6.6 APPLICATIONS OF EQUILIBRIUM in the denominator). This implies that a high value of K is suggestive of a high concentration CONSTANTS of products and vice-versa.Before considering the applications of We can make the following generalisationsequilibrium constants, let us summarise the concerning the composition of equilibriumimportant features of equilibrium constants mixtures:as follows: 1. Expression for equilibrium constant is • If Kc > 103, products predominate over applicable only when concentrations of reactants, i.e., if Kc is very large, the the reactants and products have attained reaction proceeds nearly to completion. constant value at equilibrium state. Consider the following examples: 2. The value of equilibrium constant is (a) The reaction of H2 with O2 at 500 K independent of initial concentrations of has a very large equilibrium constant, the reactants and products. Kc = 2.4 × 1047. 3. Equilibrium constant is temperature (b) H2(g) + Cl2(g) 2HCl(g) at 300K has dependent having one unique value for Kc = 4.0 × 1031. a particular reaction represented by a (c) H2(g) + Br2(g) 2HBr (g) at 300 K, balanced equation at a given temperature. Kc = 5.4 × 1018 4. The equilibrium constant for the reverse • If Kc < 10–3, reactants predominate over reaction is equal to the inverse of the products, i.e., if Kc is very small, the equilibrium constant for the forward reaction proceeds rarely. Consider the reaction. following examples: Reprint 2025-26 182 chemistry (a) The decomposition of H2O into H2 and If Qc = Kc, the reaction mixture is already O2 at 500 K has a very small equilibrium at equilibrium. constant, Kc = 4.1 × 10–48 Consider the gaseous reaction of H2 (b) N2(g) + O2(g) 2NO(g), with I2, at 298 K has Kc = 4.8 ×10–31. H2(g) + I2(g) 2HI(g); Kc = 57.0 at 700 K. • If Kc is in the range of 10 – 3 to 103, Suppose we have molar concentrations appreciable concentrations of both [H2]t=0.10M, [I2]t = 0.20 M and [HI]t = 0.40 M. reactants and products are present. (the subscript t on the concentration symbols Consider the following examples: means that the concentrations were measured at some arbitrary time t, not necessarily at(a) For reaction of H2 with I2 to give HI, equilibrium). Kc = 57.0 at 700K. Thus, the reaction quotient, Qc at this(b) Also, gas phase decomposition of N2O4 stage of the reaction is given by, to NO2 is another reaction with a value 2 –3 Qc = [HI]t / [H2]t [I2]t = (0.40)2/ (0.10)×(0.20) of Kc = 4.64 × 10 at 25°C which is neither too small nor too large. Hence, = 8.0 equilibrium mixtures contain appreciable Now, in this case, Qc (8.0) does not equal concentrations of both N2O4 and NO2. Kc (57.0), so the mixture of H2(g), I2(g) and HI(g) These generarlisations are illustrated in is not at equilibrium; that is, more H2(g) and Fig. 6.6 I2(g) will react to form more HI(g) and their concentrations will decrease till Qc = Kc. The reaction quotient, Qc is useful in predicting the direction of reaction by comparing the values of Qc and Kc. Thus, we can make the following generalisations concerning the direction of the reaction (Fig. 6.7) :Fig.6.6 Dependence of extent of reaction on Kc 6.6.2 Predicting the Direction of the Reaction The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient Q. The reaction quotient, Q (Qc with molar Fig. 6.7 Predicting the direction of the reactionconcentrations and QP with partial pressures) is defined in the same way as the equilibrium • If Qc < Kc, net reaction goes from left to constant Kc except that the concentrations right in Qc are not necessarily equilibrium values. • If Qc > Kc, net reaction goes from right to For a general reaction: left. a A + b B c C + d D (6.19) • If Qc = Kc, no net reaction occurs. Qc = [C]c[D]d / [A]a[B]b (6.20) Problem 6.7 Then, The value of Kc for the reaction If Qc > Kc, the reaction will proceed in the 2A B + C is 2 × 10–3. At a given time, direction of reactants (reverse reaction). the composition of reaction mixture is [A] = [B] = [C] = 3 × 10–4 M. In which direction If Qc < Kc, the reaction will proceed in the the reaction will proceed?direction of the products (forward reaction). Reprint 2025-26 EQUILIBRIUM 183 Solution The total pressure at equilbrium was For the reaction the reaction quotient Qc is found to be 9.15 bar. Calculate Kc, Kp and given by, partial pressure at equilibrium. Qc = [B][C]/ [A]2 Solution as [A] = [B] = [C] = 3 × 10–4M Qc = (3 ×10–4)(3 × 10–4) / (3 ×10–4)2 = 1 We know pV = nRT as Qc > Kc so the reaction will proceed in the Total volume (V ) = 1 L reverse direction. Molecular mass of N2O4 = 92 g 6.6.3 Calculating Equilibrium Number of moles = 13.8g/92 g = 0.15 Concentrations of the gas (n) In case of a problem in which we know the Gas constant (R) = 0.083 bar L mol–1K–1 initial concentrations but do not know any of Temperature (T ) = 400 K the equilibrium concentrations, the following pV = nRTthree steps shall be followed: Step 1. Write the balanced equation for the p × 1L = 0.15 mol × 0.083 bar L mol–1K–1 × 400 Kreaction. Step 2. Under the balanced equation, make p = 4.98 bar a table that lists for each substance involved N2O4 2NO2 in the reaction: Initial pressure: 4.98 bar 0 (a) the initial concentration, At equilibrium: (4.98 – x) bar 2x bar (b) the change in concentration on going to Hence, equilibrium, and ptotal at equilibrium = pN2O4 + pNO2(c) the equilibrium concentration. 9.15 = (4.98 – x) + 2x In constructing the table, define x as the 9.15 = 4.98 + xconcentration (mol/L) of one of the substances that reacts on going to equilibrium, then use x = 9.15 – 4.98 = 4.17 bar the stoichiometry of the reaction to determine Partial pressures at equilibrium are, the concentrations of the other substances in terms of x. pN2O4 = 4.98 – 4.17 = 0.81bar Step 3. Substitute the equilibrium pNO2 = 2x = 22 × 4.17 = 8.34 bar concentrations into the equilibrium equation K p = p N 2O 4  p NO 2 / for the reaction and solve for x. If you are = (8.34)2/0.81 = 85.87to solve a quadratic equation choose the mathematical solution that makes chemical Kp = Kc(RT)∆n sense. 85.87 = Kc(0.083 × 400)1 Step 4. Calculate the equilibrium Kc = 2.586 = 2.6 concentrations from the calculated value of x. Problem 6.9Step 5. Check your results by substituting them into the equilibrium equation. 3.00 mol of PCl5 kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the Problem 6.8 mixture at equilibrium. Kc= 1.80 13.8g of N2O4 was placed in a 1L reaction Solution vessel at 400K and allowed to attain PCl5 PCl3 + Cl2 equilibrium Initial N2O4 (g) 2NO2 (g) concentration: 3.0 0 0 Reprint 2025-26 184 chemistry Taking antilog of both sides, we get, Let x mol per litre of PCl5 be dissociated, K = e–∆G/RT (6.23) At equilibrium: (3-x) x x Hence, using the equation (6.23), the reaction spontaneity can be interpreted in Kc = [PCl3][Cl2]/[PCl5] terms of the value of ∆G . 1.8 = x2/ (3 – x) • If ∆G  < 0, then –∆G /RT is positive, x2 + 1.8x – 5.4 = 0 and e –∆DG /RT>1, making K >1, which x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 implies a spontaneous reaction or the x = [–1.8 ± √3.24 + 21.6]/2 reaction which proceeds in the forward direction to such an extent that the x = [–1.8 ± 4.98]/2 products are present predominantly. x = [–1.8 + 4.98]/2 = 1.59 • If ∆G  > 0, then –∆G /RT is negative, and [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M e –∆G </RT 1, that is , K < 1, which implies [PCl3] = [Cl2] = x = 1.59 M a non-spontaneous reaction or a reaction which proceeds in the forward direction

12.9 In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth 305 = 6.0 × 1024 kg.) Reprint 2025-26 Physics Chapter Thirteen NUCLEI 13.1 INTRODUCTION In the previous chapter, we have learnt that in every atom, the positive charge and mass are densely concentrated at the centre of the atom forming its nucleus. The overall dimensions of a nucleus are much smaller than those of an atom. Experiments on scattering of a-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about 104. This means the volume of a nucleus is about 10–12 times the volume of the atom. In other words, an atom is almost empty. If an atom is enlarged to the size of a classroom, the nucleus would be of the size of pinhead. Nevertheless, the nucleus contains most (more than 99.9%) of the mass of an atom. Does the nucleus have a structure, just as the atom does? If so, what are the constituents of the nucleus? How are these held together? In this chapter, we shall look for answers to such questions. We shall discuss various properties of nuclei such as their size, mass and stability, and also associated nuclear phenomena such as radioactivity, fission and fusion. 13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS The mass of an atom is very small, compared to a kilogram; for example, the mass of a carbon atom, 12C, is 1.992647 × 10–26 kg. Kilogram is not 306 a very convenient unit to measure such small quantities. Therefore, a Reprint 2025-26 Nuclei different mass unit is used for expressing atomic masses. This unit is the atomic mass unit (u), defined as 1/12th of the mass of the carbon (12C) atom. According to this definition mass of one 12 C atom 1u = 12 1.992647 × 10 −26 kg = 12 = 1.660539 × 10 −27 kg (13.1) The atomic masses of various elements expressed in atomic mass unit (u) are close to being integral multiples of the mass of a hydrogen atom. There are, however, many striking exceptions to this rule. For example, the atomic mass of chlorine atom is 35.46 u. Accurate measurement of atomic masses is carried out with a mass spectrometer, The measurement of atomic masses reveals the existence of different types of atoms of the same element, which exhibit the same chemical properties, but differ in mass. Such atomic species of the same element differing in mass are called isotopes. (In Greek, isotope means the same place, i.e. they occur in the same place in the periodic table of elements.) It was found that practically every element consists of a mixture of several isotopes. The relative abundance of different isotopes differs from element to element. Chlorine, for example, has two isotopes having masses 34.98 u and 36.98 u, which are nearly integral multiples of the mass of a hydrogen atom. The relative abundances of these isotopes are

13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u