|
Home
Types of Heat Change
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.
2. 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.
3. 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
4. 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 cm3 of 2 M sodium hydroxide are mixed with 300 cm3 of 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.
2. 64 g of anhydrous copper(II) tetraoxosulphate(VI) dissolved in 200 cm3 of 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 g-K-)
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.
|
