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Current Affairs JEE Main & Advanced

Application Of Radioactivity

Radioisotopes find numerous applications in a variety of areas such as medicine, agriculture, biology, chemistry, archeology, engineering and industry. (1) Age determination : The age of earth has been determined by uranium dating technique as follows. Samples of uranium ores are found to contain \[P{{b}^{206}}\] as a result of long series of a- and b-decays. Now if it is assumed that the ore sample contained no lead at the moment of its formation, and if none of the lead formed from \[{{U}^{238}}\] decay has been lost then the measurement of the \[P{{b}^{206}}/{{U}^{238}}\] ratio will give the value of time t of the mineral.                 \[\frac{\text{No}\text{. of atoms of }P{{b}^{206}}}{\text{No}\text{. of atoms of }{{U}^{238}}\text{ left}}={{e}^{-\lambda t-1}}\]   where l is the decay constant of uranium-238 Alternatively, \[t=\frac{2.303}{\lambda }\log \frac{\text{Initial amount of }{{U}^{238}}}{\text{Amount of }{{U}^{238}}\text{ in the mineral present till date}}\] Similarly, the less abundant isotope of uranium, \[{{U}^{235}}\] eventually decays to \[P{{b}^{207}};\,T{{h}^{232}}\] decays to \[P{{b}^{208}}\] and thus the ratios of \[P{{b}^{207}}/{{U}^{235}}\] and \[P{{b}^{208}}/T{{h}^{232}}\] can be used to determine the age of rocks and minerals. \[_{6}{{C}^{14}}\] (half-life 5760 years) was used by Willard Libby (Nobel lauret) in determining the age of carbon-bearing materials (e.g. wood, animal fossils, etc.) Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays). \[_{7}{{N}^{14}}+{{\,}_{0}}{{n}^{1}}\to {{\,}_{6}}{{C}^{14}}+{{\,}_{1}}{{H}^{1}}\] Thus carbon-14 is oxidised to \[C{{O}_{2}}\] and eventually ingested by plants and animals. The death of plants or animals puts an end to the intake of \[{{C}^{14}}\] from the atmosphere. After this the amount of \[{{C}^{14}}\] in the dead tissues starts decreasing due to its disintegration. \[_{6}{{C}^{14}}\to {{\,}_{7}}{{N}^{14}}+{{\,}_{-1}}{{e}^{0}}\] It has been observed that on an average, one gram of radioactive carbon emits about 12 b-particles per minute. Thus by knowing either the amount of C-14 or the number of b-particles emitted per minute per gram of carbon at the initial and final (present) stages, the age of carbon material can be determined by using the following formulae. \[\lambda =\frac{2.303}{t}\log \frac{{{N}_{0}}}{{{N}_{t}}}\] or \[t=\frac{2.303}{\lambda }\log \frac{{{N}_{0}}}{{{N}_{t}}}\] where t = Age of the fossil, l = Decay constant, \[{{N}_{0}}\]= Initial radioactivity (in the fresh wood), \[{{N}_{t}}\]= Radioactivity in the fossil The above formula can be modified as, \[t=\frac{2.303}{\lambda }\log \frac{\text{Initial ratio of }{{C}^{14}}/{{C}^{12}}\text{ (in fresh wood)}}{{{C}^{14}}/{{C}^{12}}\text{ ratio in the old wood}}\] Similarly,     tritium \[_{1}{{H}^{3}}\] has been used for dating purposes. (2) Radioactive tracers (use of radio–isotopes) : A radioactive isotope can be easily identified by its radioactivity. The radioactivity can, therefore act as a tag or label that allows studying the behaviour of the element or compounding which contains this isotope. An isotope added for this purpose is known as isotopic tracer. The radioactive tracer is also known as an isotopic tracer. The radioactive tracer is also known as an indicator because it indicates the reaction. Radioisotopes of moderate half-life periods are used for tracer work. The activity of radioisotopes can be detected by means of electroscope, the electrometer or the Geiger-Muller counter.  Tracers have been used in the following fields, (i) To diagnose many diseases : more...
Catgeory: JEE Main & Advanced

Hazards Of Radiations 

The increased pace of synthesis and use of radio isotopes has led to increased concern about the effect of radiations on matter, particularly in biological systems. The accident of Chernobyl occurred in 1986 in USSR is no older when radioisotopes caused a hazard there. The nuclear radiations (alpha, beta, gamma as well as X-rays) possess energies far in excess of ordinary bond energies and ionisation energies. Consequently, these radiations are able to break up and ionise the molecules present in living organisms if they are exposed to such radiations. This disrupts the normal functions of living organisms. The damage caused by the radiations, however, depends upon the radiations received. The resultant radiation damage to living system can be classified as, (1)Somatic or pathological damage : This affects the organism during its own life time. It is a permanent damage to living civilization produced in body. Larger dose of radiations cause immediate death whereas smaller doses can cause the development of many diseases such as paralysis, cancer, leukaemia, burns, fatigue, nausea, diarrhoea, gastrointestinal problems etc. some of these diseases are fatal. Many scientists presently believe that the effect of radiations is proportional to exposure, even down to low exposures. This means that any amount of radiation causes some finite risk to living civilization. (2) Genetic damage : As the term implies, radiations may develop genetic effect. This type of damage is developed when radiations affect genes and chromosomes, the body's reproductive material. Genetic effects are more difficult to study than somatic ones because they may not become apparent for several generations.
Catgeory: JEE Main & Advanced

Reversible And Irreversible Reactions

A chemical reaction is said to have taken place when the concentration of reactants decreases, and the concentration of the products increases with time. The chemical reactions are classified on the basis of the extent to which they proceed, into the following two classes; (1) Reversible reactions : Reaction in which entire amount of the reactants is not converted into products is termed as reversible reaction. (i) Characteristics of reversible reactions (a) These reactions can be started from either side,               (b) These reactions are never complete, (c) These reactions have a tendency to attain a state of equilibrium, in which Free energy change is zero (DG = 0), (d) This sign \[(\rightleftharpoons )\] represents the reversibility of the reaction, (ii) Examples of reversible reactions (a) Neutralisation between an acid and a base either of which or both are weak e.g., \[CH{}_{3}COOH+Na\,OH\]\[\rightleftharpoons \]\[C{{H}_{3}}COONa+{{H}_{2}}O\] (b) Salt hydrolysis, e.g., \[Fe\,C{{l}_{3}}+3{{H}_{2}}O\]\[ \rightleftharpoons \] \[Fe\,{{\left( OH \right)}_{3}}+3HCl\,\]\[\] (c) Thermal decomposition, e.g., \[PC{{l}_{5}}_{(g)\,}\]\[\rightleftharpoons \]\[PC{{l}_{3}}_{(g)}\,+C{{l}_{2}}_{(g)}\]\[-Q\] (d) Esterification, e.g.,  \[C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\] \[\rightleftharpoons \]\[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\] (e) Evaporation of water in a closed vessel, e.g., \[{{H}_{2}}{{O}_{(l)}}\] \[\rightleftharpoons \]\[{{H}_{2}}{{O}_{(g)}}\]\[-Q\] (2) Irreversible reactions : Reaction in which entire amount of the reactants is converted into products is termed as irreversible reaction. (i) Characteristics of irreversible reactions (a) These reactions proceed only in one direction (forward direction), (b) These reactions can proceed to completion, (c) In an irreversible reaction, DG < 0, (d) The arrow \[(\to )\] is placed between reactants and products, (ii) Examples of irreversible reactions (a) Neutralisation between strong acid and strong base e.g., \[NaOH+HCl\to NaCl+{{H}_{2}}O\]\[+13.7\ kcal\] (b) Double decomposition reactions or precipitation reactions e.g., \[BaC{{l}_{2}}_{(aq)}+{{H}_{2}}S{{O}_{4}}_{(aq)}\to BaS{{O}_{4}}_{(s)}\downarrow +2HC{{l}_{(aq)}}\] (c) Thermal decomposition, e.g., \[2KCl{{O}_{3}}_{(s)}\xrightarrow{Mn{{O}_{2}},\Delta }2KC{{l}_{(s)}}+3{{O}_{2}}\uparrow \]           (d) Redox reactions, e.g., \[SnC{{l}_{2}}_{(aq)}+2FeC{{l}_{3}}_{(aq)}\to SnC{{l}_{4}}_{(aq)}+2FeC{{l}_{2}}_{(aq)}\]
Catgeory: JEE Main & Advanced

Equilibrium And Its Dynamic Nature

"Equilibrium is the state at which the concentration of reactants and products do not change with time. i.e. concentrations of reactants and products become constant." The important characteristics of equilibrium state are, (1) Equilibrium state can be recognised by the constancy of certain measurable properties such as pressure, density, colour, concentration etc. by changing these conditions of the system, we can control the extent to which a reaction proceeds. (2) Equilibrium state can only be achieved in close vessel. (3) Equilibrium state is reversible in nature. (4) Equilibrium state is also dynamic in nature. (5) At equilibrium state, Rate of forward reaction = Rate of backward reaction (6) At equilibrium state, DG = 0, so that DH = TDS.
Catgeory: JEE Main & Advanced

Law Of Mass Action And Equilibrium Constant

On the basis of observations of many equilibrium reactions, two Norwegian chemists Goldberg and Waage suggested (1864) a quantitative relationship between the rates of reactions and the concentration of the reacting substances. This relationship is known as law of mass action. It states that “The rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants at a constant temperature at any given time.” The molar concentration i.e. number of moles per litre is also called active mass. It is expressed by enclosing the symbols of formulae of the substance in square brackets. For example, molar concentration of A is expressed as [A]. Consider a simple reversible reaction  \[aA+bB\]? \[cC+dD\]          (At a certain temperature) According to law of mass action Rate of forward reaction \[\propto {{[A]}^{a}}{{[B]}^{b}}={{k}_{f}}{{[A]}^{a}}{{[B]}^{b}}\] Rate of backward reaction \[\propto {{[C]}^{c}}{{[D]}^{d}}={{k}_{b}}{{[C]}^{c}}{{[D]}^{d}}\] At equilibrium, Rate of forward reaction = Rate of backward reaction                      \[{{k}_{f}}{{[A]}^{a}}{{[B]}^{b}}={{k}_{b}}{{[C]}^{c}}{{[D]}^{d}}\]                          \[\frac{{{k}_{f}}}{{{k}_{b}}}={{K}_{c}}=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}\] Where, \[{{K}_{c}}\] is called equilibrium constant. In terms of partial pressures, equilibrium constant is denoted by \[{{K}_{p}}\] and                          \[{{K}_{p}}=\frac{P_{C}^{c}\ P_{D}^{d}}{P_{A}^{a}\ P_{B}^{b}}\] In terms of  mole fraction, equilibrium constant is denoted by \[{{K}_{x}}\] and                          \[{{K}_{x}}=\frac{{{({{X}_{C}})}^{c}}\ {{({{X}_{D}})}^{d}}}{{{({{X}_{A}})}^{a}}\ {{({{X}_{B}})}^{b}}}\]  Relation between Kp, Kc and Kx                          \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]                          \[{{K}_{p}}={{K}_{x}}{{(P)}^{\Delta n}}\] Dn = number of moles of gaseous products – number of moles of gaseous reactants in chemical equation. As a general rule, the concentration of pure solids and pure liquids are not included when writing an equilibrium equation.  
Value of Dn Relation between Kp and Kc Units of Kp Units of Kc
0 Kp = Kc No unit No unit
>0 Kp > Kc (atm)Dn (mole l–1)Dn
<0 Kp < Kc (atm)Dn (mole l–1)Dn
  Characteristics of equilibrium constant (1) The value of equilibrium constant is independent of the original concentration of reactants. (2) The equilibrium constant has a definite value for every reaction at a particular temperature. However, it varies with change in temperature. (3) For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction. In general,  \[{{K}_{\text{forward reaction}}}=\frac{1}{{{{{K}'}}_{\text{backward reaction}}}}\] (4) The value of an equilibrium constant tells the extent to which a reaction proceeds in the forward or reverse direction. (5) The equilibrium constant is independent of the presence of catalyst. (6) The value of equilibrium constant changes more...
Catgeory: JEE Main & Advanced

Types Of Equilibria

  The equilibrium between different chemical species present in the same or different phases is called chemical equilibrium. There are two types of chemical equilibrium. (1) Homogeneous equilibrium : The equilibrium reactions in which all the reactants and the products are in the same phase are called homogeneous equilibrium reactions. \[{{C}_{2}}{{H}_{5}}OH\,(l)+C{{H}_{3}}COOH\,(l)\]? \[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}\,(l)+{{H}_{2}}O(l)\]                            \[{{N}_{2}}\,(g)+3{{H}_{2}}\,(g)\] ? \[2N{{H}_{3}}(g)\]                \[2S{{O}_{2}}\,(g)+{{O}_{2}}\,(g)\] ? \[2S{{O}_{3}}(g)\] (2) Heterogeneous equilibrium : The equilibrium reactions in which the reactants and the products are present in different phases are called heterogeneous equilibrium reactions.           \[2NaHC{{O}_{3}}\,(s)\]? \[N{{a}_{2}}C{{O}_{3}}\,(s)+C{{O}_{2}}\,(g)+{{H}_{2}}O\,(g)\] \[Ca{{(OH)}_{2}}(s)+{{H}_{2}}O\,(l)\] ? \[C{{a}^{2+}}(aq)+2O{{H}^{-}}\,(aq)\]                  \[CaC{{O}_{3}}\,(s)\] ? \[CaO\,(s)+C{{O}_{2}}\,(g)\]                    \[{{H}_{2}}O\,(l)\] ? \[{{H}_{2}}O\,(g)\]   Homogeneous equilibria and equations for equilibrium constant (Equilibrium pressure is P atm in a V L flask)
  \[\Delta n=0\,;\,\,{{K}_{p}}={{K}_{c}}\] \[\Delta n<0\] ;  \[{{K}_{p}}<{{K}_{c}}\] \[\Delta n>0;\ {{K}_{p}}>{{K}_{c}}\]
  \[\underset{(g)}{\mathop{{{H}_{2}}}}\,\]+   \[\underset{(g)}{\mathop{{{I}_{2}}}}\,\]   ?     \[\underset{(g)}{\mathop{2HI}}\,\]     \[\underset{(g)}{\mathop{{{N}_{2}}}}\,+\underset{(g)}{\mathop{3{{H}_{2}}}}\,\]? \[\underset{(g)}{\mathop{2N{{H}_{3}}}}\,\] \[\underset{(g)}{\mathop{2S{{O}_{2}}}}\,+\underset{(g)}{\mathop{{{O}_{2}}}}\,\]?\[2\underset{(g)}{\mathop{S{{O}_{3}}}}\,\] \[\underset{(g)}{\mathop{PC{{l}_{_{5}}}}}\,\]?\[\underset{(g)}{\mathop{PC{{l}_{3}}}}\,+\underset{(g)}{\mathop{C{{l}_{2}}}}\,\]
Initial mole   1               1           0     1          3              0     2             1         0    1              0        0
Mole at Equilibrium (1–x)     (1– x)           2x (1–x)      (3–3x)         2x (2–2x)     (1–x)        2x     (1–x)       x         x
Total mole at equilibrium                2           (4 – 2x)                        (3 – x)                                              (1 + x)
Active masses \[\left( \frac{1-x}{V} \right)\]  \[\left( \frac{1-x}{V} \right)\]   \[\frac{2x}{V}\] \[\left( \frac{1-x}{V} \right)\] \[3\,\left( \frac{1-x}{V} \right)\] \[\left( \frac{2x}{V} \right)\] \[\left( \frac{2-2x}{V} \right)\]  \[\left( \frac{1-x}{V} \right)\]   \[\left( \frac{2x}{V} \right)\] \[\left( \frac{1-x}{V} \right)\]      \[\left( \frac{x}{V} \right)\]    \[\left( \frac{x}{V} \right)\]
Mole fraction \[\left( \frac{1-x}{2} \right)\] \[\left( \frac{1-x}{2} \right)\]    \[\frac{2x}{2}\] \[\frac{1-x}{2\,\left( 2-x \right)}\]\[\frac{3}{2}\left( \frac{1-x}{2-x} \right)\]\[\frac{x}{(2-x)}\] \[\left( \frac{2-2x}{3-x} \right)\]     \[\left( \frac{1-x}{3-x} \right)\,\,\ \ \left( \frac{2x}{3-x} \right)\] \[\left( \frac{1-x}{1+x} \right)\]  \[\left( \frac{x}{1+x} \right)\]  \[\left( \frac{x}{1+x} \right)\]
Partial pressure \[p\,\left( more...
Catgeory: JEE Main & Advanced

Le-Chaterlier Principle and It's Application

Factors which Change the State of Equilibrium: Le-Chatelier's Principle.   Le-Chatelier and Braun (1884), French chemists, made certain generalizations to explain the effect of changes in concentration, temperature or pressure on the state of system in equilibrium. When a system is subjected to a change in one of these factors, the equilibrium gets disturbed and the system readjusts itself until it returns to equilibrium. The generalization is known as Le-Chatelier's principle. It may stated as: “Change in any of the factors that determine the equilibrium conditions of a system will shift the equilibrium in such a manner to reduce or to counteract the effect of the change.” The principle is very helpful in predicting qualitatively the effect of change in concentration, pressure or temperature on a system in equilibrium. This is applicable to all physical and chemical equilibria.   (1) Effect of change of concentration : According to Le-Chatelier's principle, “If concentration of one or all the reactant species is increased, the equilibrium shifts in the forward direction and more of the products are formed. Alternatively, if the concentration of one or all the product species is increased, the equilibrium shifts in the backward direction forming more reactants.” Thus,                         Increase in concentration of any of the reactants \[\underset{equilibrium\,\,to}{\mathop{\xrightarrow{Shifts\,\,the}}}\,\] Forward direction                         Increase in concentration of any of the products \[\underset{equilibrium\,\,to}{\mathop{\xrightarrow{Shifts\,\,the}}}\,\] Backward direction   (2) Effect of change of temperature : According to Le-Chatelier's principle, “If the temperature of the system at equilibrium is increased (heat is supplied), the equilibrium will shift in the direction in which the added heat is absorbed. In other words, the equilibrium will shift in the direction of endothermic reaction with increase in temperature. Alternatively, the decrease in temperature will shift the equilibrium towards the direction in which heat is produced and, therefore, will favour exothermic reaction.” Thus,                                     Increase in temperature \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Endothermic reaction                                     Decrease in temperature \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Exothermic reaction   (3) Effect of change of pressure : Pressure has hardly effect on the reactions carried in solids and liquids. However, it does influence the equilibrium state of the reactions that are carried in the gases. The effect of pressure depends upon the number of moles of the reactants and products involved in a particular reaction. According to Le-Chatelier's principle, “Increase in pressure shifts the equilibrium in the direction of decreasing gaseous moles. Alternatively, decrease in pressure shifts the equilibrium in the direction of increasing gaseous moles and pressure has no effect if the gaseous reactants and products have equal moles.” Thus,                                     Increase in pressure \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Decreasing gaseous moles                                     Decrease in pressure \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Increasing gaseous moles   (4) Effect of volume change : We know that increase in pressure means decrease in volume, so the effect of change of volume will be exactly reverse to that of pressure. Thus, “decreasing the volume of a mixture of gases at equilibrium shifts the equilibrium in the direction of decreasing gaseous moles while increasing the volume shifts the equilibrium in the direction of increasing more...
Catgeory: JEE Main & Advanced

Application Of Le-Chatelier's Principle

The Le-Chateliers principle has a great significance for the chemical, physical systems and in every day life in a state of equilibrium. (1) Applications to the chemical equilibrium (i) Synthesis of ammonia (Haber’s process) \[\underset{1\ vol}{\mathop{{{N}_{2}}}}\,+\underset{3\ vol}{\mathop{3{{H}_{2}}}}\,\] \[\rightleftharpoons \] \[\underset{2\ vol}{\mathop{2N{{H}_{3}}}}\,+23kcal\] (exothermic) (a) High pressure \[(\Delta n<0)\] (b) Low temperature (c) Excess of \[{{N}_{2}}\] and \[{{H}_{2}}\] (d) Removal of \[N{{H}_{3}}\] favours forward reaction. (ii) Formation of sulphur trioxide \[\underset{2\ vol}{\mathop{2S{{O}_{2}}}}\,+\underset{1\ vol}{\mathop{{{O}_{2}}}}\,\] \[\rightleftharpoons \] \[\underset{2\ vol}{\mathop{2S{{O}_{3}}}}\,+45\ kcal\]  (exothermic)
  • High pressure \[(\Delta n<0)\]
(b) Low temperature (c) Excess of \[S{{O}_{2}}\] and \[{{O}_{2}}\], favours the reaction in forward direction. (iii) Synthesis of nitric oxide \[\underset{1\ vol}{\mathop{{{N}_{2}}}}\,+\underset{1\ vol}{\mathop{{{O}_{2}}}}\,\] \[\rightleftharpoons \] \[\underset{2\ vol}{\mathop{2N{{O}_{{}}}}}\,-43.2\ kcal\] (endothermic ) (a) High temperature (b) Excess of \[{{N}_{2}}\] and \[{{O}_{2}}\] (c) Since reaction takes place without change in volume   i.e., \[\Delta n=0\], pressure has no effect on equilibrium. (iv) Formation of nitrogen dioxide \[\underset{2\ vol}{\mathop{2N{{O}_{{}}}}}\,+\underset{1\ vol}{\mathop{{{O}_{2}}}}\,\] \[\rightleftharpoons \] \[\underset{2\ vol}{\mathop{2N{{O}_{2}}}}\,+27.8\ Kcal\]
  • High pressure
  • Low temperature
(c) Excess of \[NO\] and \[{{O}_{2}}\] favours the reaction in forward direction. (v) Dissociation of phosphours pentachloride   \[\underset{1\ vol}{\mathop{PC{{l}_{5}}}}\,\] \[\rightleftharpoons \]  \[\underset{1\ vol}{\mathop{PC{{l}_{3}}}}\,+\underset{1\ vol}{\mathop{C{{l}_{2}}}}\,-15\ kcal\] (a) Low pressure or high volume of the container, \[\Delta n>0\] (b) High temperature (c) Excess of \[PC{{l}_{5}}\]. (2) Applications to the physical equilibrium (i) Melting of ice (Ice – water system) \[\underset{\text{(Greater}\ \text{Volume)}}{\mathop{\text{Ice}}}\,\] \[\rightleftharpoons \]  \[\underset{\text{(Lesser}\ \text{Volume)}}{\mathop{\text{Water}}}\,-x\ kcal\] (In this reaction volume is decreased from 1.09 c.c. to     1.01 c.c. per gm.) (a) At high temperature more water is formed as it absorbs heat. (b) At high pressure more water is formed as it is accompanied by decrease in volume. (c) At higher pressure, melting point of ice is lowered, while boiling point of water is increased. (ii) Melting of sulphur : \[{{S}_{(s)}}\] \[\rightleftharpoons \] \[{{S}_{(l)}}-x\ kcal\] (This reaction accompanies increase in volume.) (a) At high temperature, more liquid sulphur is formed. (b) At higher pressure, less sulphur will melt as melting increases volume. (c) At higher pressure, melting point of sulphur is increased. (iii) Boiling of water (water- water vapour system) \[\underset{\text{(Low}\ \text{volume)}}{\mathop{\text{Water}}}\,\]\[\rightleftharpoons \]\[\underset{\text{(Higher}\ \text{volume)}}{\mathop{\text{Water}\ \text{Vapours}}}\,-x\ kcal\] (It is accompanied by absorption of heat and increase in volume.) (a) At high temperature more vapours are formed. (b) At higher pressure, vapours will be converted to liquid as it decreases volume. (c) At higher pressure, boiling point of water is increased (principle of pressure cooker). (iv) Solubility of salts : If solubility of a salt is accompanied by absorption of heat, its solubility increases with rise in temperature; e.g., \[N{{H}_{4}}Cl,\ {{K}_{2}}S{{O}_{4}},\ KN{{O}_{3}}\] etc. \[KN{{O}_{3(s)}}+(aq)\xrightarrow{{}}KN{{O}_{3(aq)}}-x\ kcal\] On the other hand if it is accompanied by evolution of heat, solubility decreases with increase in temperature; e.g., \[CaC{{l}_{2}},\] \[Ca{{(OH)}_{2}},NaOH,\ KOH\]etc. \[Ca{{(OH)}_{2(\ s)}}+(aq)\xrightarrow{{}}Ca{{(OH)}_{2\ (aq)}}+x\ kcal\]
Catgeory: JEE Main & Advanced

Relation Between Vapour Density And Degree Of Dissociation

In the following reversible chemical equation.                                      \[A\]  \[\rightleftharpoons \]   \[yB\] Initial mole              1              0 At equilibrium  (1–x)                   yx        x = degree of dissociation Number of moles of \[A\] and \[B\] at equilibrium \[=1-x+yx=1+x(y-1)\] If initial volume of 1 mole of A is V, then volume of equilibrium mixture of \[A\] and \[B\] is,\[=[1+x(y-1)]V\] Molar density before dissociation,  \[D=\frac{\text{molecular}\ \text{weight}}{\text{volume}}=\frac{m}{V}\] Molar density after dissociation, \[d=\frac{m}{[1+x(y-1)]V}\];\[\frac{D}{d}=[1+x(y-1)]\] ; \[x=\frac{D-d}{d(y-1)}\] \[y\] is the number of moles of products from one mole of reactant. \[\frac{D}{d}\] is also called Van’t Hoff factor. In terms of molecular mass,\[x=\frac{M-m}{(y-1)\,m}\] Where \[M=\] Initial molecular mass,    \[m=\] molecular mass at equilibrium  Thus for the equilibria (I) \[PC{{l}_{5(g)}}\] \[\rightleftharpoons \]  \[PC{{l}_{3(g)}}+C{{l}_{2(g)}},y=2\] (II) \[{{N}_{2}}{{O}_{4(g)}}\] \[\rightleftharpoons \]  \[2N{{O}_{2(g)}},\ y=2\] (III) \[2N{{O}_{2}}\] ? \[{{N}_{2}}{{O}_{4}},\ y=\frac{1}{2}\] \[\therefore \] \[x=\frac{D-d}{d}\] (for I and II)  and \[x=\frac{2(d-D)}{d}\] (for III) Also \[D\times 2=\] Molecular weight (theoretical value) \[d\times 2=\] Molecular weight (abnormal value) of the mixture.
Catgeory: JEE Main & Advanced

Electrical Conductors

Substances, which allow electric current to pass through them, are known as conductors or electrical conductors. Conductors can be divided into two types, (1) Conductors which conduct electricity without undergoing any chemical change are known as metallic or electronic conductors. (2) Conductors which undergo decomposition (a chemical change) when an electric current is passed through them are known as electrolytic conductors or electrolytes. Electrolytes are further divided into two types on the basis of their strengths, (i) Substances which almost completely ionize into ions in their aqueous solution are called strong electrolytes. Degree of ionization for this type of electrolyte is one i.e., \[\alpha \approx 1\]. For example : \[HCl,\ {{H}_{2}}S{{O}_{4}},\ NaCl,\ HN{{O}_{3}},\ KOH,\ \]\[NaOH,\ \] \[HN{{O}_{3}},AgN{{O}_{3}},\ CuS{{O}_{4}}\] etc. means all strong acids, bases and all types of salts. (ii) Substances which ionize to a small extent in their aqueous solution are known as weak electrolytes. Degree of ionization for this types of electrolytes is \[\alpha <<<1\]. For example :\[{{H}_{2}}O,\ C{{H}_{3}}COOH,\ N{{H}_{4}}OH,\ \]HCN, \[Liq.\ S{{O}_{2}}\], \[HCOOH\] etc. means all weak acids and bases.
Catgeory: JEE Main & Advanced

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