# TYPES OF HEAT REACTIONS

There are several types of heat change, these include:

1. Heat of Solution

When a substance is dissolved in a pure solvent or in a solution, or when a given solution is diluted with more solvent, heat is usually either absorbed or released.

Heat involved in this type of processes at constant temperature and pressure is dependent on the concentration of the solution. There are three processes involved in the dissolution of solutes, these are:

(a). Separation of solute particles – this requires heat energy (endothermic) equal to the lattice energy of the crystals.

(b). Separation of solvent particles – energy is required (endothermic) to separate the dipole- dipole forces holding solvent particles close.

(c). Attraction of solute to solvent (i.e. water) molecules (hydration) – heat energy, called the heat of hydration is given off (exothermic). Thus, the heat of solution of a substance in a solvent will either be positive or negative – depending on the extent of each of the processes above.

Notice that all three processes occur simultaneously. If the sum of lattice energy and the energy to separate solvent particles is more than the heat of hydration, then the heat of solution will be endothermic (+ve enthalpy).

But if the heat of hydration is more than the sum of the lattice energy and the energy to separate solvent particles, the heat of solution will be exothermic (-ve enthalpy). Example, the dissolution of NaOH in water – heat is given off to the surroundings.

(I.e. ∆H = -ve).

The dissolution of NH4Cl in water is endothermic (i.e. ∆H = +ve) – the side of the containing vessel becomes cool to touch.

The dissolution of NaCl in water releases heat to the surroundings, although not much to be felt by touch.

Addition of Na or K to water results in a chemical reaction, i.e. new bonds are formed between the metal, and oxygen of water to form oxides, while hydrogen is liberated. The reaction is vigorous and great amount of heat is liberated (exothermic).

Notice that this process is not dissolution – dissolution does not involve formation of new bonds. Dissolution is a physical change, while chemical reactions are chemical Changes.

The dissolution of gases in water is generally an exothermic process – the total enthalpy of the system is decreased.

Example, HCl(g) + water → HCl(aq) ∆H298 = – 75 kJ mol-1.

Solubility of gases in solvents decreases with increase in temperature of the solution, and vice versa, while the solubility of solids in solvents increases with temperature. The difference in the above occurence can be explained in two ways: 1. The heat required to separate solute particles is far higher for solid solutes than for gaseous solutes; 2. The heat of hydration given off is far less for solid solutes than for gaseous solutes.

The overall effect is that for solid solutes the heat of solution is generally +ve, i.e., more heat is required to dissolve more solute, while for gaseous solutes the heat of solution is generally -ve, i.e., less heat is required to dissolve more solute, and vice versa.

The heat of solution can be defined as the amount of heat absorbed or evolved when one mole of a substance is dissolved in so much water that further dilution results in no detectable heat change.

1. Heat of Combustion

The heat of combustion of a substance is the heat evolved when one mole of a substance is burnt in excess oxygen at a given temperature. Excess oxygen is important for complete combustion of the substance.

1. Heat of Neutralization

The heat of neutralization of an acid or a base is the heat evolved when that amount of acid or base needed to form one mole of water is neutralized. It is found that heat of neutralization of a strong dilute acid by a strong dilute base is almost constant at – 57.3 kJ – this is because the reaction is essentially

H+(aq) (from acid) + OH(aq) (from base) → H2O(l) (unionized )

Notice that to form 2 moles of water, the heat produced will be 2 x (-57.3) kJ = 114.6 kJ

1. Heat of Formation

The standard enthalpy or heat change of formation ∆HƟf of any compound is the heat change in forming one mole of a substance in its standard state from its elements in their standard states. By convention, the enthalpy of all elements in their standard state is zero.

Example, the standard enthalpies of formation of HI, HCl and SO2 are given below:

(a). H2(g) + I2(g) → 2HI(g) – ∆HƟf = + 51.8 kJ

(b). H2(g) + Cl2(g) → 2HCl(g) – ∆HƟf = -184.4 kJ

(c). S(g) + O2(g) → SO2(g) – ∆HƟf = – 296.9 kJ

There are other types of heat change – depending on the process. Example, heat of vaporization, and heat of fusion.

Notice that heat change is based on 1 mole of a specific substance involved in the change. Example, in neutralization reactions, it is based on the formation of 1 mole of water; in heat of formation, it is based on the formation of 1 mole of the substance; and in heat of solution, it is based on the dissolution of 1 mole of the substance.

Note: standard state refers to standard temperature of 298 K and standard pressure of 1.013 x 105 N m-2. Heat change measured under this condition is regarded as standard enthalpy, ∆HƟ.

The heat evolved or gained during a chemical process can be calculated by allowing it to change the temperature of a known mass of another substance. The formula below is then applied:

Q = mcƟ Where Q = quantity of heat absorbed ; m = mass of the substance; Ɵ = temperature change and c = specific heat capacity of the substance.

Calculations

1. Calculate the heat given out when 200 cm3of 2 M sodium hydroxide are mixed with 300 cm3of 2 M hydrochloric acid. (Heat of neutralization = – 57 kJ per mole).

Solution:

From the equation of reaction: HCl + NaOH → NaCl + H2O -57 kJ mol

Heat of neutralization, -57 kJ mol is the heat given off when 1 mole of water is formed.

Number of moles of HCl in reaction = molarity x volume (dm3) = 3 x 0.2 = 0.6 mole.

Number of moles of NaOH in reaction = molarity x volume (dm3) = 2 x 0.2 = 0.4 mole.

From the above equation of reaction: 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of H2O.

HCl is in excess, so we use the concentration of NaOH, which is in limited concentration. 0.4 mole of HCl reacted with 0.4 mole of NaOH to form 0.4 mole of H2O.

Since 1 mole of water formed produces 57 kJ (heat of neutralization), 0.4 mole of water formed produced 0.4 x 57 kJ = 22.8 kJ.

1. 64 g of anhydrous copper(II) tetraoxosulphate(VI) dissolved in 200 cm3of water caused a rise in temperature of 31oC,

Calculate (i). The heat produced from the dissolution. (ii). The heat of solution. (Cu = 64, S =32, O = 16, specific heat capacity of water is 4.2 J gK)

Solution:

The heat produced from the dissolution is the heat which caused the temperature of the water to rise. This heat is same as

Q = mcƟ

where m is the mass of water (200 cm3 = 200 g); c is the specific heat capacity of water (4.2 J g K ); and Ɵ is the rise in temperature (31oC) = 200 x 4.2 x 31

= 26040 J or 26.04 kJ

(ii). The heat of solution is the heat given off from 1 mole of the salt dissolving.

Number of mole of salt present 64/160 = 0.4 mole of salt dissolved to produced 26.04 kJ

Therefore, 1 mole of salt will dissolve to produce 26.04 0.4 Heat of solution of the salt is -65.1 kJ

# Types of Enthalpy reaction

(i)  Enthalpy of Formation

It may be defined as, “The quantity of heat evolved or absorbed when one mole of the compound is formed from its elements”. It is expressed as $\Delta Hf$

E.g.

$\dfrac{1}{2} H_2 (g) + \dfrac{1}{2}I_2 (s) \to HI(g) \hspace{5mm} \Delta H_f = +6.2 kcal$

$H_2 (g) + \dfrac{1}{2} O_2(g) \to H_2 O (l) \hspace{5mm} \Delta H_f = -68 kcal$

$\dfrac{1}{2} N_2 (g) + \dfrac{3}{2} H_2 (g) \to NH_3 (g) \hspace{5mm} \Delta H_f = -11 kcal$

It is important to note that the thermochemical equation should be balanced in such a way that it represents the formation of one mole of the substance only.

The value of heat of formation at 298 K and 1 atm pressure is called standard heat of formation. $(\Delta H^0)\Delta H^0$ for free state of elements and standard state is taken as Zero.

(ii)  Enthalpy of Combustion

It may be defined as, “The quantity of heat evolved when one mole of the substance is completely oxidized”.

Eg:

$(Delta H = -ve)$ always exothermic

$\underset{diamond}{C} + O_2 \to CO_2 + 95.5 kcal$

$CH_4 (g) + 2O_2 (g) \to CO_2 (g) + 2H_2 O (l) \hspace{2mm} \Delta H = -213 kcal$

$C(diamond) + O_2 (g) \to CO_2 (g) \hspace{10mm} \Delta H = -95.5 kcal$

$C(graphite) + O_2 (g) \to CO_2 (g) \hspace{10mm} \Delta H = -94.0 kcal$

$H_2 (g) + \dfrac{1}{2} O_2 (g) \to H_2O (l) \hspace{10mm} \Delta H = -68.4 kcal$

The heat of combustion is very useful in:

(a)    Calculating the heat of formation which is otherwise not possible in some cases.

(b)   Calculating the calorific value of fuels.

$\text{calorific value} = \dfrac{\Delta H}{Atomic wt. or molar mass}$

E.g.: Combustion of butane evolves -2878.8 k] per mole of heat so it’s calorific value (C.V.) is

$C.V = \dfrac{-287.8}{58} = -49.62 kJ$

Substance                                   Calorific value in

Keroscene                                        48

L.P.G. j                                              55

Milk                                                  3.11

Butter                                              30.5

(c)    Elucidating the structure of organic compounds.

(iii)   Enthalpy of Solution

It may be defined as, “The quantity of heat evolved or absorbed when one mole of a solute is dissolved completely in large excess of water, so that further dilution of solution does not produce any heat change”.

E.g.

$KCl(s) + nH_2O \to KCl(aq) \Delta H = +4.40 kcal$

$HCl (g) + nH_2O \to HCl (aq) \Delta H = -39.2 kcal$

(iv) Enthalpy of Neutralisation

It may be defined as, “the quantity of heat evolved when one equivalent (or equivalent mass) of an acid is completely neutralised by one equivalent (or equivalent If mass) of a base in dilute solution”.

E.g.

$HNO_3 (aq) + NaOH (aq) \to NaNO_3 (aq) + H_2O (l) \Delta H = -13.69 kcal$

$HCl (aq) + NaOH (aq) \to NaCl(aq) + H_2O (l) \Delta H = -13.68 kcal$

The heat of neutralisation of strong acid and a strong base is taken as 13.7 kcal. or 57kJ.

On the basis of electrolytic dissociation theory, it has been clearly explained that this heat of neutralisation is merely the heat of formation of water from $H^+$ of an acid and $OH^-$ of a base.

$H^+(aq) + OH^- (aq) \to H_2O (l) \Delta H = -13.7 kcal$

The heat of neutralisation in case of a weak acid or a weak base is somewhat less than 13.7 kcal because some energy is used up in dissociating the weak electrolyte. The difference in the values gives the dissociation energy of the weak acid or a weak base.

$NH_4OH(w. base) + HCl(Strong acid) \leftrightharpoons NH_4Cl + H_2O \Delta H = -12.3 kcal$

Here 1.4 kcal heat is absorbed in the dissociation of weak base $NH_4OH$

$CH_3COOH (weak acid (aq)) + NH_4OH(weak base(aq)) \leftrightharpoons CH_3COOHNH_4 (aq) + H_2O(l) \Delta H = -11.9 kcal$

Here 1.8 kcal heats are absorbed in the dissociation of both the weak electrolytes.

Heat of neutralisation is measured in lab by using pothythene or polystyrene bottles.

(v)   Enthalpy of Neutralisation

It may be defined as, “The quantity of heat absorbed when one mole of a substance completely dissociates into its ions”.

E.g.

$H_2O(l) \to H^+ + oh^- \Delta H = 13.7 kcal$

(vi)  Enthalpy of Dilution

It may be defined as, “The quantity of heat evolved or absorbed when solution containing one mole of a solution is diluted from one concentration to another”.

$KCl (S) + 20H_2O \to KCl(20 H_2O) \Delta H_1 = +3.8 kcal$

$KCl(s) + 200 H_2O \to KCl(200 H_2O) \Delta H_2 + 4.44 kcal$

$\therefore$ Heat of dilution = $\Delta H_2- \Delta H_1 = 4.44- 3.80 = 0.64 kcal$

(vii)   Enthalpy of Precipitation

It may be defined as, “The quantity of heat evolved in the precipitation of one mole of a sparingly soluble substance on mixing dilute solutions of suitable electrolytes” .

E.g.

$Ba^{2+} (aq) + SO^{2-}_4 (aq) \to BaSO_4 (s), \Delta H = -4.66 kcal$

(viii)    Enthalpy of Hydration

It may be defined as, “The quantity of heat evolved or absorbed? when one mole of an anhydrous or a partially hydrated salt is combined with the required number of moles of water to form a specific hydrated substance”.

E.g.

$CuSO_4 (s) + 5H_2O (l) \to CuSO_4. 5H_2O (s) \Delta H = -18.7 kcal$

$CaCl_2(s) + 6H_2O (l) \to CaCl_2. 6H_2O (s) \Delta H = -18.8 kcal$

(ix)     Enthalpy of fusion

It is change in enthalpy during conversion of 1 mole of a substance from solid to liquid state at its melting point (mostly endothermic)

$H_2O (solid) \overset{\Delta}{\underset{m.p.} \rightarrow} H_2O(liquid)- 1.44 kcal$

$\Delta H_{fus.} = 1.44 (kcal)$

(x)   Enthalpy of vaporisation

It is the change in enthalpy when 1 mole of a substance is converted from liquid to gaseous state at it’s boiling point.

$H_2O(l) \underset{B.P}{\overset{\Delta} \rightarrow} H_2O- 10.5 kcal$ mostly endothermic.

(xi)  Enthalpy of Sublimation

It is change in enthalpy when I mole of a substance is converted directly from solid to Vapour (gaseous) state.

$I_2(s) \overset{\Delta}{\rightarrow} I_2- 14.9 kcal$

$\Delta H = 14.9 kcal$

$\Delta H_{sub} = \Delta H_{vap.} + \Delta_{fus.}$

## Enthalpy of Solution (Heat of Solution) Example

In an experiment, 1.2 g of sodium hydroxide pellets, NaOH(s), were dissolved in 100 mL of water at 25°C.

The temperature of the water rose to 27.5°C.

Calculate the enthalpy change (heat of solution) for the reaction in kJ mol-1 of solute.

1. Calculate the heat released, q, in joules (J), by the reaction:q = mass(water) × specific heat capacity(water) × change in temperature(solution)
q = m(H2O(l)) × cg(H2O(l)) × (Tf – Ti)
q = 100 × 4.184 × (27.5 – 25) = 1046 J
2. Calculate the moles of solute (NaOH(s)):moles = mass ÷ molar mass
moles (NaOH) = 1.2 ÷ (22.99 + 16.00 + 1.008)
n(NaOH) = 0.030 mol
3. Calculate the enthalpy change, ΔH, in kJ mol-1 of solute:ΔH = -q/1000 ÷ n(solute) = -1046/1000 ÷ 0.030 = -35 kJ mol-1
ΔH is negative because the reaction is exothermic (energy is released causing the temperature of the solution to increase).

### ASSIGNMENT : TYPES OF HEAT REACTIONS ASSIGNMENTMARKS : 10  DURATION : 1 week, 3 days

#### Girls in STEM

We at FAWE have built this platform to aid learners, trainers and mentors get practical help with content, an interactive platform and tools to power their teaching and learning of STEM subjects

## STEM Elearning

We at FAWE have built this platform to aid learners, trainers and mentors get practical help with content, an interactive platform and tools to power their teaching and learning of STEM subjects, more