58 g of butane on complete combustion give heat = 2658 kJ

14 x 10³ g of butane on complete combustion gives heat =

The family needs 20,000 kJ of heat for cooking per day.

641586 kJ of heat will be used for cooking by a family in

The cylinder will last for 82 days

(b) 25 per cent of the gas is wasted due to inefficiency. This means that only 75% of butane is combusted. Therefore,

the energy produced by 75% combustion of butane

13. The enthalpy change involved in the oxidation of glucose is - 2880 kJ mol^-1. Twenty five percent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometer, what is the maximum distance that a person will be able to walk after eating 120 g of glucose.

Solution

DH_comb of Glucose (C₆H₁₂O₆) = - 2880 kJ mol^-1

14. Calculate the enthalpy change of combustion of cyclopropane at 298 K. The enthalpy of formation of CO_2(g) , H₂O_(l) and propane_(g) are -393.5, -285.8 and 20.42 kJmol^-1 respectively. The enthalpy of

isomerisation of cyclopropane to propene is -33.0 kJ mol^-1.

Solution

The required DH is

The given equations are:

Multiply equation (i) and (ii) by 3 and add them. Now subtract equation (iii) and subsequently add equation (iv) from the resulting expression.

DH = 3DH₁ + 3DH₂ - DH₃ - DH₄

= 3(- 393.5) + 3(- 285.8) - (20.42) + (-33.0) = - 2091.32 kJ.

Heat or Enthalpy of Neutralisation

The reaction in which an acid and a base react to give a salt and water is called neutralization reaction. Neutralization reactions are exothermic in nature. The heat change when one gram equivalent of an acid is completely neutralised by a base or vice versa in dilute solution, is called heat of neutralization.

Examples of heat of neutralization are:

Neutralization of HCl with NaOH

Neutralization of CH₃COOH with NaOH

DH =-55.9 kJ

It is important to note that the term gram equivalent is used in the definition of heat of neutralization. This is because neutralization involves 1 mole of H⁺ ions and 1 mole of OH^- ions to form 1mole of water and 57.1 kJ of heat is liberated.

Now, one gram equivalent of various acids on complete dissociation liberates one mole of H⁺ ions. But one mole of the acid may produce more than one mole of H⁺ ions in solution depending upon its basicity; for example 1mol of H₂SO₄ gives 2 mol of H⁺ ions and 1mol of H₃PO₄ gives 3 mol of H⁺ ions on complete dissociation. But 1gram equivalent of both (H₂SO₄ or H₃PO₄) produces only 1 mol of H⁺ions.

Thus, it is more appropriate to use the term gram equivalent in the definition of enthalpy of neutralization.

The average enthalpy of neutralization of any strong acid by a strong base is found to be - 57.7 kJ (- 13.7 kcal) irrespective of the nature of acid or the base. This suggests that the net chemical reaction in all neutralization reactions is the same, viz.,

This is because strong acids and strong bases are completely ionized in aqueous solutions. The aqueous solution of one gram equivalent of all strong acids contains the same number of H⁺ ions. Similarly, aqueous solution of one gram equivalent of all strong bases also contains same number of OH^-. The neutralization reactions between strong acids and strong bases in aqueous solutions involve simply the combination of H⁺ ions (from an acid) and OH^- ions (from a base) to form unionized water molecules.

For example, in the reaction between hydrochloric acid and sodium hydroxide ion. The neutralization can be represented as:

DH = -57.1kJ

Cancelling common ions:

Neutralization of weak acids and weak bases

The heat of neutralization of a weak acid or a weak base is less than

-57.1 kJ and is also different for different weak acids or bases.

For example for acetic acid the enthalpy of neutralization is -54.9 kJ. This can be explained as follows:

Acetic acid (CH₃COOH) is a weak acid. Weak acids (or weak bases) are ionised to a small extent in solutions. So acetic acid is only partially ionised in solution. Since the neutralization involves a reaction between H⁺(from the acid) and OH^- (from the base), hence acetic acid must be fully ionised as per the reaction,

DH = +1.2 kJ

The ionization reaction is endothermic reaction. So, during ionization of acetic acid a small amount of heat (1.2 kJ) is absorbed. As a result, the enthalpy of neutralization of acetic acid is 1.2 kJ less than that for a strong acid-strong base pair.

Therefore, the aqueous solutions containing one gram equivalent of different weak acids do not contain 1 gm equivalent of H⁺ ions. Similarly, the aqueous solutions containing 1 gram equivalent of different weak bases do not contain 1 gram equivalent of OH^- ions. The weak acids, weak bases therefore, have to be dissociated to give 1 gram equivalent of H⁺ or OH^- ions but, neutralisation of weak acid and strong base (or a weak base and strong acid) not only involves the combination of H⁺ and OH^- ions but also the dissociation of a weak acid (or a weak base). The dissociation process is accompanied by the absorption of energy. This energy is called the heat of dissociation. Therefore, the overall liberated energy is less than 57.1 kJ (i.e., 57.1 of dissociation of acid or base).

The neutralization of acetic acid with sodium hydroxide can be explained as follows:

DH = -55.9kJ

Thus, heat of neutralization of acetic acid and sodium hydroxide is

-55.9 kJ, because 1.2 kJ of heat energy is used up in dissociating acetic acid. Similarly, heat of neutralization of ammonium hydroxide and hydrochloric acid is -51.5 kJ as 5.6 kJ is the heat of dissociation of NH₄OH.

Problems

15. 100 ml of 1N of an acid and 100 ml of 1N of a base are mixed at 298K. During the experiment, the temperature arose to 298.0067 K. Calculate the heat of neutralization.

Solution

Heat capacity of solution = Mass of solution x Specific heat capacity

Total mass of solution = 100 + 100 = 200 ml

Heat capacity of solution = 200 x 4.2 = 840 JK^-1

Heat change in the reaction = Heat capacity x Rise in temperature

= (840 JK^-1) (298.0067 - 298)K

= 840 x 0.0067 J = 5.63 J

Now, one gram equivalent of acid = 1N HCl in 1000 ml

100 ml of 1N acid on neutralization gives heat = 5.63 J

= 56.3J

Heat of neutralization = -56.3 J

16. Whenever an acid is neutralized by a base, the net reaction is:

Calculate the heat evolved for the following experiments:

(a) 0.60 mol of HNO₃ solution is mixed with 0.30 mol of KOH solution.

(b) 400 cm³ of 0.2 M H₂SO₄ is mixed with 600 cm³ of 0.1 M NaOH solution.

Solution

According to the reaction,

When 1 mole of H⁺ ions and 1 mol of OH^- ions are neutralized to form 1 mol of water, 57.1 kJ of energy is released.

(a) Heat evolved on mixing 0.60 mol of HNO₃ with 0.30 mol of KOH solution.

Since HNO₃ and KOH are strong acids and bases,

O.60 mol of HNO₃ 0.60 mol of H⁺ ions

0.30 mol of KOH 0.30 mol of OH^- ions

In this case, out of 0.60 mol of H⁺ ions (from HNO₃) only 0.30 mol will be neutralised (equal to OH^- ions present) by the base. 0.3 mol of H⁺ ions of HNO₃ will remain unreacted. The net reaction is:

Now, heat evolved during the formation of 1 mol of H₂O = 57.1kJ

Heat evolved in the formation of 0.3 mol of H₂O=57.1 x 0.3=17.13 kJ.

DH_sol)">

Enthalpy of solution (DH_sol)

When a solute is dissolved in a solvent a solution is formed. During dissolution of a solute in any solvent, a certain amount of heat is either absorbed or evolved. Such heat changes under constant pressure conditions are known as the enthalpy of solution. 'The change in enthalpy when one mole of a solute is dissolved in a specified quantity of a solvent at a given temperature is called enthalpy of solution'.

To avoid the amount of solvent, heat of solution is usually defined for an infinite dilute solution. Thus, heat of solution at infinite dilution is the heat change when one mole of a substance is dissolved in such a large quantity of solvent so that further dilution does not give any further heat change.

For example, dissolution of sodium chloride

Here 'aq' represents aqueous meaning a large excess of water.

For substances, which dissolve with the absorption of heat (endothermic), the enthalpy of solution is positive while for the substances which dissolve by liberating heat (exothermic), the enthalpy of solution is negative.

For example, when KCl is dissolved in water, heat is absorbed. Thus, the enthalpy of solution of KCl is positive. For a 200 times dilution

(water : KCl = 200 : 1), the enthalpy change during the process,

So, the enthalpy of solution of KCl at a dilution of 200 is 18.6 kJ mol^-1. The dissolution of CaCl₂(s) in water is an exothermic process. So, the enthalpy of solution of calcium chloride (CaCl₂) is negative. At a dilution of 400, the enthalpy change for the reaction,

So, the enthalpy of solution of CaCl₂(s) at a dilution of 400 is

-75.3 kJ mol^-1.

DH_fus)">

Enthalpy of fusion (DH_fus)

The enthalpy of fusion of a substance is defined as 'the change in enthalpy when one mole of a solid substance is melted at its melting temperature'. For example, the enthalpy change of the reaction,

is the enthalpy of fusion of ice. Enthalpy of fusion for some common substances are given below:

Substance: Ethanol Oxygen Hydrogen sulphide Sodium chloride

(C₂H₅OH) (O₂) (H₂S) (NaCl)

H_fus kJmol^-1: +4.8 +0.45 +2.0 +29

Melting point: 156 K 55 K 188 K 1074 K

From the data given above, we see that the enthalpies of fusion for ionic substances are very high. This is due to strong Coulombic forces between the constituent ions in ionic solids. The solids such as O₂and H₂S, which are molecular solids, have low heats of fusion because the forces of attraction between their molecules are weak forces. Thus, the heats of fusion of substances give an idea of nature of solid and the magnitude of forces acting between the particles constituting the solids._vap)">

Enthalpy of Vapourisation (_vap)

'The change in enthalpy when one mole of a liquid is converted into vapours at its boiling temperature is called enthalpy of vapourisation' (DHvap).

Thus, the enthalpy change of the reaction

is the enthalpy of vapourisation of water.

The enthalpies of vapourisation for certain common liquids are:

Substance	He(l)	H2(l)	O2(l)	HCl(l)	H2O(l)	NaCl(l)
Boiling point	4K	20K	90K	188K	373K	1738 K
ΔHvap kJmol -1	0.1	0.9	6.7	16.2	40.6	170

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From the data given above, we can say

DH_vap (ionic liquids) > DH_vap (polar liquids) > DH_vap (non-polar liquids)

Thus, the DH_vap depends upon the strength of intermolecular forces in any liquid.

Problem

17. Determine the value of DH and DE for the reversible isothermal evaporation of 90.0 g of water at 100°C. Assume that water vapours behave as ideal gas and heat of evaporation of water is 540 cal g ^-1.

(R = 2 cal mol^-1 K^-1)

Solution

Heat of evaporation of 1 g of water = 540 cal

Heat of evaporation of 90 g of water = 540 x 90 = 48600 Cal.

D H = 48600 Cal

The evaporation of 5 mol of water is represented as

Dn = (5 - 0) = 5

DH = DE + DnRT or DE = DH - DnRT

= 48600 - (5) (2.0) (373) = 44870 cal

DH_sub)">

Enthalpy of Sublimation (DH_sub)

Sublimation is a process in which a solid substance directly changes into its vapours at any temperature below its melting point. Enthalpy of sublimation is defined as follows:

The change in enthalpy when one mole of a solid substance is converted into its vapours without melting at a temperature below its melting point is called the enthalpy of sublimation.

For example, when one mole of solid iodine is converted into its vapours at room temperature, heat equal to 62.4 kJ is absorbed. So, the enthalpy of sublimation of iodine is + 62.4 kJ mol^-1, i.e.,

Compounds, which sublime on heating are camphor, dry ice, ammonium chloride etc.

The heat of sublimation is related to heat of fusion and heat of vaporization as:

DH_sublimation = DH_fusion + DH_vaporization

Problem

18. When 1 g of liquid naphthalene (C₁₀H₈) solidifies, 149 J of heat is evolved. Calculate the heat of fusion of naphthalene.

Solution

The molecular mass of naphthalene is C₁₀H₈, = 10 x 12 + 8 x 1 =128

Heat evolved when 1 g of naphthalene solidifies = 149 J

Heat evolved when 128 g of naphthalene solidifies

= 149 x 128 = 19072 J

For the fusion reaction,

This reaction is the reverse of the above solidification reaction so that

DH_fusion = - DH_{solidification}

DH_fusion = 19072 J or = 19.072 kJ

Enthalpy of hydration

This is defined as the heat change (evolved or absorbed) when one mole of the anhydrous salt combines with the required number of moles of water to form the specific hydrated salt.