Law of Constant Proportion / Definite Proportion / Fixed Proportion

Louis Proust in 1799 after analysis of large number of compounds arrived at the generalized statement of the law of constant proportion. It states that, “A pure chemical compound, irrespective of its source of method of preparation, always contains the same elements combined together in same definite proportions by weight.”
Thus, if the element ‘A’ and ‘B’ combine chemically to form the compound AB, then in whatever manner AB is formed, it is always composed of same two elements ‘A’ and ‘B’ combined together in the same fixed ratio or proportion by weight.

For example: Sulphur dioxide can be obtained when
         (i).     Sulphur is burnt in air,    
                                    S   +    O₂------>SO₂

         (ii).    Copper is heated with conc. sulphuric acid

                                    Cu  +   2H₂SO₄ ------------>  CuSO₄   +   2H₂O   +   SO₂

         (iii).   Dilute hydrochloric acid is added to sodium bisulphate

                                    NaHSO₃   +   HCl  ---------->   NaCl   +   H₂O   +   SO₂

In each case, sulphur and oxygen in the sulphur dioxide obtained are found to be in the same ratio of 32 : 32 or 1 : 1 by weight.

Law of Conservation of Mass

Antoino Lavoisier (Known as Father of Chemistry) in 1774 established this law, which states, “Whenever a chemical change or physical change takes place the total mass of reacting species (reactants) is exactly equal to the total mass of the products of the reaction.”

 Thus, according to this law there is no increase or decrease in total mass of matter during a chemical or a physical change. In other words, “Matter can neither be created nor destroyed.” Hence this law is also known as “Law of Indestructibility of Matter”.

Experimental Verification              

Landolt Experiment

                  Landolt verified the law by various experiments conducted by him. In one of the experiment, Landolt took two solutions (i) sodium chloride, NaCl and (ii) Silver nitrate AgNO₃. Separately in the two limbs of H – shaped glass tube known as Landolt Tube. The two limbs were sealed and the tube was weighed. The tube was tilted to mix two solutions. Sodium chloride reacts with silver nitrate and precipitate of silver chloride was formed.

                        AgNO₃  +  NaCl    AgCl¯  +  NaNO₃

The tube was again weighed, Landolt observed that total mass remained practically constant after the reaction.

Limitations
                  In all the chemical reactions, energy is evolved or absorbed which would be at the expense of change in mass. In ordinary chemical reactions, this change in mass is so small that it can not be registered on the most sensitive balance. This suggests that some matter of the reaction mixture gets converted into energy such as light, heat etc. Thus mass and energy are interconvertible. The mass is converted to energy by Einstein’s relation E = mc².

Concept of ORBITALS

According to Heisenberg's Uncertainty Principle, it is not possible to determine precisely the position and momentum of an electron in the atom simultaneously. Therefore, Bohr's concept of well defined orbits is ruled out. According to quantum mechanics the probability of finding an electron does not become zero even at large distances from the nucleus. Therefore it is not possible to draw a boundary that will enclose the region of 100% probability. However, for the sake of simplicity arbitrary boundaries are drawn which encloses the region where probability of finding the electron is maximum, this region is known as an Orbital. Thus, An orbital may be defined as the region of space around the nucleus where the probability of finding an electron is maximum.

Absolute or Kelvin Scale of Temperature

The hypothetical temperature of –273°c at which volume of gas becomes zero is known as Absolute Zero. This temperature i.e. –273°c cannot be achieved by gas under any condition in practical life, so it is theoretical i.e. hypothetical.

Lord Kelvin

suggested a new scale of temperature with –273^oc as lowest temperature. The size of degree on this scale was same as on centigrade scale. Thus new scale was known as Kelvin Scale or Absolute Zero Scale. The temperature on Kelvin scale is denoted by T K or T. The zero of centigrade scale corresponds to 273K of Kelvin scale.

The Kelvin and Celsius scale temperatures are related to each other by following relation.

T_K  =  t°c  +  273.

Significance of Charles’s Law

According to this law, the gases expand on heating. Since the mass of gas remains same, the number of molecules per unit volume (density) decreases on heating. In other words, hot air is less dense than cold air. This enables the hot air balloons to rise up by displacing the cooler air of the atmosphere.

Charle’s Law

The relationship between temperature of a gas and its volume was studied by Jacques Charles in 1787 and was further modified by Gay-Lussac in 1802. They gave a generalization known as Charle’s Law. The Law states that, “At constant pressure, the volume of a given mass of gas increases or decreases by 1/273 of its volume at 0^oc for every one degree rise or fall in temperature.”

Mathematically, the law can be represented as fallows;

Let ‘V₀’ be the volume of given mass of gas at 0°c when temperature is increased by one degree i.e. 1°c, then volume of  gas according to Charles law would be

V₁  =  V₀  + (1/273) V₀         where V₁ is volume of gas at 1°c

If temperature is further increased by 1°c  i.e. 1°c   +   1°c   =   2°c

Then,    V₂   =   (V₀  +(1/273) V₀)   + (1/273)  V₀

=  V₀   + (2/273) V₀

Similarly at 3°c

V₃   =  V₀   + (3/273)  V₀

At t^oc

V_t   =   V₀   +  (t/273) V₀

Similarly if temperature is reduced to −1°c, then

V_–1   =   V₀  −(1/273) V₀

At  −2°c

V_–2   =   V₀  − (2/273) V₀

At  −273°c

V_–273    =    V₀  − (273/273) V₀

=    V₀ − V₀  =  0

This means that the volume of a given mass of a gas becomes zero at −273°c. In other words, the gas will cease to exist at this temperature.

The same conclusion can also be drawn graphically. Keeping the pressure of the gas constant, a graph is plotted between temperature and volume. Upon extrapolation (produce), the graph meets temperature at −273°c. This means volume of every gas becomes zero at −273°c.

Charle’s law, may be put in another form as fallows;

V_t   =  V₀   +  (t/273)  V₀

Or   V_t    = (273 V₀ +t V₀ )/ 273

V_t    =   V₀ (273+t) / 273

If 273 + t   =   T

& 273   =    T₀ (on Kelvin scale)

Then,     V_t   = V₀ T / T₀

Or,      V/T= V₀ / T₀

Thus   V / T  =   constant

V / T    =  K

V  =   KT

Or     V  α  T  (at constant pressure)

Thus, Charle’s law may also be defined as, “The volume of a given mass of a gas at constant pressure is directly proportional to the absolute or Kelvin temperature.”

Let V₁ be the volume of gas at temperature T₁, keeping pressure constant, if temperature is increased to T₂, the volume will change to V₂. Then according to Charles’s law;

     V₁ / T₁ = V₂ / T₂

Significance / Importance of Boyle’s Law

According to Boyle’s law, at constant temperature the density of gas is directly proportional to pressure. This is because density depends on volume, smaller the volume higher the density. So when pressure is applied, volume is reduced i.e. density is increased.

Thus, at higher altitudes, as the atmospheric pressure is low, the air is less dense. As a result less oxygen is available for breathing. That is why mountaineers have to carry oxygen cylinders with them.

Graphical Representation of Boyles’s Law

Boyles law can be represented graphically in number of ways.

1).Volume- Pressure Graph

When a graph is drawn between volume of gas and pressure on it. A hyperbola is formed which indicates that on increasing pressure, volume is decreased.

2).Pressure - Volume Vs Pressure Graph

When a graph is drawn between PV and Pressure (P) at constant temperature a straight line parallel to x-axis is obtained. The line shows PV is constant at different pressures.

Boyle’s Law

A characteristic property of gases is their great compressibility. The relationship between pressure of gas and its volume was given by Robert Boyle in 1662.
Boyle’s law states that,
“At constant temperature, the volume of given mass of gas (V) is inversely proportional to the pressure of the gas (P).”
This means if the pressure on the gas is increased the volume of the gas is reduced and if the pressure is reduced, the volume of gas is increased.
Mathematically, Boyle’s law can be expressed as;
V ∝ 1/P
Or V = K (1/P) where ‘K’ is proportionality constant.
Or PV = K
Thus, PV = Constant
From this expression, Boyle’s law can also be stated as, “At constant temperature, the product of the pressure and volume of given mass of gas is constant.”
Let ‘V1’ be the volume of given mass of gas at pressure P1. By keeping the temperature constant, if pressure be increased to P2 then volume will decrease to V2. According to Boyles law
P₁V₁ = K ––––––– (1)
P₂V₂ = K ––––––– (2)
Equating 1 & 2 we get
P₁V₁ = P₂V₂

Measurement of Pressure Exerted by the Gas

The pressure exerted by a gas can be measured by a device called Manometer
. Two types of Manometers are commonly used. These are Open End Manometer and Closed End Manometer.

1.  OPEN END MANOMETER

It consists of a U-shaped glass tube partially filled with mercury. One limb of the tube is shorter than the other. The longer limb is open while shorter limb is connected to glass bulb containing a gas whose pressure is to be determined. The mercury inside of longer limb is under atmosphere pressure while as mercury on the side of shorter limb is under the pressure of gas. There are three possibilities;

 CASE 1:

If Hg level in the two limbs is same, then

                Gas pressure   =   Atmosphere pressure.

CASE 2:

                If Hg level in longer limb is higher, then

                Gas pressure = Atmosphere pressure + h, (Where ‘h’ is difference in two levels).

CASE 3:

If Hg level in shorter limb is higher, then

Gas pressure = Atmosphere pressure – h, (Where ‘h’ is difference in two levels).

Open limb manometer is used to find the pressure of gases of all type i.e. whose gas pressure is greater than atmospheric pressure or less than atmospheric pressure.

2. CLOSED-END MANOMETER

The closed end manometer is similar to open–end Manometer except that the longer limb is closed. The space above the mercury level in longer limb is perfect vacuum. The gas present in the bulb exerts pressure on mercury level in shorter limb. As a result, it moves down and level of mercury in longer limb rises. The gas pressure is equal to the difference in the levels of mercury in two limbs.   

Thus, Gas pressure   =   Difference in mercury level in two limbs.

Calculation of Pressure by Mercury Column

Let ‘V’ be the volume of mercury inside the column and ‘ρ’ be its density

Density (ρ)   =  Mass / Volume

Mass of mercury column (m)   =    density  × volume

m    =     ρV         ––––––– (1)

Since tube is cylindrical in shape,

Therefore,

Volume (V)   =   Area of cross-section    ×   height

Thus, Eq. (1) becomes,                                 

m    =    ρ × A × h            where A= Area of cross-section

But,   Pressure  =   Force / Area        

=  (mass × acceleration) / Area

P      =  (ρ × A × h × g) / A               

P =  ρ × h × g         

where ρ = density of mercury, h = height of column,  g = acceleration due to gravity,

For 1 atmosphere,

ρ   =  13.596 × 10^-3 Kg m^-3

h    =  0.76 m

g   =  9. 81 ms^-2

Thus,      1atm.  =   13.596 × 10 ^–3 Kg m ^–3 × 0.76 m × 9. 81 ms ^–2

1atm   =   101325 Kg m^–2 s^–2 = 101325 Nm^–2

=  101325 Pascal (Pa)            [ 1 Pa = 1 Kg m^–1 s^–2]

=  1.01325 × 10⁵ Pa

=  1.01325 bar                        ( 1 bar = 10⁵  )

Therefore,  1 atm.  =  101325 Pa  = 1.01325 bar  = 760 mm  Hg = 760 Torr.

Measurement of Atmospheric Pressure

Earth is surrounded by a thick layer of air. Due to gravity of the earth, air is pulled towards it and exerts pressure on the surface of earth. The pressure exerted by the air layer is known as Atmospheric Pressure. The atmospheric pressure is measured by Barometer.

The barometer, in simple form can be made by taking a glass tube about 80 cm long. The glass tube is filled with mercury and then inverted in a vessel which also contains mercury. The level of mercury in the tube falls and then becomes static at certain height. At this stage, the pressure exerted by the column of mercury in tube is equal to pressure exerted by air (i.e. atmosphere pressure) on the mercury in vessel. The space above the mercury in the tube is perfect vacuum. Therefore, pressure due to mercury column is equal to atmospheric pressure. At sea level the height of column is observed to be 76 cm or 760 mm at 273 K.

CTM *(Mercury is generally used in barometer, because it has very high density and does not  stick to glass walls and is also not volatile at room temperature)

Gaseous State–Measurable Properties

The measurable properties of gases are:-

1. Mass  

2. Volume  

3. Temperature  

4. Pressure.

1.   MASS

The mass of gas can be determined by weighing the container in which the gas is enclosed and weighing the container again when vaccumised (empty).The difference between two weights will give the mass of the gas. From mass, the number of moles can be calculated using the relation

Number of moles = mass of gas in grams / gram molecules mass

and

Number of molecules  =  Number of moles ´ Avogadro’s Number (N)

2.   VOLUME

The Volume of any substance is the space occupied by that substance. For gases, the volume of sample is the same as the volume of the container in which it is held. Generally, the volume is specified in units of Litre (L),  Milliliters (ml) , Cubic decimeter (dm³) .

\            1L =1dm³ =   1000ml

3.   TEMPERATURE

Temperature is a measure of the degree of hotness of a substance. The measurement of temperature is based upon the principle that substances expand on heating. The most common substances whose expansion is made in measurement of temperature is Mercury (Hg). There are three different scales on which temperature are measured. These are;

·                    Centigrade or Celsius scale

·                    Fahrenheit scale

·                    Kelvin scale

In Celsius scale, the freezing point of water is taken as 0°C and boiling of water as 100°C at normal atmosphere pressure & then dividing the space in range into 100 equal parts.

In Fahrenheit scale, the freezing point of water is taken as 32°F and the boiling point of water as 212°F and range is divided into 180 degrees.

*(C.T.M) Temperature in Celsius can be converted to Fahrenheit scale by following relation;

                            F^o  = 9/5 C^o + 32          or   C^o  =   5/9 F^o – 32

The Kelvin scale of temperature takes the lowest temperature of –273.15° as 0k. It is a better scale than Celsius scale because it does not involve negative values. The two scales are related by following relation.  

K  = C^o + 273  

4.     PRESSURE

Pressure is force per unit area. A gas exerts uniform pressure in all the directions on the walls of the container in which it is kept. Thus,

Pressure of gas = force of gas exerts on walls of container / surface area of container

Introduction to State of Matter

There are three states of matter namely solid, liquid and gas. Depending upon the temperature and pressure, a substance may exist in any one of the three states. Thus, water exists as solid (ice), liquid (water) or gaseous (steam) depending upon the temperature and pressure. The three states of matter can be defined as follows:-

A substance is said to be Solid if its melting point is above room temperature under atmosphere pressure.

A substance is said to be Liquid if its melting point is below room temperature under atmosphere pressure.

A substance is said to be Gas if its boiling point is below room temperature under atmosphere pressure.

Calculation of Enthalpy of Reaction

The enthalpy of reaction can be found out by finding the difference in the bond energies of reactants and products.
ΔH   =   Σ Bond Energy of Reactants – Σ Bond energy of Products.
Example:
Calculate DH for the following reaction.
H₂ (g)  +  ½ O₂ (g) --------->  H2O
Given
H – H  =  433 KJ/mol.
O – O =   499 KJ/mol.
and        O – H =   460 KJ/mol.
Solution:
ΔH  =   Σ B.E of H₂ and ½ O₂ - Σ B.E of H₂O
=  (B. E of H₂ + ½  B. E. of O₂ ) – (B. E. of H2O)
=  (B. E. of H – H + ½  B. E. of O – O) – (B. E. of 2 .O – H)
= (433 + ½ x 499) – (2  x 460)
=  (433 + 249.5) – (920)
=   682.5 – 920
ΔH  =  – 237.5 KJ/Mol.

Bond Energy and Bond Dissociation Energy

Whenever a bond is formed energy is evolved and bond is broken, energy is absorbed. Bond energy may be defined as, “The amount of energy released when 1 mole of bonds are formed from isolated gaseous atoms.”
Whereas Bond Dissociation energy may be defined as, “The amount of energy required to break one mole bond present between the atoms of a gaseous molecule.”

Application of Hess Law

Hess’s law finds its application in determining the heat of changes for reactions for which experimental determination is not possible. The various applications of Hess’s law can be understood from following headings;

1.   Calculation of Enthalpy of Formation

The enthalpy of formation of certain compounds can be determined by Hess’s Law, where direct determination is not possible experimentally.

Example: Calculate the enthalpy of formation of methane, given that the enthalpies of combustion of methane, graphite and hydrogen are 890.2 KJ, 393.4 KJ and 285.7 KJ mol^–1 respectively.

Solution:

Given

(i).          CH₄  +  2O₂   ––––––––>  CO₂  +  2H₂O  ;  DH₁  =  – 890.2 KJ/mol^–1

(ii).         C   +   O₂        ––––––––>  CO₂  ;  DH₂  =  393.4 KJ/mol^–1    

(iii).        H₂  +  ½ O_2­    ––––––––>  H₂O   ;  DH₃  =  285.7 KJ/mol^-1

To find

C  +   2H₂   ––––––––>  CH₄  ;  DH  =  ?

Multiplying (iii) by 2 and then add to (ii) we get,

C  +  O₂ + 2 (H₂  +  ½ O₂)    ––––––––>  CO₂ + 2H₂O  ;  DH  =  DH₂  +  2DH₃  =   – 393.4  +  2(– 285.7)  

C  +  O₂ + 2 H₂  + O₂             ––––––––>  CO₂ + 2H₂O    ;  DH  =  – 393.4 – 571.4

C  +  2 H₂  +  2O₂                    ––––––––>  CO₂ + 2H₂O    ;  DH  =  – 964.8 –––––––– (iv)

Subtract (i) from (iv), we get.    

C  +  2H₂ – CH₄  ––––––––> 0             ;  DH  =  – 964 + 890.2 KJ         

C  +  2H₂                     ––––––––> CH₄        ;  DH  =   – 74.6 KJ      

2.     Calculation of Enthalpy of Allotropic Transformation

      The transformation of an element from one allotropic form to another form involves a small change of heat. This heat can be calculated using Hess’s Law.

Example:  Calculate enthalpy change of transformation of sulphur (rhombic) to sulphur (monoclinic). Given than combustion of rhombic sulphur and monoclinic sulphur as – 71.1 K Cal. and – 71.7 K Cal respectively.

Solution:

Given

 (i).     S _(Rhombic)   +    O₂  (g) ––––––––>   SO₂ (g)  ;  DH  =  – 71.1 K Cal.

 (ii).    S _(monoclinic) +    O₂ (g)  ––––––––>   SO₂ (g)  ; DH   =  – 71.7 K Cal.

Subtract (ii) from (i) we get           

S _(Rhombic)   –   S _(monoclinic)  ––––––––>     0   ;   DH    =    – 71.1  +  71.7           

S _(Rhombic)   ––––––––> S _(monoclinic)    ;   DH  =  0.6 K Cal.

3.            Calculation of Calorific Value

Example:   Calculate the heat of glucose and its calorific value from following data:

(i).          C _(graphite)  +  O₂ (g)   ––––––––>  CO₂ (g)  ;  DH  =  – 395 KJ

(ii).         H₂ (g)  +     ––––––––>  H₂O (l)   ;  DH  =  – 269.4 KJ.

(iii).        6C _(graphite)  +  6H₂ (g)  +  3O₂ (g) ––––––––>  C₆H₁₂O₆ (s)  ;  DH  =  –1169.8 KJ.

Solution:

We have to find

C₆H₁₂O₆ (s) +  6O₂ (g)   ––––––––>   6CO₂ (g)   +   6H₂O (l)  ;  DH   =   ?

Multiplying (i) by 6 and (ii) by 6, then add.

 6C _(graphite) + 6O₂ (g) + 6H2 (g)  +  6 ´ ½ O₂ (g) ––––––––> 6CO₂ + 6H₂O (l) ; DH = –395 ´ 6 + (– 269.4 ´ 6)

(iv) 6C_graphite + 6H2 (g)  + 9O₂ (g)  ––––––––> 6CO₂ + 6H₂O (l)  ;  DH =–2370–1616.4 = – 3986.4 KJ

Subtract eq. (iii) from (iv), we get.

C₆H₁₂O₆ (s)  +  6O₂     ––––––––>  6CO₂  +  6H₂O (l)    ;     DH  =  2816.6 KJ.

Thus,

1 mole of C₆H₁₂O₆ liberates energy = – 2816.6 KJ.

i.e.   180 gm of C₆H₁₂O₆ liberates energy = – 2816.6 KJ.

\            1 gm of C₆H₁₂O₆ liberates energy =  -2816.6/180= 15.64 KJ.

Thus, calorific value of C₆H₁₂O₆ is 15.64 KJ.

Proof of Hess’s Law

Hess’s law can be proved on the basis of first law of thermodynamics, according to which heat can neither be created nor destroyed. Let us suppose Q₁ Joules is heat change when A changes to B directly and Q₂ Joules is when A changes B indirectly via C. Let us suppose Q₁ > Q₂. Now if we move from A to B directly and then come back via C, then (Q₁ – Q₂) will be the heat evolved. This means that when we move from A to B and then back from B to A certain amount of heat is created which is against first law of thermodynamics. Thus we conclude, that Q₁ cannot be more than nor less than Q₂, but are equal, thus verifies Hess’s Law.

Hess’s Law of Constant Heat Summation

Hess’s Law of Constant Heat Summation is also known as Second Law of Thermochemistry. The law states, “That the total heat change accompanying a chemical reaction is the same whether the reaction takes place in one step or more steps.”  

Example: Carbon dioxide can be formed directly from carbon and also from carbon via carbon monoxide. The heat change involved in both the process are found to be same.

Formation of CO₂ directly:

C (s)    +    O₂ (g)     ––––––––>  CO₂ (g)   ;    DH   =   – 94 K Cal.

Formation of CO₂ Via Carbon Monoxide:

C (s)     +   ½ O₂ (g)  ––––––––>     CO (g)    ;         DH₁  =  – 26.4 K Cal.

CO (g)  +   ½ O₂ (g)  ––––––––>   CO₂ (g)   ;    DH₂  =   – 67.6 K Cal.

DH  =  DH₁  +  DH₂      

DH  =  –26.4 + (– 67.6) K Cal.

DH  =  – 26.4 – 67.6 K Cal.

DH  =  – 94 K Cal..