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	<title>IIT JEE Blogs &#187; Chemistry</title>
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		<title>Introduction to Chemical Equilibrium and Ionic Equilibrium</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/introduction-to-chemical-equilibrium-and-ionic-equilibrium/</link>
		<comments>http://smartlearnwebtv.com/iitjee/IITJEE/introduction-to-chemical-equilibrium-and-ionic-equilibrium/#comments</comments>
		<pubDate>Mon, 28 Dec 2009 03:30:04 +0000</pubDate>
		<dc:creator>janani</dc:creator>
				<category><![CDATA[Chemical Equilibrium and Ionic Equilibrium]]></category>

		<guid isPermaLink="false">http://smartlearnwebtv.com/iitjee/?p=909</guid>
		<description><![CDATA[Out of the many chemical reactions certain reactions are reversible reaction i.e. they proceed in forward as well as backward direction i.e. reactants combine to give products and products gives back the reactants. In such reactions after some time of commencement of reaction system a stage is reached, when the properties of the system remain [...]]]></description>
			<content:encoded><![CDATA[<p>Out of the many chemical reactions certain reactions are reversible reaction i.e. they proceed in forward as well as backward direction i.e. reactants combine to give products and products gives back the reactants. In such reactions after some time of commencement of reaction system a stage is reached, when the properties of the system remain constant with time, then the reaction system is said to be in equilibrium. There are many kinds of equilibrium but we will focus on Dynamic equilibrium only, then we will study the important laws governing equilibrium, Le Chatelier&#8217;s Principle &amp; Law of Mass Actions. Then we will see some important constants associated with the equilibrium &amp; relationship between them. Then we will end with some special features of Equilibrium.</p>
<p><strong>OBJECTIVE</strong></p>
<p>The stage of equilibrium can be reached starting from either reactants or products only or with any amounts of reactants and products taken together. At equilibrium, the processes of conversion of reactant(s) into product(s) and vice versa do not cease but both of them occur simultaneously at equal rates. Thus an equilibrium is actually a dynamic equilibrium. Our aim in this chapter is to learn about Chemical Equilibrium &amp; its properties so that we can manipulate the reactants or product concentrations along with the reaction conditions in order to get a desired effect.</p>
<p><strong>PRE-REQUISITE</strong></p>
<p>The knowledge of the following will help you understand the chapter easily &amp; effectively.</p>
<p>Þ            Gaseous State and Gas laws.</p>
<p>Þ            Stoichiometry</p>
<p>Þ            Mole concept.</p>
<p><strong>Core Concept</strong></p>
<p>Certain reactions are reversible reaction i.e. they proceed in forward as well as backward direction i.e. reactants combine to give products and products give back reactants. In such reactions after some time of commencement of reaction a stage is reached, when the properties of the system remain constant with time, then the reaction system is said to be in equilibrium. The stage of equilibrium can be reached starting from either reactants or products only or with any amounts of reactants and products taken together. At equilibrium, the processes of conversion of reactant(s) into product(s) and vice versa do not cease but both of them occur simultaneously at equal rates. Thus an equilibrium is actually a dynamic equilibrium. Our aim in this chapter is to learn about Chemical Equilibrium.</p>
<p><strong>The concept of Equilibrium:</strong></p>
<p>It is a common observation that clothes dry quicker when there is a breeze or when we keep shaking it. Have you ever thought why does this happen? Well, we can get an answer to this query if we know what an equilibrium is.</p>
<p>In the study of any chemical reaction, two questions are of vital importance viz. How far a reaction would go and how fast will it reach the goal.</p>
<p>Well, equilibrium is that state of a system in which it has minimum energy. If we disturb the equilibrium state (by changing T, p, volume, concentration of reactants, products or adding an inert gas), the system always tries to come back to an equilibrium state. This tendency can be exploited for getting maximum production in industry.</p>
<p><strong>Equilibrium could be of two types</strong></p>
<p><strong> </strong></p>
<p><strong>Physical Equilibrium  <em> </em></strong></p>
<p>It is found between two different states of a substance, e.g., water at its boiling point has got two states, water (liquid) and water vapour (gas). Both of these are in equilibrium with each other.</p>
<p><strong>Chemical Equilibrium </strong></p>
<p>When a reaction starts, reactants are present at some definite concentrations. As reaction proceeds, concentration of reagents decreases and that of products increases. Sooner or later, however, concentration levels off and becomes constant. A state in which concentration no longer changes with time (this state which persists as long as the system is free of external perturbations) is known as the state of chemical equilibrium</p>
<p>We use the symbol to denote equilibrium in contrast to the symbol → where the reaction goes to completion.</p>
<p>The reaction in which products recombine to give back reactants or the reaction which proceeds in forward as well as backward direction is termed as <strong>reversible reaction </strong>or in other words it is a dynamic state.</p>
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		<title>Physical state of reactants and products</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/physical-state-of-reactants-and-products/</link>
		<comments>http://smartlearnwebtv.com/iitjee/IITJEE/physical-state-of-reactants-and-products/#comments</comments>
		<pubDate>Sat, 19 Dec 2009 03:30:22 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[Chemical Energetics]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[IIT JEE]]></category>
		<category><![CDATA[physical state]]></category>
		<category><![CDATA[products]]></category>
		<category><![CDATA[reactants]]></category>

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		<description><![CDATA[Heat energy is involved for changing the physical state of a chemical substance. For example in the conversion of water into steam, heat is absorbed and heat is evolved when steam is condensed.
 Example
Consider the following two reactions –

It is observed that there is difference in the value of ΔH if water is obtained in [...]]]></description>
			<content:encoded><![CDATA[<p>Heat energy is involved for changing the physical state of a chemical substance. For example in the conversion of water into steam, heat is absorbed and heat is evolved when steam is condensed.</p>
<p><strong> Example</strong></p>
<p>Consider the following two reactions –</p>
<p style="text-align: center"><img class="size-full wp-image-896 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che5.2.png" alt="che5.2" width="376" height="125" /></p>
<p>It is observed that there is difference in the value of Δ<em>H</em> if water is obtained in gaseous or liquid state. Δ<em>H</em> value in second case is higher because heat is evolved when steam condenses. Hence, physical state always affects the heat of reaction.</p>
<p><strong> </strong></p>
<p><strong>Allotropic forms of the elements</strong></p>
<p>Heat energy is also involved when one allotropic form of an element is converted into another. Thus, the value of Δ<em>H</em> depends on the allotropic form used in the reaction.</p>
<p><strong>Example:</strong></p>
<p>The  value of Δ<em>H</em> is different when carbon in the form of diamond or graphite is used.</p>
<p style="text-align: center"><img class="size-full wp-image-897 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che5.3.png" alt="che5.3" width="292" height="96" /></p>
<p>The difference between two values is equal to the heat absorbed when 12 g of diamond is converted into 12 g of graphite</p>
<p style="text-align: center"><img class="size-full wp-image-898 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che5.4.png" alt="che5.4" width="238" height="46" /></p>
<p><strong>Reaction carried out at constant pressure or constant volume</strong></p>
<p>When a chemical reaction  occurs at constant volume, the heat change is called the internal energy. However, most of the reactions are carried out at constant pressure and the enthalpy change is termed as the energy of reaction at constant pressure.</p>
<p>The relation between Δ<em>H</em> (Enthalpy change) and Δ<em>E</em> (Internal energy change) is given as follows:</p>
<p>Δ<em>E</em> + Δ<em>n<sub>g</sub>RT</em> = Δ<em>H</em></p>
<p>D<em>n<sub>g</sub></em> =             (Total number of moles of products) – (total number of moles of  reactants).</p>
<p><em>R</em> = Gas Constant</p>
<p><em>T</em> = Temperature (in Kelvin)</p>
<p>The difference between ΔH and ΔE value is negligible when solids and liquids are involved in a chemical change. But, in reactions which involve gases, the difference in two values is considerable.</p>
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		<title>Introduction &#8211; Chemical Energetics</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/introduction-6/</link>
		<comments>http://smartlearnwebtv.com/iitjee/IITJEE/introduction-6/#comments</comments>
		<pubDate>Sat, 12 Dec 2009 09:56:15 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[Chemical Energetics]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[Endothermic Reaction]]></category>
		<category><![CDATA[IIT JEE]]></category>

		<guid isPermaLink="false">http://smartlearnwebtv.com/iitjee/?p=861</guid>
		<description><![CDATA[In chemistry, we deal with processes, is which are invariably associated with transfer of energy between the system under study and its surroundings. For example, heat is evolved when an acid is neutralised by a base. The heat transfer is basically due to the conservation of energy or first law of thermodynamics. It is of [...]]]></description>
			<content:encoded><![CDATA[<p>In chemistry, we deal with processes, is which are invariably associated with transfer of energy between the system under study and its surroundings. For example, heat is evolved when an acid is neutralised by a base. The heat transfer is basically due to the conservation of energy or first law of thermodynamics. It is of prime importance to a chemist to understand these energy changes &amp; use this knowledge in his study of the subject.</p>
<p><strong> </strong></p>
<p><strong>Objective</strong></p>
<p>The branch of physical chemistry, which deals with the study of heat changes, accompanying a chemical reaction is termed as Thermochemistry. Thermo (heat) Dynamics (work) is the study of those interactions among various materials which involve the transfer of heat, and the performance of work. Our aim in this chapter will be [to mathematically &amp; conceptually understand] the changes that the system &amp; surrounding undergo when exchange of energy takes place.</p>
<p><strong> PRE-REQUISITE</strong></p>
<p>Þ    Since this subject is incomplete without the use of mathematical relationships, it is important that we must understand the basic operations of Logarithms, Ratios etc.</p>
<p>Þ    The units &amp; dimensions come in extremely handy in dealing with unknown variables &amp; constants.</p>
<p>Þ    Basic stoichiometry &amp; Mole concept are also important.</p>
<p><strong>CORE CONCEPTS</strong></p>
<p>The heat transfer is basically due to the conservation of energy or first law of thermodynamics. Thermo (heat) Dynamics (work) is the study of those interactions among various materials which involve the transfer of heat, and the performance of work.</p>
<p>Thermochemistry basically deals with the transfer of heat between a chemical system and its surrounding.</p>
<p>A system is defined as a specified part of the universe or specified portion of the matter which is under experimental investigation and the rest of the universe i.e. all other matter which can interact with the system, is surrounding.</p>
<p>Note</p>
<p>Þ      To calculate the heat transferred, the reactants and the products must be at the same temperature.</p>
<p>Depending on the heat transferred the reactions can be classified as</p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Exothermic Reaction</strong></p>
<p>The reaction in which heat is transferred to the surroundings from the system.</p>
<p><strong>Endothermic Reaction</strong></p>
<p>The reaction in which heat is transferred to the system from the surroundings.</p>
<p>By SI convention, the heat transferred is taken as negative and positive for exothermic and endothermic reactions, respectively. In other words, the process which increases the energy of the system is taken as +ve and which decrease the energy of the system is taken as –ve.</p>
<p>The molar enthalpy H<sub>m</sub> of any substance is a function of temperature and pressure, i.e. H<sub>m</sub> = H<sub>m</sub>(T, p). The pressure dependence is removed by defining the standard molar enthalpy H<sup>°</sup><sub>m</sub>, which is the enthalpy of the substance at the standard pressure of 101.325 k Pa.</p>
<p>Note</p>
<p>Þ      It is impossible to determine the absolute value of enthalpy.</p>
<p>It is impossible to determine absolute value of enthalpy. The values we observe are based on the SI convention. However relative enthalpies of substances can be determined if the enthalpy of free elements at 25 °C and 1 atmosphere pressure are taken arbitrary as zero or in other words, the enthalpy of every element in its stable state of aggregation at 101.325 k Pa (or 1 atmosphere pressure) and at 25 °C is assigned a zero value.</p>
<p>At 101.325 kPa and 298.15 K, the stable state of aggregation of Nitrogen is the gaseous state, hence <em>H</em><sup>°</sup><sub>m</sub> (N2<sub>g</sub>) = 0.</p>
<p>If an element exists in more than one allotropic forms, the most stable allotrope is assigned zero value.</p>
<p><strong>Example</strong></p>
<p>Solid sulphur (rhombic) and solid carbon (graphite) are assigned a zero standard molar enthalpy. i.e., H°.</p>
<p><strong>Example:</strong></p>
<p>To find out the standard molar enthalpies of various substances, the above conventions are used. For example, consider the following reaction:</p>
<p style="text-align: center"><img class="size-full wp-image-862 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che5.1.png" alt="che5.1" width="475" height="227" /></p>
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		<title>Grahams Law Of Diffusion</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/grahams-law-of-diffusion/</link>
		<comments>http://smartlearnwebtv.com/iitjee/IITJEE/grahams-law-of-diffusion/#comments</comments>
		<pubDate>Wed, 09 Dec 2009 09:44:04 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[States of Matter – Gaseous and Liquid State]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[Grahams Law Of Diffusion]]></category>
		<category><![CDATA[IIT]]></category>
		<category><![CDATA[IIT JEE]]></category>
		<category><![CDATA[Kinetic Theory of Gases]]></category>

		<guid isPermaLink="false">http://smartlearnwebtv.com/iitjee/?p=848</guid>
		<description><![CDATA[According to the Graham&#8217;s Law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of it&#8217;s density or molecular weights. If r1 and r2 are the rates of diffusion of two gases, whose densities under the given conditions are d1 and d2 respectively, then we have

(M1 and M2 are the respective molecular [...]]]></description>
			<content:encoded><![CDATA[<p>According to the Graham&#8217;s Law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of it&#8217;s density or molecular weights. If <em>r</em>1 and <em>r</em>2 are the rates of diffusion of two gases, whose densities under the given conditions are d1 and d2 respectively, then we have</p>
<p style="text-align: center"><img class="size-full wp-image-854 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che3.2.png" alt="che3.2" width="135" height="150" /></p>
<p>(<em>M</em><sub>1</sub> and <em>M</em><sub>2</sub> are the respective molecular weights of the two gases) and <em>d</em><sub>1</sub> and <em>d</em><sub>2</sub> are their respective vapour densities.)</p>
<p>Knowing the experimental gas laws, it is of interest to develop a theoretical model based on the structure of gases, which can co-relate the experiment. Fortunately, such a theory has been developed and is known as kinetic theory of gases.</p>
<p><strong>Kinetic Theory of Gases</strong></p>
<p>The word &#8216;Kinetic&#8217; means &#8216;Motion&#8217;. Gaseous molecules are assumed to be in constant motion. A theory with the help of which the various gas laws can be derived mathematically is known as the <strong>kinetic theory of gases</strong>.<strong> </strong></p>
<p><strong> </strong></p>
<p>The main postulates of the theory are as follows:</p>
<p>Þ      A gas is made of extremely tiny particles called molecules. The molecules of any given gas are identical and have the same mass,</p>
<p>and the molecules are assumed to be dispersed in a lot of vacant space.</p>
<p>Þ      The individual molecules are relatively far apart from each other and they exert very little attraction for each other except under collision of molecules and near the <strong>liquification point</strong>. The real volume of the gas molecules at ordinary temperatures and pressures is very small in comparison to the total volume of the gas. Here we are talking about real gases or non-ideal gases since ideal gases cannot be liquefied.</p>
<p>Þ      The gaseous molecules are in continuous random, straight line motion with very high speeds in all directions. They collide frequently and this may bring about a change in the direction of movement and a redistribution of energy between the colliding molecules. The collisions are perfectly elastic (i.e. no loss of energy) but only redistribution of energy may occur.</p>
<p>Þ      The force of gravity has negligible effect on the speed of the gas molecules.</p>
<p>Þ      The pressure exerted by a gas is due to collisions made by gas molecules on the walls of the container. Gases not only distribute themselves throughout the total volume of the container but also exert uniform pressure on every point of the container.</p>
<p>Þ      The average K.E. of the molecules is directly proportional to the absolute temperature of the gas.</p>
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		<title>Ernest Rutherford&#8217;s Model (1911)</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/ernest-rutherfords-model-1911/</link>
		<comments>http://smartlearnwebtv.com/iitjee/IITJEE/ernest-rutherfords-model-1911/#comments</comments>
		<pubDate>Fri, 04 Dec 2009 03:30:30 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[Atomic Structure]]></category>
		<category><![CDATA[Bohr's atomic model]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[IIT]]></category>
		<category><![CDATA[IIT JEE]]></category>
		<category><![CDATA[IIT JEE Chemistry]]></category>
		<category><![CDATA[Planck's quantum theory]]></category>
		<category><![CDATA[Rutherford's model]]></category>

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		<description><![CDATA[Rutherford placed thin sheets of metal in the path of a-particles in order to see how various metals would affect the a-particle trajectory.
a-particles are actually helium atoms from which electrons have been removed. Each a-particle consists of a mass equal to about 4 times that of hydrogen atom and carries a positive charge of 2 [...]]]></description>
			<content:encoded><![CDATA[<p>Rutherford placed thin sheets of metal in the path of a-particles in order to see how various metals would affect the a-particle trajectory.</p>
<p>a-particles are actually helium atoms from which electrons have been removed. Each a-particle consists of a mass equal to about 4 times that of hydrogen atom and carries a positive charge of 2 units. It is represented by symbol <img class="alignleft size-full wp-image-826" src="http://smartlearnwebtv.com/iitjee/files/che2.8.png" alt="che2.8" width="43" height="36" /> .</p>
<p style="text-align: center"><img class="size-full wp-image-827 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che2.9.png" alt="che2.9" width="405" height="319" /></p>
<p>If Thomson&#8217;s explanation were correct, a-particles would have been deflected at very small angles only</p>
<p>from a straight line path. But Rutherford found that maximum a-particles go straight, some get deflected at small angles, a few at large angles and in rare cases the deflection is 180° as shown in fig. 1.4. He hypothesised that deflection at 180° can arise only if an intense positive electric field is present inside atoms. Observations showed that a positive charge spread throughout a sphere of radius 10<sup>-8</sup> cm would be incapable of producing this field. Calculations showed that this radius should be of the order 10<sup>-13</sup> cm to account for scattering data. Based on these observations Rutherford presented following model for atom.</p>
<p>Þ        Atom consists of a nucleus which contains protons making it positively charged &amp; mass being centered here in a small space of radius 10<sup>-13</sup> cm.</p>
<p>Þ        There is a lot of empty space around nucleus in which electrons are present. The total size of the atom is of radius 10<sup>-8</sup> cm.</p>
<p>Þ        Electrons can&#8217;t be stationary as they would be pulled by nucleus. Instead they are revolving around nucleus, the necessary centripetal force for revolutions is provided by attractive forces between nucleus &amp; electrons</p>
<p style="text-align: center"><img class="size-full wp-image-828 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che2.10.png" alt="che2.10" width="371" height="340" /></p>
<p>Bohr also argued the same that the electron  (being a charged particle) should also lose energy while moving in a circle (i.e. with an acceleration). As a result its orbit should become smaller and smaller and finally it should drop into the nucleus. But the fact is that atom is stable.</p>
<p>Niel&#8217;s Bohr supplied a solution to this problem by applying Planck&#8217;s quantum theory. Let us first study the Planck&#8217;s quantum theory.</p>
<h3>Planck&#8217;s quantum theory (1901)</h3>
<p>It states</p>
<p>Þ        Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles of energy called Quanta.</p>
<p>Þ    Each quantum is associated with a definite amount of energy E which is proportional to frequency of radiation.</p>
<p style="text-align: center"><img class="size-full wp-image-830 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che2.11.png" alt="che2.11" width="92" height="52" /></p>
<p>where,             h = Planck&#8217;s constant = 6.626 * 10^-34 Joule-sec.</p>
<p>v = Frequency of the light radiation</p>
<p>Þ        A body can emit or absorb radiations only in whole multiples of quantum i.e. E = nhv where<br />
n = 1, 2, 3, &#8230;&#8230;.</p>
<h3><strong>Bohr&#8217;s atomic model</strong></h3>
<p style="text-align: center"><strong><img class="size-full wp-image-831 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che2.12.png" alt="che2.12" width="332" height="262" /></strong></p>
<p style="text-align: left">
<p>The postulates of Bohr&#8217;s atomic theory stability of an atom are as follows</p>
<p>Electron revolves in only allowed stationary orbits. Energy of different stationary states vary. An electron can be excited from a lower state to higher state with the absorption of a quantum of energy, or can come down from a higher to lower state with emission of a radiation of energy (as shown in figure 1.5) equal to energy to quantum ΔE = E2 - E1 = hv. E2 &amp; E1 are energies of the electron associated with stationary orbits.</p>
<p>The stability of the circular motion of an electron requires that the electrostatic force (due to the attraction between the nucleus and the electron) provides the necessary centrepetal force for the motion of electron.</p>
<p style="text-align: center"><img class="size-full wp-image-832 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che2.13.png" alt="che2.13" width="196" height="58" /></p>
<p>where Z &#8211; atomic number</p>
<p>e &#8211; charge on electron</p>
<p>ε0 - permittivity of the charge in vaccum</p>
<p>r &#8211; distance between positive charge &amp; electron</p>
<p>Angular momentum of electron is quantised i.e. electron can revolve only in those orbits where its angular momentum is an integral multiple of h/2Π.</p>
<p style="text-align: center"><img class="size-full wp-image-833 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che2.14.png" alt="che2.14" width="281" height="38" /></p>
<p style="text-align: left">
<p>where, v &#8211; velocity of electron</p>
<p>m &#8211; mass of electron</p>
<p>h &#8211; Planck&#8217;s constant</p>
<p>n = 1, 2, 3, &#8230;. are known as Principal quantum number.</p>
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		<title>Avogadro&#8217;s law</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/avogadros-law/</link>
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		<pubDate>Wed, 02 Dec 2009 03:30:08 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[States of Matter – Gaseous and Liquid State]]></category>
		<category><![CDATA[avagadro's law]]></category>
		<category><![CDATA[charle's law]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[IIT]]></category>
		<category><![CDATA[IIT JEE]]></category>
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		<description><![CDATA[Equal number of molecules of different gases under identical conditions of temperature and pressure occupy the same volume.
Hence, the volume occupied by one mole of an ideal gas at standard temperature (273.15 K) and pressure (101.325 K Pa) has a fixed volume (22.414 dm3). This indicates that the number of molecules contained in one mole [...]]]></description>
			<content:encoded><![CDATA[<p>Equal number of molecules of different gases under identical conditions of temperature and pressure occupy the same volume.</p>
<p>Hence,<strong><em> </em></strong>the volume occupied by one mole of an ideal gas at standard temperature (273.15 K) and pressure (101.325 K Pa) has a fixed volume (22.414 dm<sup>3</sup>). This indicates that the number of molecules contained in one mole of any real gas should be a constant quantity. This number is found to be 6.023 × 10<sup>23</sup> and is known as Avogadro number.<strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Important</strong></p>
<p>Ideal gas is a gas which follows all the above gas laws under all conditions of temperature and pressure.</p>
<p>Real gases generally do not obey the gas laws, exactly, under all conditions of temperature and pressure.</p>
<p><strong> The Ideal Gas Equation</strong></p>
<p>Combination of Boyle&#8217;s and Charle&#8217;s laws. When temp. (<em>T</em>1) is kept constant and pressure is changed from <em>p</em>1 to <em>p</em>2, Let the new volume be V <sup>&#8216;</sup>.</p>
<p style="text-align: center"><em><img class="size-full wp-image-807 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/phy3.15.png" alt="phy3.15" width="482" height="388" /></em></p>
<p><strong>Important</strong></p>
<p><em>Value of the proportionality constant k depends on:</em></p>
<p>(a)      Quantity of gas and</p>
<p>(b)         Units, in which <em>p</em>, <em>V</em> and <em>T,</em> are expressed.</p>
<p>On the basis of Avogadro&#8217;s hypothesis, 1 mole of all gases under similar conditions of temp. and  pressure occupies the same volume. Hence k will have the same value for 1 mole of any gas taken.</p>
<p><em>pV = kT. </em> &#8230;(5)</p>
<p>(k is replaced by <em>R</em> called the molar gas constant).</p>
<p>For n moles of gas considered  (5) becomes</p>
<p><em>PV = nRT.</em><strong> </strong>&#8230;(6)</p>
<p>Eq. (6) is called the <strong>ideal gas equation </strong>showing the effect on the third variable when two of the three variables are changed simultaneously for a given amount of a gas. The units of <em>R</em> varies with the units of the other parameters (<em>p, V, T</em>).</p>
<p>e.g. R has the following values</p>
<p>0.0821 litre atm/ K /mole</p>
<p>5.28 × 10<sup>19</sup> ev / K / mole</p>
<p>8.314 Joules / K / mole</p>
<p>1.99 cal / K / mole</p>
<p>0.002 k cal / K / mole</p>
<p>8.314 × 10<sup>7</sup> erg / K / mole</p>
<p><strong> </strong></p>
<p><strong>Illustrations</strong></p>
<p>1.   A two litre flask, containing O2 at 1 atm pressure is at a constant temperature at 27°C. The gas pressure is reduced to 10^–6 atm by attaching the flask to a vacuum pump. Assuming ideal behaviour, answer the following:</p>
<p>(a) What will be the volume of the gas which is left behind?</p>
<p>(b) What will be the no of molecule given in the problem?</p>
<p><strong>Sol. </strong>Given that <em>V</em><sub>1</sub> = 2 <em>l</em>, <em>p</em><sub>1</sub> = 1 atm, <em>T</em> = 27°C = 300 K</p>
<p>We have the following results</p>
<p>(a)         The volume of oxygen left behind will be the same i.e. 2 <em>l</em>.</p>
<p>(b)         The number of moles of oxygen left behind is given by</p>
<p style="text-align: center"><img class="size-full wp-image-808 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/phy3.16.png" alt="phy3.16" width="279" height="123" /></p>
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		<title>Gaseous state</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/gaseous-state/</link>
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		<pubDate>Thu, 26 Nov 2009 11:51:08 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[States of Matter – Gaseous and Liquid State]]></category>
		<category><![CDATA[boyle's law]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[gas laws]]></category>
		<category><![CDATA[gaseous state]]></category>
		<category><![CDATA[IIT]]></category>
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		<description><![CDATA[Various kinds of substances that constitute matter can be roughly divided into three categories namely- gases, liquids and solids. The existence of matter in either of these three forms is a result of the competition between two opposing intermolecular forces
(1)     The forces of attraction, which hold the molecules together.
(2)     The thermal energy of these molecules [...]]]></description>
			<content:encoded><![CDATA[<p>Various kinds of substances that constitute matter can be roughly divided into three categories namely- gases, liquids and solids. The existence of matter in either of these three forms is a result of the competition between two opposing intermolecular forces</p>
<p>(1)     The forces of attraction, which hold the molecules together.</p>
<p>(2)     The thermal energy of these molecules which tend to increase the intermolecular distances.</p>
<p>If the thermal energy of the molecules is much greater than the forces of attraction, the state of matter that result is called the gaseous state. On the other hand, if the forces of attraction are greater than the thermal energy, we have the matter in the liquid state. When these forces of attraction are much more greater than the thermal energy compared to the liquid state, we have matter in its solid state condition. However, on the application of heat, the thermal energy of the molecules can be increased and as a result the intermolecular forces of attraction would relatively decrease simultaneously.</p>
<p>Molecules in the gaseous state possess high energy and have almost no force of attraction. They are far apart and show a great uniformity in behaviour, irrespective of their chemical nature, colour or odour. They are highly compressible and can also be expanded without limit. They also produce pressure on the walls of any container uniformly in all directions. They diffuse rapidly through one other to form a homogeneous mixture, and their separation is also not very easy.</p>
<p><strong> </strong></p>
<h2>Gas laws</h2>
<p><strong>Boyle&#8217;s law</strong></p>
<p>At constant temperature, the volume of a sample of gas of definite mass varies inversely with its pressure.</p>
<p>i.e    when temperature is kept constant for a given mass of gas</p>
<p>where V = volume and  p = pressure</p>
<p>Introducing a constant <em>k</em>, we have</p>
<p><em>pV</em> =<em> k</em> = constant                                …(1)</p>
<p>The value of the proportionality constant k depends upon the following factors:</p>
<p>(1)           Nature of a gas</p>
<p>(2) Temperature of the gas, and</p>
<p>(3)           The mass of the gas</p>
<p>Hence, at constant temperature, for a given mass of a gas, Boyle&#8217;s Law states</p>
<p><em>p</em><sub>1</sub><em>V</em><sub>1</sub> = <em>p</em><sub>2</sub><em>V</em><sub>2</sub> = <em>k</em> = constant                  …(2)</p>
<p>Eq. (1) can be represented graphically as shown in figure given below.</p>
<p>The general term <strong>isotherm</strong> (i.e. at const. Temp.) is used to describe the above curves.</p>
<p style="text-align: center"><img class="size-full wp-image-770 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che3.1.png" alt="che3.1" width="228" height="209" /></p>
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		<title>Introduction</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/introduction-5/</link>
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		<pubDate>Wed, 25 Nov 2009 03:30:50 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[States of Matter – Gaseous and Liquid State]]></category>
		<category><![CDATA[Chemistry]]></category>
		<category><![CDATA[gaseous state]]></category>
		<category><![CDATA[IIT]]></category>
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		<category><![CDATA[particles]]></category>

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		<description><![CDATA[Have you ever realized that whenever you are sitting alone somewhere, actually, you are not alone (surprised). Well, you are always surrounded by air around you. In fact, air surrounds everything like a blanket. Now, what is air? Well, air is nothing but a mixture of gases.  The gaseous state results when the forces of [...]]]></description>
			<content:encoded><![CDATA[<p>Have you ever realized that whenever you are sitting alone somewhere, actually, you are not alone (surprised). Well, you are always surrounded by air around you. In fact, air surrounds everything like a blanket. Now, what is air? Well, air is nothing but a mixture of gases.  The gaseous state results when the forces of attraction between the particles of matter are very low. In this state the molecules are far apart from one another and their positions are not fixed. Hence gases have neither definite shape nor definite volume but a gas occupies fully its container. It was correctly stated by Robert Boyle <strong>&#8220;Imagine the air to be such a heap of little bodies, lying one upon another, as may be resembled to fleece of wool.&#8221;</strong></p>
<p><strong>OBJECTIVE</strong></p>
<p>Gaseous state is the simplest state of matter and shows the greatest uniformity in behaviour. Let us in this chapter know more about this ever accompanying companion. In this chapter we will study the behaviour of different gases at different physical condition of pressure and temperature. The study includes different laws related to behaviour of gases e.g., Boyle’s law, Charles’ law, Ideal gas law etc. to comprehend the ideal gas equation. We shall also</p>
<ul>
<li>know the meaning of absolute scale of temperature</li>
<li>learn about Dalton&#8217;s law of partial pressures and Graham&#8217;s law of diffusion</li>
<li>understand the kinetic gas equation and average kinetic energy of gaseous molecules</li>
<li>learn the definition of root-mean-square, average and most probable speeds</li>
<li>learn about the compression factor ?!</li>
<li>know about van der Waals equation of state</li>
</ul>
<p><strong>PRE-REQUISITE</strong></p>
<p>Relative atomic mass of an element</p>
<p>The ratio of average mass per atom of the natural isotopic composition of the element to of the mass of an atom of nuclide <sup>12</sup>C is known as the relative atomic mass of the element.</p>
<p>Relative molecular mass of Compound</p>
<p>The ratio of average mass per molecule of the natural isotopic composition of the compound to  mass of an atom of nuclide <sup>12</sup>C is known as relative molecular mass of the compound.</p>
<p>Mole of a substance</p>
<p>One mole of a substance contains as many particles (atoms, molecules, ions) as there are atoms in exactly 12 gm of the nuclide <sup>12</sup>C. This number is approximately equal to 6.023 × 10<sup>23</sup>.<em> </em></p>
<p><em>or</em></p>
<p>The minimum weight of any chemical species which contains 6.023 × 10<sup>23 </sup>molecules is called one mole.</p>
<p><em>or</em></p>
<p>At NTP or STP (Normal/standard temperature and pressure are 0°C and 1 atm, respectively) one mole occupies 22.4 litre of volume.</p>
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		<title>Standard Hydrogen Electrode (SHE)</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/standard-hydrogen-electrode-she/</link>
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		<pubDate>Mon, 16 Nov 2009 03:30:04 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[Mole Concepts and Chemical Arithmetics]]></category>
		<category><![CDATA[IIT]]></category>
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		<category><![CDATA[SHE]]></category>

		<guid isPermaLink="false">http://smartlearnwebtv.com/iitjee/?p=721</guid>
		<description><![CDATA[The SHE is the universal reference for reporting relative half-cell potentials. It is a type of gas electrode and was widely used in early studies as a reference electrode, and as an indicator electrode for the determination of pH values. The SHE could be used as either an anode or cathode depending upon the nature [...]]]></description>
			<content:encoded><![CDATA[<h2><span style="font-weight: normal;font-size: 13px">The SHE is the universal reference for reporting relative half-cell potentials. It is a type of gas electrode and was widely used in early studies as a reference electrode, and as an indicator electrode for the determination of pH values. The SHE could be used as either an anode or cathode depending upon the nature of the half-cell it is used with. The SHE consists of a platinum electrode immersed in a solution with a hydrogen ion concentration of 1.00M and H<sub>2</sub> (g) at 1 atm pressure kept at 25<sup>o</sup> C.</span></h2>
<p>The platinum electrode is made of a small square of platinum foil which is platinized (known as platinum black). Hydrogen gas, at a pressure of 1 atmosphere, is bubbled around the platinum electrode. The platinum black serves as a large surface area for the reaction to take place, and the stream of hydrogen keeps the solution saturated at the electrode site with respect to the gas. It is interesting to note that even though the SHE is the universal reference standard, it exists only as a theoretical electrode which scientists use as the definition of an arbitrary reference electrode with a half-cell potential of 0.00 volts.</p>
<p>(Because half-cell potentials cannot be measured, this is the perfect electrode to allow scientists to perform theoretical research calculations.) The reason this electrode cannot be manufactured is due to the fact that no solution can be prepared that yields a hydrogen ion activity of 1.00M.</p>
<p style="text-align: center"><img class="size-full wp-image-722 aligncenter" src="http://smartlearnwebtv.com/iitjee/files/che1.11.png" alt="che1.11" width="296" height="267" /></p>
<p>Hydrogen electrode is made by adding platinum black to platinum wire or a platinum plate. It is immersed in the test solution and an electric charge is applied to the solution and platinum black with hydrogen gas. The hydrogen-electrode method is a standard among the various methods for measuring pH. The values derived using other methods become trustworthy only when they match those measured using hydrogen electrode method. However, this method is not appropriate for daily use because of the effort and expense involved, with the inconvenience of handling hydrogen gas and great influence of highly oxidizing or reducing substances in the test solution.</p>
<p>E<sup>0</sup> (H<sub>2</sub>/H<sup>+</sup>) = 0 = E<sup>0</sup> (H<sup>+</sup>/H<sub>2</sub>)</p>
<p>Now the other half cell can be divided into two categories:</p>
<p>Þ            one which will act as anode</p>
<p>Þ            the others which will acts as cathode</p>
<p>Each type of cell arrangement will give an EMF value which will be actually the EMF value of unknown electrodes as EMF value of SHE is ‘0’ volts.</p>
<h4>Example</h4>
<p>Cu electrode (half cell) acts as cathode with SHE i.e., as Cu<sup>2+</sup>/Cu [Cu<sup>2+</sup> (1.0 M) + 2e --&gt; Cu]. The experimental measurement of EMF value for this cell arrangement give 0.34 volts. Since Cu electrode shows reduction with SHE, the given value of EMF represents the reduction potential of Cu half-cell.</p>
<p>E<sup>0</sup> (Cu<sup>2+</sup>/Cu) = + 0.34 volts</p>
<p>The oxidation potential of Cu half-cell is just the negative of this value.</p>
<p>E<sup>0</sup> (Cu/Cu<sup>2+</sup>) = &#8211; 0.34 volts</p>
<p><em><br />
</em></p>
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		<title>Gravimetric Analysis</title>
		<link>http://smartlearnwebtv.com/iitjee/IITJEE/gravimetric-analysis/</link>
		<comments>http://smartlearnwebtv.com/iitjee/IITJEE/gravimetric-analysis/#comments</comments>
		<pubDate>Tue, 10 Nov 2009 03:30:54 +0000</pubDate>
		<dc:creator>Prathap</dc:creator>
				<category><![CDATA[Mole Concepts and Chemical Arithmetics]]></category>
		<category><![CDATA[gravimetric analysis]]></category>
		<category><![CDATA[IIT]]></category>
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		<description><![CDATA[ü        There is no set of rules to distinguish between gravimetric and volumetric analysis and all the laws used in one are equally applicable to the other. For making our calculations simple, we divide stoichiometry into two parts
Þ      simple calculation based on mole concept (Gravimetric Analysis)
Þ      complex calculation based on volume of solutions and their [...]]]></description>
			<content:encoded><![CDATA[<h1><span style="font-weight: normal;font-size: 13px">ü        There is no set of rules to distinguish between gravimetric and volumetric analysis and all the laws used in one are equally applicable to the other. For making our calculations simple, we divide stoichiometry into two parts</span></h1>
<p>Þ      simple calculation based on mole concept (<em>Gravimetric Analysis</em>)</p>
<p>Þ      complex calculation based on volume of solutions and their concentrations (<em>Volumetric analysis</em>)</p>
<p>ü        While studying gravimetric analysis, we have three types of relationships in a chemical reaction.</p>
<p>Þ      Calculations involving mass – mass relations</p>
<p>Þ      Calculation involving mass – volume relations</p>
<p>Þ      Calculation involving volume – volume relations</p>
<h1><em>Mass – Mass Relationship</em></h1>
<p>ü        Following steps are followed while making necessary calculations</p>
<p>Þ      Write down balanced molecular equation for the chemical change</p>
<p>Þ      Write down the number of moles below the formula of each of the reactant and product</p>
<p>Þ      Write down the relative masses of the reactants and the products with the help of formula below the respective formula. These are the theoretical amounts of reactants and products.</p>
<p>Þ      By the applications of unitary method, mole concept or proportionality method, the unknown factor or factors can be determined.</p>
<h2><strong><em>Illustrations</em></strong></h2>
<p><strong>1.</strong> How many gms of oxygen are required to burn completely 570 gm of octane.</p>
<p><img class="alignnone size-full wp-image-682" src="http://smartlearnwebtv.com/iitjee/files/che1.5.png" alt="che1.5" width="176" height="25" /></p>
<p>Now balancing the chemical equation, we get</p>
<p><img class="alignnone size-full wp-image-685" src="http://smartlearnwebtv.com/iitjee/files/che1.61.png" alt="che1.6" width="258" height="53" /></p>
<p>Then let us understand what this reaction means.</p>
<p>Does it mean,</p>
<p><img class="alignnone size-full wp-image-686" src="http://smartlearnwebtv.com/iitjee/files/che1.7.png" alt="che1.7" width="640" height="79" /></p>
<p>Well, actually, the reaction does not mean either of the two. The first interpretation is wrong because the numbers (stoichiometric coefficients) appearing before substances in a reaction are not the number of atoms or molecules. If it were so, what do you think would be the interpretation of this reaction.</p>
<p><img class="alignnone size-full wp-image-688" src="http://smartlearnwebtv.com/iitjee/files/che1.8.png" alt="che1.8" width="641" height="194" /></p>
<h2><strong><em>Limiting Reagent</em></strong></h2>
<p>ü        The reagent that gives the least no of moles of the product is the limiting reagent.</p>
<h2><strong><em>Illustrations</em></strong></h2>
<p>1.      1g of Mg is burnt in a closed vessel which contain 0.5g of O<sub>2</sub>.</p>
<p>i)    Which reactant is left in excess?</p>
<p>ii)      Find the mass of excess reactant?</p>
<p><img class="alignnone size-full wp-image-689" src="http://smartlearnwebtv.com/iitjee/files/che1.9.png" alt="che1.9" width="569" height="230" /></p>
<h2><strong><em>Percent Yield</em></strong></h2>
<p>Generally when a reaction is carried out in the laboratory we do not obtain the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield the percentage yield can be calculated as</p>
<p><img class="alignnone size-full wp-image-690" src="http://smartlearnwebtv.com/iitjee/files/che1.10.png" alt="che1.10" width="664" height="299" /></p>
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