LAW OF DEFINITE OR CONSTANT COMPOSITION:

A chemical compound always contains same elements in definite proportion by mass and it does not depend on the source of compound.

For example:

The composition of CO₂ obtained by different means always having same hence Law of definite proportion is proving.

ILLUSTRATIVE EXAMPLE (2): Ammonia contains 82.65 % N₂ and 17.65% H₂. If the law of constant proportions is true, then the mass of zinc required to give 10 g Ammonia will be:

                        (A)       8.265 g                                    (B) 0.826 g

                        (C)       82.65 g                                     (D) 826.5 g

SOLUTION: The mass of zinc required to give 10 g ammonia will be

ILLUSTRATIVE EXAMPLE (3): Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of:

                        (A) Conservation of mass                    (B) Constant composition

                        (C) Multiple proportion                      (D) Constant volume

SOLUTION: As in water

Both values always constant. Obey law of constant composition. Hence (B) is correct.

ILLUSTRATIVE EXAMPLE (4): 6.488 g of lead combine directly with 1.002 g of oxygen to form lead peroxide PbO₂. Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxide is 13.38 percent. Use these data to illustrate the law of constant composition

SOLUTION:

Step 1: To calculate the percentage of oxygen in first experiment.

Step 2: To compare the percentage of oxygen in both the experiments.

            Percentage of oxygen in PbO₂ in the first experiment = 13.38

            Percentage of oxygen in PbO₂ in the second experiment = 13.38

Since the percentage composition of oxygen in both the samples of PbO₂ is identical, the above data illustrate the law of constant composition.

ILLUSTRATIVE EXAMPLE (5): Copper oxide was prepared by the following methods:

(A) In one case, 1.75 g of the metal was dissolved in nitric acid and igniting the residual copper nitrate yielded 2.19 g of copper oxide.

(B) In the second case, 1.14 g of metal dissolved in nitric acid was precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying and heating yielded 1.43 g of copper oxide.

(C) In the third case, 1.45 g of copper when strongly heated in a current of air yielded 1.83 g of copper oxide. Show that the given data illustrate the law of constant composition.

SOLUTION:

Step 1:  In the first experiment.

            2.19 g of copper oxide contained 1.75 g of Cu.

            100 g of copper oxide contained = 1.75/2.19*100=79.91 %

Step 2:  In the second experiment.

            1.43 g of copper oxide contained 1.14 g of copper.

            100 g of copper oxide contained =1.14/1.43*100=79.72 %.

Step 3: In the third experiment.

            1.83 g of copper oxide contained 1.46 g of copper

            100 g of copper oxide contained =1.46/1.83*100 = 79.78 %

Thus the percentage of copper in copper oxide derived from all the three experiments is nearly the same. Hence, the above data illustrate the law of constant composition.

ILLUSTRATIVE EXAMPLE (6): 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO was produced in the process. Show that these results illustrate the law of constant proportions.

SOLUTION: The ratio of copper and oxygen is 1: 0.25. Hence, the law of constant proportions is illustrated.

ILLUSTRATIVE EXAMPLE (7): In an experiment, 2.4 g of iron oxide on reduction with hydrogen yield 1.68 g of iron. In another experiment 2.9 g of iron oxide give 2.03 g of iron on reduction with hydrogen. Show that the above data illustrate the law of constant proportions.

SOLUTION: Ratio of oxygen and iron in both the experiment is 1:2.33

ILLUSTRATIVE EXAMPLE (8): 2.8 g of calcium oxide (CaO) prepared by heating limestone were found to contain 0.8 g of oxygen. When one gram of oxygen was treated with calcium, 3.5 g of calcium oxide were obtained. Show that the results illustrate the law of definite proportions.

SOLUTION:  try yourself……

Limitation of law definite proportions: Discovery of isotopes created some limitations to this law. Isotopes of an element have different atomic mass hence they form same chemical compounds with different compositions;

For example:

LAW OF CONSERVATION OF MASS:

For any chemical change total mass of active reactants are always equal to the mass of the product formed. It is a derivation of Dalton’s atomic theory ‘atoms neither created nor destroyed’.

Total masses of reactants = Total masses of products + Masses of unreacted reactants

ILLUSTRATIVE EXAMPLE (1): 5.2 g of CaCO₃ when heated produced 1.99 g of Carbon dioxide and the residue (CaO) left behind weighs 3.2g. Show that these results illustrate the law of conservation of mass.

SOLUTION: Weight of CaCO₃ taken = 5.2 g

            Total weight of the products (CaO +CO₃) = 3.20+ 1.99 = 5.19 g

            Difference between the wt. of the reactant and the total wt. of the products

            = 5.20 – 5.19 =0.01 g.

            This small difference may be due to experimental error.

            Thus law of conservation of mass holds good within experimental errors.

Limitation of Law of conservation of mass: Nuclear reactions do not follow the law of conservation of mass because some of the mass of reactants is converted into energy according to Einstein equation E=mc²     where c is the velocity of light.

LAW OF CHEMICAL COMBINATIONS:

In order to understand the composition of the compounds, it is necessary to have a theory which accounts for both qualitative and quantitative observations during chemical changes. Observations of chemical reactions were most significant in the development of a satisfactory theory of the nature of matter. These observations of chemical reactions are summarized in certain statements known as laws of chemical combination.

(1) Law of conservation of mass:

(2) Law of constant composition

(3) Law of multiple proportion:

(4) Law of reciprocal proportion:

(5) Gay Lussac’s law of Combining Volumes:

SOLUBILITY OF GASES AND HENRY’S LAW:

Solubility:

Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specific temperature. It depends upon the nature of solute and solvent as well as temperature and pressure.

Unit: Unit of Solubility is gm/litre or Mole/Litre

(A) Solubility of Solid in a Liquid:
Every solid does not dissolve in a given liquid. While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not. On the other hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not.

(1) It is clear observed that polar solutes dissolve in polar solvents and non polar solutes in non-polar solvents.

(2) In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.

(3) Dissolution: When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution.

(4) Crystallisation: Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation.

“A stage is reached when the two processes (dissolution and crystallisation) occur at the same rate. Under such conditions, number of solute particles going into solution will be equal to the solute particles separating out and a state of dynamic equilibrium is reached.

At this stage the concentration of solute in solution will remain constant under the given conditions, i.e., temperature and pressure. Similar process is followed when gases are dissolved in liquid solvents.

(5) Saturated solution: Such a solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution.

(6) Unsaturated solution:  The Solution in which more solute can be dissolved at the same temperature.

(7)  The solution which is in dynamic equilibrium with undissolved solute is the saturated solution and contains the maximum amount of solute dissolved in a given amount of solvent. Thus, the concentration of solute in such a solution is its solubility.

Factors affecting Solubility:
Earlier we have observed that solubility of one substance into another depends on the nature of the substances. In addition to these variables, two other parameters, i.e., temperature and pressure also control this phenomenon.
(1) Effect of temperature:
The solubility of a solid in a liquid is significantly affected by temperature changes. Consider the equilibrium exist between dissolution and crystallisation. This, being dynamic equilibrium, must follow Le Chateliers Principle. In general…….

(i) If in a nearly saturated solution, the dissolution process is endothermic (Δsol H > 0), the solubility should increase with rise in temperature and

(ii) If it is exothermic (Δsol H > 0) the solubility should decrease. These trends are also observed experimentally.
(2) Effect of Pressure:
Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

(2) Solubility of gas in Liquid:
Many gases dissolve in water. Oxygen dissolves only to a small extent in water. It is this dissolved oxygen which sustains all aquatic life. On the other hand, hydrogen chloride gas (HCl) is highly soluble in water. Solubility of gases in liquids is greatly affected by pressure and
temperature.

Factors affecting Solubility:

(1) Effect of Pressure:

The solubility of gases increase with increase of pressure. For solution of gases in a solvent, consider a solution is act as system and that system to be in a state of dynamic equilibrium, i.e., under these conditions rate of gaseous particles entering and leaving the solution phase is the same. Now increase the pressure over the solution phase by compressing the gas to a smaller volume, this will increase the number of gaseous particles per unit volume over the solution and also the rate at which the gaseous particles are striking the surface of solution to enter it. The solubility of the gas will increase until a new equilibrium is reached resulting in an increase in the pressure of a gas above the solution and thus its solubility increases.

Henry’s Law:

(1) The solubility of a gas in a liquid is determined by several factors. In addition to the nature of the gas and the liquid, solubility of the gas depends on the temperature and pressure of the system.

(2) The solubility of a gas in a liquid is governed by Henry’s law which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

(3) Dalton, a contemporary of Henry, also concluded independently that the solubility of a gas in a liquid solution is a function of the partial pressure of the gas. If we use the mole fraction of the gas in the solution as a measure of its solubility, then: Mole fraction of the gas in a solution is proportional to the partial pressure of the gas.

Or, partial pressure of the gas in solution = K_H ´ mole fraction of the gas in solution

Here K_H is Henry’s law constant

                                  p = K_H X _(Solute)

If we draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the straight line passing through origin.

Experimental result for the solubility of HCl gas in Cyclohexane at 93 K the slope of line is the Henry’s law constant

Different gases have different K_H values at the same temperature. This suggests that K_H is a function of the nature of the gas. Table gives K_H values of some common gases at specified temperature

Values of Henry’s law constant (K_H) for some selected gases in water:

It is obvious from figure that the higher the value of K_H at a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from table that K_H value for both N₂ and O₂ increases with increase in temperature indicating that solubility of gases decreases with increase of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than warm waters.

Application of Henry’s Law:
Henry’s law finds several applications in industry and explains some biological phenomena Notable among these are:

 (1) To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

(2)  To minimize the painful effects accompanying the decompression of deep sea divers, oxygen diluted with less soluble helium gas is used as breathing gas.

(3)  In lungs, where oxygen is present in air with high partial pressure, haemoglobin combines with oxygen to form oxyhaemoglobin. In tissues where partial pressure of oxygen is low, oxyhaemoglobin releases oxygen for utilization in cellular activities.

(1) At High Pressure:

Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life.

 To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
(2) At Low Pressure: At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia.

(2) Effect of temperature:
Solubility of gases in liquids decreases with rise in temperature. When dissolved, the gas molecules are present in liquid phase and the process of dissolution can be considered similar to condensation and heat is evolved in this process. We have known that dissolution process involves dynamic equilibrium and thus must follow Le Chatelier’s Principle. As dissolution is an exothermic process, the solubility should decrease with increase of temperature.

ILLUSTRATIVE EXAMPLE (1): If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water. Assume that N­₂ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N₂ at 293 K is 76.84 kbar.

SOLUTION: The solubility of gas is related to its mole fraction in the aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus,

As 1litre water contains 55.5 mol of it, therefore, if n represents number of moles of N₂ in solution,

DALTON'S LAW VERSES RAOULT'S LAW:

Determination of composition in vapour phase:

The composition of the vapour in equilibrium with the solution can be calculated applying Daltons’ law of partial pressures. Let the mole fractions of vapours A and B be Y_A and Y_B respectively. Let P_A and P_B be the partial pressure of vapours A and B respectively and total pressure P_T.

In Vapours phase:

Y_A= mole fraction of A in vapour phase

Y_B = mole fraction of B in vapour phase

                   (Y_A+Y_B =1)

In liquid solution phase:

X_A = mole fraction of A in liquid phase

X_B = mole fraction of B in liquid phase

                  (X_A + X_B = 1)

According to Raoult’s Law: The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution.

      Where X_A­ and X_B is the mole fraction of the component A and B in liquid phase respectively

According to Dalton’s Law:

The vapour behaves like an ideal gas, then according to Dalton’s law of partial pressures, the total pressure P_T is given by:

 Partial pressure of the gas = Total pressure x Mole fraction

                                        P_A = P_T Y_A and P_B =P_T Y_B

Where Y_A­ and Y_B is the mole fraction of the component A and B in gas phase respectively

Combination of Raoult’s and Dalton’s Law:

(3) Thus, in case of ideal solution the vapour phase is phase is richer with more volatile component i.e., the one having relatively greater vapour pressure

Graph Between 1/Y_A Vs 1/X_A:

According to Dalton’s law of partial pressures, the total pressure P_T is given by:

 Partial pressure of the gas = Total pressure x Mole fraction

Where Y_A­ and Y_B is the mole fraction of the component A and B in gas phase respectively

According to Raoult’s law:

On rearrangement of this equation we get a straight line equation: