9.20 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. f1 = 30 cm and u1 = – , give v1 = + 30 cm. This image becomes a virtual object for the second lens. f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives, v2 = – 220 cm. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm. This image becomes a real object for the second lens: f2 = + 30 cm, u2 = – (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f2, and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system. (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); 112 × 20 f2 = – 20 cm which gives v2 = − cm 92 Magnitude of magnification due to the second (concave) 347 Reprint 2025-26 Physics lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm 9.21 If the refracted ray in the prism is incident on the second face at the critical angle ic, the angle of refraction r at the first face is (60°–ic). Now, ic = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° 1 1 1 9.22 (a) + = v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. Each square in the virtual image has an area 10 × 10 × 1 mm2 = 100 mm2 = 1 cm2 (b) Magnifying power = 25/9 = 2.8 (c) No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is |(v/u)| and magnifying power is (25/ |u|). Only when the image is located at the near point |v| = 25 cm, are the two quantities equal. 9.23 (a) Maximum magnifying power is obtained when the image is at the near point (25 cm) u = – 7.14 cm. (b) Magnitude of magnification = (25/ |u|) = 3.5. (c) Magnifying power = 3.5 Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. 9.24 Magnification = ( 6.25 / 1) = 2.5 v = +2.5u 1 1 1    2.5u u 10 i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly. 9.25 (a) Even though the absolute image size is bigger than the object size, the angular size of the image is equal to the angular size of the object. The magnifier helps in the following way: without it object would be placed no closer than 25 cm; with it the object can be placed much closer. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved. (b) Yes, it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The Reprint 2025-26 Answers effect is negligible if the image is at a very large distance away. [Note: When the eye is separated from the lens, the angles subtended at the eye by the first object and its image are not equal.] (c) First, grinding lens of very small focal length is not easy. More important, if you decrease focal length, aberrations (both spherical and chromatic) become more pronounced. So, in practice, you cannot get a magnifying power of more than 3 or so with a simple convex lens. However, using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which increases if fe is smaller. Further, magnification of the objective v O 1 = is given by | u O | (| u O |/ f O ) − 1 which is large when |u O | is slightly greater than fO. The micro- scope is used for viewing very close object. So |u O | is small, and so is fO. (e) The image of the objective in the eye-piece is known as ‘eye-ring’. All the rays from the object refracted by objective go through the eye-ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. If we position our eyes on the eye-ring and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece. When you view through a microscope by placing your eyes on one end,the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument.

6.7 Relationship between to such a small degree that only a very Equilibrium Constant K, minute quantity of product is formed. Reaction Quotient Q and Gibbs Energy G Problem 6.10 The value of Kc for a reaction does not depend The value of ∆G  for the phosphorylation of on the rate of the reaction. However, as you glucose in glycolysis is 13.8 kJ/mol. Find have studied in Unit 5, it is directly related the value of Kc at 298 K. to the thermodynamics of the reaction and Solutionin particular, to the change in Gibbs energy, ∆G. If, ∆G  = 13.8 kJ/mol = 13.8 × 103J/mol • ∆G is negative, then the reaction is Also, ∆G  = – RT lnKc spontaneous and proceeds in the forward Hence, ln Kc = –13.8 × 103J/mol direction. (8.314 J mol–1K–1 × 298 K) • ∆G is positive, then reaction is considered ln Kc = – 5.569 non-spontaneous. Instead, as reverse reaction would have a negative ∆G, the Kc = e–5.569 products of the forward reaction shall be Kc = 3.81 × 10–3 converted to the reactants. Problem 6.11• ∆G is 0, reaction has achieved equilibrium; Hydrolysis of sucrose gives, at this point, there is no longer any free energy left to drive the reaction. Sucrose + H2O Glucose + Fructose A mathematical expression of this Equilibrium constant Kc for the reaction is thermodynamic view of equilibrium can be 2 ×1013 at 300K. Calculate ∆G  at 300K. described by the following equation: Solution ∆G = ∆G + RT lnQ (6.21) ∆G  = – RT lnKcwhere, G is standard Gibbs energy. ∆G  = – 8.314J mol–1K–1× At equilibrium, when ∆G = 0 and Q = Kc, 300K × ln(2×1013) the equation (6.21) becomes, ∆G  = – 7.64 ×104 J mol–1 ∆G = ∆G + RT ln K = 0 6.8 FACTORS AFFECTING EQUILIBRIA ∆G = – RT lnK (6.22) One of the principal goals of chemical lnK = – ∆G / RT synthesis is to maximise the conversion of the Reprint 2025-26 EQUILIBRIUM 185 reactants to products while minimising the “When the concentration of any of the expenditure of energy. This implies maximum reactants or products in a reaction at yield of products at mild temperature and equilibrium is changed, the composition pressure conditions. If it does not happen, of the equilibrium mixture changes so as then the experimental conditions need to be to minimize the effect of concentration adjusted. For example, in the Haber process changes”. for the synthesis of ammonia from N2 and Let us take the reaction, H2, the choice of experimental conditions is of real economic importance. Annual world H2(g) + I2(g) 2HI(g) production of ammonia is about hundred If H2 is added to the reaction mixture million tones, primarily for use as fertilisers. at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, Equilibrium constant, Kc is independent the reaction proceeds in a direction whereinof initial concentrations. But if a system at equilibrium is subjected to a change in the H2 is consumed, i.e., more of H2 and I2 react to form HI and finally the equilibrium shiftsconcentration of one or more of the reacting in right (forward) direction (Fig.6.8). This is insubstances, then the system is no longer at accordance with the Le Chatelier’s principleequilibrium; and net reaction takes place in which implies that in case of addition of asome direction until the system returns to reactant/product, a new equilibrium willequilibrium once again. Similarly, a change be set up in which the concentration of thein temperature or pressure of the system may reactant/product should be less than what italso alter the equilibrium. In order to decide was after the addition but more than what itwhat course the reaction adopts and make was in the original mixture.a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to all physical and chemical equilibria. We shall now be discussing factors which can influence the equilibrium. 6.8.1 Effect of Concentration Change In general, when equilibrium is disturbed by the addition/removal of any reactant/ products, Le Chatelier’s principle predicts that: • The concentration stress of an added reactant/product is relieved by net Fig. 6.8 Effect of addition of H2 on change reaction in the direction that consumes of concentration for the reactants the added substance. and products in the reaction, • The concentration stress of a removed H2(g) + I2 (g) 2HI(g) reactant/product is relieved by net reaction in the direction that replenishes The same point can be explained in terms the removed substance. of the reaction quotient, Qc, or in other words, Qc = [HI]2/ [H2][I2] Reprint 2025-26 186 chemistry Addition of hydrogen at equilibrium concentration of [Fe(SCN)]2+ decreases, the results in value of Qc being less than Kc . Thus, intensity of red colour decreases. in order to attain equilibrium again reaction Addition of aq. HgCl2 also decreases redmoves in the forward direction. Similarly, colour because Hg2+ reacts with SCN– ions to we can say that removal of a product also form stable complex ion [Hg(SCN)4]2–. Removalboosts the forward reaction and increases of free SCN– (aq) shifts the equilibrium the concentration of the products and this in equation (6.24) from right to left to has great commercial application in cases replenish SCN– ions. Addition of potassium of reactions, where the product is a gas or a thiocyanate on the other hand increases the volatile substance. In case of manufacture of colour intensity of the solution as it shift the ammonia, ammonia is liquified and removed equilibrium to right. from the reaction mixture so that reaction keeps moving in forward direction. Similarly, 6.8.2 Effect of Pressure Change in the large scale production of CaO (used A pressure change obtained by changing the as important building material) from CaCO3, volume can affect the yield of products in constant removal of CO2 from the kiln drives case of a gaseous reaction where the total the reaction to completion. It should be number of moles of gaseous reactants and remembered that continuous removal of a total number of moles of gaseous products are product maintains Qc at a value less than Kc different. In applying Le Chatelier’s principle and reaction continues to move in the forward to a heterogeneous equilibrium the effect direction. of pressure changes on solids and liquids can be ignored because the volume (and Effect of Concentration – An experiment concentration) of a solution/liquid is nearly This can be demonstrated by the following independent of pressure. reaction: Consider the reaction, Fe3+(aq)+ SCN–(aq) [Fe(SCN)]2+(aq) (6.24) CO(g) + 3H2(g) CH4(g) + H2O(g)yellow colourless deep red Here, 4 mol of gaseous reactants (CO + 3H2) become 2 mol of gaseous products (CH4 + H2O). Suppose equilibrium mixture (for above (6.25) reaction) kept in a cylinder fitted with a piston at constant temperature is compressed to A reddish colour appears on adding two one half of its original volume. Then, totaldrops of 0.002 M potassium thiocynate solution pressure will be doubled (according to to 1 mL of 0.2 M iron(III) nitrate solution due pV = constant). The partial pressure and to the formation of [Fe(SCN)]2+. The intensity therefore, concentration of reactants and of the red colour becomes constant on products have changed and the mixture is no attaining equilibrium. This equilibrium can be longer at equilibrium. The direction in which shifted in either forward or reverse directions the reaction goes to re-establish equilibrium depending on our choice of adding a reactant can be predicted by applying the Le Chatelier’s or a product. The equilibrium can be shifted principle. Since pressure has doubled, in the opposite direction by adding reagents the equilibrium now shifts in the forward that remove Fe3+ or SCN– ions. For example, direction, a direction in which the number oxalic acid (H2C2O4), reacts with Fe3+ ions of moles of the gas or pressure decreases (we to form the stable complex ion [Fe(C2O4)3]3–, know pressure is proportional to moles of the thus decreasing the concentration of free gas). This can also be understood by using Fe3+(aq). In accordance with the Le Chatelier’s reaction quotient, Qc. Let [CO], [H2], [CH4] principle, the concentration stress of removed and [H2O] be the molar concentrations at Fe3+ is relieved by dissociation of [Fe(SCN)]2+ equilibrium for methanation reaction. When to replenish the Fe3+ ions. Because the volume of the reaction mixture is halved, the Reprint 2025-26 EQUILIBRIUM 187 partial pressure and the concentration are Production of ammonia according to the doubled. We obtain the reaction quotient by reaction, replacing each equilibrium concentration by N2(g) + 3H2(g) 2NH3(g);double its value. ∆H= – 92.38 kJ mol–1  CH 4 ( g )  H 2 O ( g ) is an exothermic process. According to Qc = 3  CO ( g )  H 2 ( g ) Le Chatelier’s principle, raising the temperature shifts the equilibrium to left As Qc < Kc , the reaction proceeds in the and decreases the equilibrium concentration forward direction. of ammonia. In other words, low temperature is favourable for high yield of ammonia, but In reaction C(s) + CO2(g) 2CO(g), when practically very low temperatures slow downpressure is increased, the reaction goes in the the reaction and thus a catalyst is used.reverse direction because the number of moles of gas increases in the forward direction. Effect of Temperature – An experiment Effect of temperature on equilibrium can6.8.3 Effect of Inert Gas Addition be demonstrated by taking NO2 gas (brown If the volume is kept constant and an inert gas in colour) which dimerises into N2O4 gas such as argon is added which does not take (colourless). part in the reaction, the equilibrium remains 2NO2(g) N2O4(g); ∆H = –57.2 kJ mol–1undisturbed. It is because the addition of an inert gas at constant volume does not NO2 gas prepared by addition of Cu change the partial pressures or the molar turnings to conc. HNO3 is collected in two 5 mL test tubes (ensuring same intensityconcentrations of the substance involved in of colour of gas in each tube) and stopperthe reaction. The reaction quotient changes sealed with araldite. Three 250 mL beakersonly if the added gas is a reactant or product 1, 2 and 3 containing freesing mixture, waterinvolved in the reaction. at room temperature and hot water (363K), 6.8.4 Effect of Temperature Change respectively, are taken (Fig. 6.9). Both the test tubes are placed in beaker 2 for 8-10 minutes.Whenever an equilibrium is disturbed by After this one is placed in beaker 1 and thea change in the concentration, pressure or other in beaker 3. The effect of temperaturevolume, the composition of the equilibrium on direction of reaction is depicted very wellmixture changes because the reaction in this experiment. At low temperatures inquotient, Qc no longer equals the equilibrium beaker 1, the forward reaction of formation ofconstant, Kc. However, when a change in temperature occurs, the value of equilibrium N2O4 is preferred, as reaction is exothermic, and thus, intensity of brown colour dueconstant, Kc is changed. to NO2 decreases. While in beaker 3, high In general, the temperature dependence temperature favours the reverse reaction of of the equilibrium constant depends on the sign of ∆H for the reaction. • The equilibrium constant for an exothermic reaction (negative ∆H) decreases as the temperature increases. • The equilibrium constant for an endothermic reaction (positive ∆H) increases as the temperature increases. Temperature changes affect the Fig. 6.9 Effect of temperature on equilibrium for the reaction, 2NO2 (g) N2O4 (g)equilibrium constant and rates of reactions. Reprint 2025-26 188 chemistry formation of NO2 and thus, the brown colour Similarly, in manufacture of sulphuric intensifies. acid by contact process, Effect of temperature can also be seen in 2SO2(g) + O2(g) 2SO3(g); Kc = 1.7 × 1026 an endothermic reaction, though the value of K is suggestive of reaction [Co(H2O)6]3+(aq) + 4Cl–(aq) [CoCl4]2–(aq) + going to completion, but practically the 6H2O(l) oxidation of SO2 to SO3 is very slow. Thus, pink colourless blue platinum or divanadium penta-oxide (V2O5) is used as catalyst to increase the rate of the At room temperature, the equilibrium reaction.mixture is blue due to [CoCl4]2–. When cooled Note: If a reaction has an exceedingly smallin a freesing mixture, the colour of the mixture K, a catalyst would be of little help.turns pink due to [Co(H2O)6]3+. 6.9 IONIC EQUILIBRIUM IN SOLUTION6.8.5 Effect of a Catalyst Under the effect of change of concentrationA catalyst increases the rate of the chemical on the direction of equilibrium, you havereaction by making available a new low energy pathway for the conversion of reactants to incidently come across with the following products. It increases the rate of forward equilibrium which involves ions: and reverse reactions that pass through the Fe3+(aq) + SCN–(aq) [Fe(SCN)]2+(aq) same transition state and does not affect There are numerous equilibria that involve equilibrium. Catalyst lowers the activation ions only. In the following sections we will energy for the forward and reverse reactions study the equilibria involving ions. It is well by exactly the same amount. Catalyst does known that the aqueous solution of sugar not affect the equilibrium composition of does not conduct electricity. However, when a reaction mixture. It does not appear in common salt (sodium chloride) is added the balanced chemical equation or in the to water it conducts electricity. Also, the equilibrium constant expression. conductance of electricity increases with an Let us consider the formation of NH3 increase in concentration of common salt. from dinitrogen and dihydrogen which is Michael Faraday classified the substances highly exothermic reaction and proceeds into two categories based on their ability with decrease in total number of moles to conduct electricity. One category of formed as compared to the reactants. substances conduct electricity in their Equilibrium constant decreases with increase aqueous solutions and are called electrolytes in temperature. At low temperature rate while the other do not and are thus, referred to decreases and it takes long time to reach at as non-electrolytes. Faraday further classified equilibrium, whereas high temperatures give electrolytes into strong and weak electrolytes. satisfactory rates but poor yields. Strong electrolytes on dissolution in water German chemist, Fritz Haber discovered are ionized almost completely, while the weak that a catalyst consisting of iron catalyse electrolytes are only partially dissociated. the reaction to occur at a satisfactory rate For example, an aqueous solution of at temperatures, where the equilibrium sodium chloride is comprised entirely of concentration of NH3 is reasonably favourable. sodium ions and chloride ions, while that Since the number of moles formed in the of acetic acid mainly contains unionized reaction is less than those of reactants, the acetic acid molecules and only some acetate yield of NH3 can be improved by increasing ions and hydronium ions. This is because the pressure. there is almost 100% ionization in case Optimum conditions of temperature of sodium chloride as compared to less and pressure for the synthesis of NH3 using than 5% ionization of acetic acid which is catalyst are around 500°C and 200 atm. a weak electrolyte. It should be noted Reprint 2025-26 EQUILIBRIUM 189 that in weak electrolytes, equilibrium is exists in solid state as a cluster of positively established between ions and the unionized charged sodium ions and negatively charged molecules. This type of equilibrium involving chloride ions which are held together due to ions in aqueous solution is called ionic electrostatic interactions between oppositely equilibrium. Acids, bases and salts come charged species (Fig.6.10). The electrostatic under the category of electrolytes and may act forces between two charges are inversely as either strong or weak electrolytes. proportional to dielectric constant of the medium. Water, a universal solvent, possesses

2.4 POTENTIAL DUE TO AN ELECTRIC DIPOLE As we learnt in the last chapter, an electric dipole consists of two charges q and –q separated by a (small) distance 2a. Its total charge is zero. It is characterised by a dipole moment vector p whose magnitude is q × 2a and which points in the direction from –q to q (Fig. 2.5). We also saw that the electric field of a dipole at a point with position vector r depends not just on the magnitude r, but also on the angle between r and p. Further, 49 Reprint 2025-26 Physics the field falls off, at large distance, not as 1/r 2 (typical of field due to a single charge) but as 1/r3. We, now, determine the electric potential due to a dipole and contrast it with the potential due to a single charge. As before, we take the origin at the centre of the dipole. Now we know that the electric field obeys the superposition principle. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q 1  q q  V = − (2.9)FIGURE 2.5 Quantities involved in the calculation 4 πε0  r1 r2  of potential due to a dipole. where r1 and r2 are the distances of the point P from q and –q, respectively. Now, by geometry, r12 = r 2 + a 2 − 2ar cosq r22 = r 2 + a 2 + 2ar cosq (2.10) We take r much greater than a ( r  a ) and retain terms only upto the first order in a/r 2 2  2a cosθ a 2  r1 = r 1 − + 2  r r  2  2a cosθ (2.11) ≅ r  1 − r  Similarly, 2 2  2a cosθ (2.12) r2 ≅ r 1 + r  Using the Binomial theorem and retaining terms upto the first order in a/r ; we obtain, 1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 − ≅ 1 + [2.13(a)] r1 r  r  r  r  1 a 1  2a cos θ − 1 / 2 1   cos θ ≅ 1 + ≅ 1 − [2.13(b)] r2 r  r  r  r  Using Eqs. (2.9) and (2.13) and p = 2qa, we get q 2 acosθ p cos θ V = = 4 πε0 r 2 4 πε0r 2 (2.14) 50 Now, p cos q = p.rˆ Reprint 2025-26 Electrostatic Potential and Capacitance where ˆr is the unit vector along the position vector OP. The electric potential of a dipole is then given by 1 p.rˆ V = 2 ; (r >> a) (2.15) 4 πε0 r Equation (2.15) is, as indicated, approximately true only for distances large compared to the size of the dipole, so that higher order terms in a/r are negligible. For a point dipole p at the origin, Eq. (2.15) is, however, exact. From Eq. (2.15), potential on the dipole axis (q = 0, p ) is given by 1 p V = ± 2 (2.16) 4 πε0 r (Positive sign for q = 0, negative sign for q = p.) The potential in the equatorial plane (q = p/2) is zero. The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Eqs. (2.8) and (2.15): (i) The potential due to a dipole depends not just on r but also on the angle between the position vector r and the dipole moment vector p. (It is, however, axially symmetric about p. That is, if you rotate the position vector r about p, keeping q fixed, the points corresponding to P on the cone so generated will have the same potential as at P.) (ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as 1/r, characteristic of the potential due to a single charge. (You can refer to the Fig. 2.5 for graphs of 1/r 2 versus r and 1/r versus r, drawn there in another context.)