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Sunday, May 12, 2019

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.

Wednesday, May 1, 2019

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 = KH ´ mole fraction of the gas in solution
Here KH is Henry’s law constant
                                  p = KH 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 KH values at the same temperature. This suggests that KH is a function of the nature of the gas. Table gives KH values of some common gases at specified temperature
Values of Henry’s law constant (KH) for some selected gases in water:


It is obvious from figure that the higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from table that KH 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,


Tuesday, April 30, 2019

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 YA and YB respectively. Let PA and PB be the partial pressure of vapours A and B respectively and total pressure PT.
In Vapours phase:
YA= mole fraction of A in vapour phase
YB = mole fraction of B in vapour phase
                   (YA+YB =1)
In liquid solution phase:
XA = mole fraction of A in liquid phase
XB = mole fraction of B in liquid phase
                  (XA + XB = 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 XA­ and XB 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 PT is given by:
 Partial pressure of the gas = Total pressure x Mole fraction
                                        PA = PT YA and PB =PT YB
Where YA­ and YB 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/YA Vs 1/XA:
According to Dalton’s law of partial pressures, the total pressure PT is given by:
 Partial pressure of the gas = Total pressure x Mole fraction
Where YA­ and YB 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:

Monday, April 29, 2019

VAPOUR PRESSURE AND RAOULT’S LAW:

VAPOUR PRESSURE:
(1) If a sample of water in its liquid phase is placed in an empty container, some of it will vaporize to form gaseous of water. This change is called evaporation.
(2) The pressure exerted by the vapour (molecules in the vapour phase) over the surface of the liquid at the equilibrium at given temperature is called the vapour pressure of the liquid.
OR
(3) It is the pressure exerted by the vapour when vapours are equilibrium with the liquid.
(4) The pressure exerted by vapours is called unsaturated vapour pressure or partial vapour at non equilibrium condition. 
Factors affecting vapour pressure:
(A) Temperature:.
(1) The temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure is called its boiling point.
(2) Vapour pressure is directly proportional to the Temperature so that on increasing temperature the rate of evaporation increases and rate of condensation decreases and hence vapour pressure increases.
(3) The dependence of vapour pressure and temperature is given by CLASIUS CLAPERON equation.

(4) Vapour pressure of a particular liquid system is only the function of temperature only. It is independent from all other factors like surface area, amount of liquid, available space etc.

(A) Nature of liquid:
Vapour pressure of liquid =1/the strength of intermolecular forces acting between molecules 
For example: CCl4 has higher vapour pressure because of the weak intermolecular forces acting between its molecules than water which has stronger intermolecular forces acting between water molecules of volatile liquid has lower boiling point than a non-volatile liquid.
 Note:
(1) Relative lowering of vapour pressure of a solvent is a colligative property equal to the vapour pressure of the pure solvent minus the vapour pressure of the solution.
(2) For example: water at 20°C has a vapour pressure of 17.54 mmHg. Ethylene glycol is a liquid whose vapour pressure at 20°C is relatively low, an aqueous solution containing 0.010 mole fraction of ethylene glycol has a vapour pressure of 17.36 mmHg. Thus the vapour pressure lowering, DP = 17.54 mmHg ¾ 17.36 mmHg = 0.18 mmHg.

RAOULT’S LAW:
(1) Vapour pressure of a number of binary solutions of volatile liquids such as benzene and toluene at constant temperature gave the following generalization which is known as the Raoult’s law.
(2) Raoult’s law states thatThe 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
(A) Vapour pressure of liquid-liquid Solution:
(3) Suppose a binary solution contains nA moles of a volatile liquid A and nB moles of a volatile liquid B, if PA and PB are partial pressure of the two liquid components, the according to Raoult’s law
(4) If the vapour behaves like an ideal gas, then according to Dalton’s law of partial pressures, the total pressure P is given by 
Graphical representation of Raoult’s law:
(5) The relationship between vapour pressure and mole fraction of an ideal solution at constant temperature is shown. The dashed lines 1 and 2 represent the partial pressure of the components. The total vapour pressure is given by 3rd line in the above figure.

(B) Vapour pressure of Solid-liquid Solution:
(1) Vapor pressure, when a small amount of a non-volatile solute (solid) is added to the liquid (solvent). It is found that the vapour pressure of the solution is less than that of the pure solvent.
(2) The lowering of vapour pressure is due to the fact that the solute particles occupy a certain surface area and evaporation takes place from the surface only. and
(3) The particles of the solvent will have a less tendency to change into vapour i.e. the vapour pressure of the solution will be less than that of the pure solvent and it is termed as lowering of vapour pressure.
For a solution of non-volatile solute with volatile solvent.
ILLUSTRATIVE EXAMPLE (1): The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate total vapour pressure of the solution.
SOLUTION:     
ILLUSTRATIVE EXAMPLE (2): What is the composition of the vapour which is in equilibrium at 30°C with a benzene-toluene solution with a mole fraction of benzene of 0.400?                    
SOLUTION: 
ILLUSTRATIVE EXAMPLE (3): The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If XA    and YA are the mole-fraction of A in the liquid and vapour, respectively find the value of XA for which YA-X has a minimum. What is the value of the pressure at this composition?
SOLUTION
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ILLUSTRATIVE EXAMPLE (4): One mole of a non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is
SOLUTION:
Mole fraction of solute in solution Or
                                                                 
Raoult's Law  v/s Dalton's Law: Determination of composition in vapour phase: Coming soon..