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

3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

1.3 SIGNIFICANT FIGURES figures in a measurement. This important remark makes most of the followingAs discussed above, every measurement observations clear:involves errors. Thus, the result of (1) For example, the length 2.308 cm has fourmeasurement should be reported in a way that significant figures. But in different units, theindicates the precision of measurement. same value can be written as 0.02308 m or 23.08Normally, the reported result of measurement is a number that includes all digits in the mm or 23080 µm. number that are known reliably plus the first All these numbers have the same number of digit that is u Reprint 2025-26 4 PHYSICS This shows that the location of decimal point is negative exponent (or power) of 10. In order to of no consequence in determining the number get an approximate idea of the number, we may of significant figures. round off the number a to 1 (for a ≤5) and to 10 The example gives the following rules : (for 5<a ≤10). Then the number can be • All the non-zero digits are significant. expressed approximately as 10b in which the • All the zeros between two non-zero digits exponent (or power) b of 10 is called order of magnitude of the physical quantity. When only are significant, no matter where the an estimate is required, the quantity is of the decimal point is, if at all. order of 10b. For example, the diameter of the • If the number is less than 1, the zero(s) earth (1.28×107m) is of the order of 107m with on the right of decimal point but to the the order of magnitude 7. The diameter of left of the first non-zero digit are not hydrogen atom (1.06 ×10–10m) is of the order of significant. [In 0.00 2308, the underlined 10–10m, with the order of magnitude zeroes are not significant]. –10. Thus, the diameter of the earth is 17 orders • The terminal or trailing zero(s) in a of magnitude larger than the hydrogen atom. It is often customary to write the decimal after number without a decimal point are not the first digit. Now the confusion mentioned in significant. (a) above disappears : [Thus 123 m = 12300 cm = 123000 mm has 4.700 m = 4.700 × 102 cm three significant figures, the trailing zero(s) = 4.700 × 103 mm = 4.700 × 10–3 km being not significant.] However, you can also The power of 10 is irrelevant to the see the next observation. determination of significant figures. However, all • The trailing zero(s) in a number with a zeroes appearing in the base number in the decimal point are significant. scientific notation are significant. Each number [The numbers 3.500 or 0.06900 have four in this case has four significant figures. significant figures each.] Thus, in the scientific notation, no confusion (2) There can be some confusion regarding the arises about the trailing zero(s) in the base trailing zero(s). Suppose a length is reported to number a. They are always significant. be 4.700 m. It is evident that the zeroes here (4) The scientific notation is ideal for reporting are meant to convey the precision of measurement. But if this is not adopted, we use measurement and are, therefore, significant. [If the rules adopted in the preceding example : these were not, it would be superfluous to write • For a number greater than 1, without anythem explicitly, the reported measurement decimal, the trailing zero(s) are notwould have been simply 4.7 m]. Now suppose we change units, then significant. • For a number with a decimal, the trailing4.700 m = 470.0 cm = 4700 mm = 0.004700 km zero(s) are significant. Since the last number has trailing zero(s) in a number with no decimal, we would conclude (5) The digit 0 conventionally put on the left of a erroneously from observation (1) above that the decimal for a number less than 1 (like 0.1250) number has two significant figures, while in is never significant. However, the zeroes at the fact, it has four significant figures and a mere end of such number are significant in a change of units cannot change the number of measurement. significant figures. (6) The multiplying or dividing factors which are (3) To remove such ambiguities in neither rounded numbers nor numbers determining the number of significant representing measured values are exact and figures, the best way is to report every have infinite number of significant digits. For measurement in scientific notation (in the dpower of 10). In this notation, every number is example in r = or s = 2πr, the factor 2 is an expressed as a × 10b, where a is a number 2 between 1 and 10, and b is any positive or exact number and it can be written as 2.0, 2.00 Reprint 2025-26 UNITS AND MEASUREMENT 5 decimal place. The final result should, therefore, t or 2.0000 as required. Similarly, in T = , n is be rounded off to 663.8 g. n Similarly, the difference in length can be an exact number. expressed as : 1.3.1 Rules for Arithmetic Operations with 0.307 m – 0.304 m = 0.003 m = 3 ×10–3 m. Significant Figures Note that we should not use the rule (1) applicable The result of a calculation involving approximate for multiplication and division and write 664 g as measured values of quantities (i.e. values with the result in the example of addition and limited number of significant figures) must 3.00 × 10–3 m in the example of subtraction. They reflect the umeasured values. It cannot be more accurate properly. For addition and subtraction, the rule than the original measured values themselves is in terms of decimal places. on which the result is based. In general, the final result should not have more significant 1.3.2 Rounding off the Ufigures than the original data from which it was The result of computation with approximate obtained. Thus, if mass of an object is measured numbers, which contain more than one to be, say, 4.237 g (four significant figures) and uits volume is measured to be 2.51 cm3, then its for rounding off numbers to the appropriate density, by mere arithmetic division, is significant figures are obvious in most cases. A 1.68804780876 g/cm3 upto 11 decimal places. number 2.746 rounded off to three significant It would be clearly absurd and irrelevant to figures is 2.75, while the number 1.743 would record the calculated value of density to such a be 1.74. The rule by convention is that the precision when the measurements on which the preceding digit is raised by 1 if the value is based, have much less precision. The insignificant digit to be dropped (the following rules for arithmetic operations with underlined digit in this case) is more than significant figures ensure that the final result 5, and is left unchanged if the latter is less of a calculation is shown with the precision that than 5. But what if the number is 2.745 in is consistent with the precision of the input which the insignificant digit is 5. Here, themeasured values : convention is that if the preceding digit is(1) In multiplication or division, the final even, the insignificant digit is simplyresult should retain as many significant dropped and, if it is odd, the preceding digitfigures as are there in the original number with the least significant figures. is raised by 1. Then, the number 2.745 rounded Thus, in the example above, density should off to three significant figures becomes 1.74. On be reported to three significant figures. the other hand, the number 2.735 rounded off to three significant figures becomes 1.74 since 4.237g -3 Density = = 1.69 g cm the preceding digit is odd. 3 2.51 cm In any involved or complex multi-step Similarly, if the speed of light is given as calculation, you should retain, in intermediate 3.00 × 108 m s-1 (three significant figure) and steps, one digit more than the significant digits one year (1y = 365.25 d) has 3.1557 × 107 s (five and round off to proper significant figures at the significant figures), the light year is 9.47 × 1015 m end of the calculation. Similarly, a number (three significant figures). known to be within many significant figures, such as in 2.99792458 × 108 m/s for the speed (2) In addition or subtraction, the final result of light in vacuum, is rounded off to anshould retain as many decimal places as are approximate value 3 × 108 m/s , which is oftenthere in the number with the least employed in computations. Finally, rememberdecimal places. that exact numbers that appear in formulae like For example, the sum of the numbers 436.32 g, 227.2 g and 0.301 g by mere arithmetic L addition, is 663.821 g. But the least precise 2 π in T = 2π , have a large (infinite) number measurement (227.2 g) is correct to only one g Reprint 2025-26 6 PHYSICS of significant figures. The value of π = = 16.2 cm ± 0.6 %. 3.1415926.... is known to a large number of significant figures. You may take the value as Similarly, the breadth b may be written as 3.142 or 3.14 for π, with limited number of b = 10.1 ± 0.1 cm significant figures as required in specific = 10.1 cm ± 1 % cases. Then, the error of the product of two (or more)⊳ Example 1.1 Each side of a cube is experimental values, using the combination of measured to be 7.203 m. What are the errors rule, will be total surface area and the volume of the l b = 163.62 cm2 + 1.6% cube to appropriate significant figures? = 163.62 + 2.6 cm2 Answer The number of significant figures in the measured length is 4. The calculated area This leads us to quote the final result as and the volume should therefore be rounded off l b = 164 + 3 cm2 to 4 significant figures. Here 3 cm2 is the uSurface area of the cube = 6(7.203)2 m2 estimation of area of rectangular sheet. = 311.299254 m2 (2) If a set of experimental data is specified = 311.3 m2 to n significant figures, a result obtained by combining the data will also be valid to n Volume of the cube = (7.203)3 m3 significant figures. = 373.714754 m3 However, if data are subtracted, the number of = 373.7 m3 ⊳ significant figures can be reduced. ⊳ Example 1.2 5.74 g of a substance For example, 12.9 g – 7.06 g, both specified to occupies 1.2 cm3. Express its density by three significant figures, cannot properly be keeping the significant figures in view. evaluated as 5.84 g but only as 5.8 g, as u in a different fashion (smallest number ofmeasured mass whereas there are only 2 decimal places rather than the number of significant figures in the measured volume. significant figures in any of the number added Hence the density should be expressed to only or subtracted). 2 significant figures. (3) The relative error of a value of number 5.74 −3 specified to significant figures depends not Density = g cm 1.2 only on n but also on the number itself. = 4.8 g cm--3 . ⊳ For example, the accuracy in measurement of mass 1.02 g is ± 0.01 g whereas another 1.3.3 Rules for Determining the U in the Results of Arithmetic The relative error in 1.02 g is Calculations = (± 0.01/1.02) × 100 % = ± 1%The rules for determining the u Similarly, the relative error in 9.89 g iserror in the number/measured quantity in = (± 0.01/9.89) × 100 %arithmetic operations can be understood from = ± 0.1 %the following examples. Finally, remember that intermediate results in(1) If the length and breadth of a thin rectangular sheet are measured, using a metre a multi-step computation should be scale as 16.2 cm and, 10.1 cm respectively, there calculated to one more significant figure in are three significant figures in each every measurement than the number of measurement. It means that the length l may digits in the least precise measurement. be written as These should be justified by the data and then l = 16.2 ± 0.1 cm the arithmetic operations may be carried out; Reprint 2025-26 UNITS AND MEASUREMENT 7 otherwise rounding errors can build up. For mass, one dimension in length, and –2 example, the reciprocal of 9.58, calculated (after dimensions in time. The dimensions in all other rounding off) to the same number of significant base quantities are zero. figures (three) is 0.104, but the reciprocal of Note that in this type of representation, the magnitudes are not considered. It is the quality0.104 calculated to three significant figures is of the type of the physical quantity that enters. 9.62. However, if we had written 1/9.58 = 0.1044 Thus, a change in velocity, initial velocity, and then taken the reciprocal to three significant average velocity, final velocity, and speed are figures, we would have retrieved the original all equivalent in this context. Since all these value of 9.58. quantities can be expressed as length/time, This example justifies the idea to retain one their dimensions are [L]/[T] or [L T–1]. more extra digit (than the number of digits in the least precise measurement) in intermediate 1.5 DIMENSIONAL FORMULAE AND steps of the complex multi-step calculations in DIMENSIONAL EQUATIONS order to avoid additional errors in the process The expression which shows how and which of of rounding off the numbers. the base quantities represent the dimensions of a physical quantity is called the dimensional