MODULE – I

 

Contents:

 

1. Stoichiometry 
2. Atomic Structure
3. Gaseous State
4. Chemical Kinetics
5. Chemical Equilibrium

1. Ionic Equilibirum

 

STOICHIOMETRY

 

The stoichiometric coefficients of reactant(s) and product(s) in the chemical equation of any reaction indicate the molar ratio in which they are reacting/formed, and hence by converting mole into mass in the desired unit or volume (in the case of gaseous species) it is possible to obtain the following relationships between the various species taking part in the reaction.

1. i) Mass-mass relationship
2. ii) Mass-volume relationship

1. Volume-volume relationship

 

The branch of science dealing with such chemical calculations is called stoichiometry. For stoichiometric calculations following types of chemical analysis are generally carried out.

 

1. a) Gravimetric Analysis
2. b) Volumetric Analysis
3. c) Eudiometric analysis

 

Before giving into core of the subject matter we must always know the following simple points.

 

6. i) Mole represents a group of 6.023 × 1023 molecules, atoms or ions and its mass whose numerical value in gram is equal to the molecular mass, atomic mass or ionic mass of the species concerned is called one gram-molecule, one-gram atom or one gram-ion, or simply gram – molecule, gram-atom or gram-ion, respectively.

No. of mole(s) =

=

=

 

1. ii) Chemical reaction involves loss-gain/donation/acceptance sharing of electrons.

The mass of a substance that can lose/gain/donate/accept one mole of electrons, i.e. 6.023 × 1023 electrons during any reaction is called its equivalent mass or weight (E) of the substance in that very reaction.

 

E =

 

Where n-factor is the valency of an element, basicity of an acid, acidity of a base, change in O.N. of the active element per molecule or no. of electron(s) lost or gained per molecule of an oxdising or a reducing agent; or no. of cationic unit charge or anionic unit charge per molecule in the case of salt provided that O.N of no any element of the salt undergoes any change.

 

As we already know the basicity of an acid is the number of dissociable H-atom(s) i.e. proton(s) present per molecule of the acid while acidity of a base is the number of OH– ions that it can furnish in solution or number of molecule(s) of monobasic acid needed to neutralise (react) completely (with) one molecule of the base.

Sometimes we may come across a redox reaction in which more than one element in the oxidising agent, taking part are getting reduced and/or more than one element in reducing agent are getting oxidised. In such a case n-factor is the sum of the number of electrons lost or gained per molecule or alternatively, sum of the changes in O.N of all the elements undergoing O.N change as illustrated below.

n = [increase in 0.N of Fe + increase in 0.N of carbon atoms] = 1 + 1 × 2 = 3

n = decrease in 0.N of two atoms of Fe + Decrease in O.N. of three atoms of S

= 1 × 2 + 2 × 3 = 8

Note that the above two reactions are not disproportionation reactions but just half reactions (oxidation in the former case and reduction in the latter case). In disproportionation reaction the above summation concept cannot be applied).

 

iii) Equivalent weight expressed in gram is called one gram-equivalent or simply gram-equivalent or equivalent. 1/1000th part of equivalent is the milli equivalent or milligram equivalent.

 

1. iv) a) Normality of a solution = ,    V is in litre.

= = , V is in milli litre.

So, Number of equivalent(s) = Normality × Volume in litre =

If the volume is taken in mL

Number of milli equivalent(s) = Normality × Volume in mL

Also, Number of equivalent(s)=

1. b) Molarity of a solution =

So, Number of mole(s) = Molarity × volume in litre

1. c) Normality = n × Molarity
2. d) Milli equivalent = milli moles × n

 

1. v) Law of Equivalence:
2. a) A chemical reaction between two or more substances always occur in the ratio of their equivalent masses i.e., in a chemical reaction, number of equivalents or milli-equivalents of each species (reactant product) remains the same.

 

1. b) In any titration the end point is reached when

N1V1 = N2V2

Where N1 and N2 are the normalities of titrant and titre, respectively and V1 and V2 are their respective volumes. The above equation is also called law of normalities which is stated as follows:

Equal volumes of equinormal solutions will completely react with one another

 

1. vi) The stoichiometric problems can be solved by applying one of the following two concepts.
2. a) Mole concept
3. b) Equivalent concept
4. a) Mole Concept: In order to apply mole concept we require chemical equation representing the reaction between the species involved in the problem. From the stoichiometric coefficients we derive information that how many moles of one react with how many moles of other. For example, from the balanced chemical equation for the thermal decomposition of KClO3 expressed as.

2KClO3 2KCl + 3O2↑

we come to understand that 2 mole i.e. 245g of KClO3 upon complete thermal decomposition yields 2 moles i.e. 149 g of KCl and 3 moles of oxygen gas. Since molar volume at NTP of any gas assumed to behave ideally is 22.4 L. So one can also say that by the thermal decomposition of 245 g of KClO3, 3× 22.4 i.e. 67.2 L of O2 gas will evolve at NTP. Thus, the chemical equation gives mass-mass as well as mass-volume relationship. Now let us consider the chemical equation for the formation of CO2(g) from the combustion of CO(g) as shown below.

CO(g) + ⎯→ CO2(g)

From the stoichiometric coefficients and since number of mole(s) varies directly as the volume of the gas at a given pressure and temperature, so we can say that one part by volume of CO(g) combines with ½ part by volume of O2(g) to form one part by volume of CO2 i.e. 10 L CO(g) will combine with 5 L O2(g) give 10 L CO2(g) at a given pressure and temperature. This is actually volume – volume relationship that is reflected from a chemical equation.

 

1. b) Equivalent Concept: Using equivalent concept requires the knowledge of n-factors of the species concerned so that their equivalents may be calculated. We do not need chemical equation representing the reaction between them. Irrespective of the stoichiometry of the reaction, one equivalent of any reactant must react completely with one equivalent of another to give one equivalent of each of the products formed during the reaction.

 

Since it is not possible to recall chemical equations for all the reactions whose number may run in several millions, so equivalent concept is more easier to adopt.

 

It is needless to mention that in any balanced chemical equation n-factor ratio between two compounds is the reverse of their molar ratio i.e., ratio of their stoichiometric coefficients as is evident in the reaction.

GRAVIMETRIC ANALYSIS

 

In this analysis mass-mass type of calculation is mainly done. A weighed quantity of the given sample is subjected to some chemical change like one of those mentioned below and the weight of the residual mass or the precipitate, if any formed, is usually taken, till the weight assumes constant value 

CaCO3 CaO + CO2↑

CuSO4⋅ 5H2O CuSO4 + 5H2O↑

(aq) + BaCl2(aq) BaSO4↓ + 2Cl–(aq)

Cl– (aq) + AgNO3 (aq) AgCl↓ + (aq)

The calculation is done using mole concept.

 

VOLUMETRIC ANALYSIS (Titrimetry)

 

There are mainly three types of titration:

1. i) Acid-Base titration
2. ii) Redox Titration

iii) Precipitation titration (not included in IIT-JEE Syllabus)

 

Types of Acid-Base titration (alkalimetry or acidimetry)

 

1. i) Strong acid vs. strong base

[pH jump occurring at the end point: pH = 4 to 10 so both methyl orange (pH range: 3.1 to 4.4] Phenopthalin (pH range: 8 to 9.8) can be used as successful indicator for this titration

 

1. ii) Strong acid (HCl) vs. weak base (NH4OH)

[pH jump occurring at the end point: pH = 4 to 6; end point lies in acidic medium due to hydrolysis of the salt formed. Only M.O., not phenolphthalein can be used as an indicator for this titration]

 

iii) Weak acid vs. strong base

(CH3COOH) (NaOH)

[pH jump occurring at the end point: pH = 8 to 10, end point lies in alkaline medium due to hydrolysis of the salt formed. Since pH range of phenolphthalein, lies in alkaline medium so only phenolphthalein, not M.O. can be used as a successful indicator for this titration.

 

The pH –range of an indicator is from pKin-1 to pKin+1 where Kin is the dissociation constant of the indicator also called indicator constant.

 

Types of Redox Titration:

1. i) Permanganometry

Mn++ + 4H2O (redn)

EW of KMnO4 =

Reducing agents like Fe++, oxidate ion, H2O2 etc., are estimated here in this type of redox titration.

1. ii) Dichrometry

+ 14H+ + 6e 2Cr3+ + 7H2O (Redn)

EW of K2Cr2O7 =

Reducing agents like Fe++, N2H4, Na2S2O3 etc. may be estimated.

iii) Iodimetry and iodometry

I2 or liberated iodine upon oxidation of KI in acid medium by some oxidising agent like O3, Cu++ etc., is determined by titration with Na2S2O3

+ 2e

EW of Na2S2O3 =

Some important titration’s:

1. a) Back Titration: Let we have a sample of CaCO3 which is impure. We want to check its purity. For this we may proceed as follows. A known weight of CaCO3 is dissolved in excess of HCl (known amount). The excess of HCl may be determined by titration of the resulting solution with any alkali say NaOH. Knowing amount of HCl remaining unreacted, the amount of HCl having reacted and hence amount of CaCO3 may be calculated.

Again suppose we have to find out the amount of Ca2+ ions in a given solution of it. Ca2+ ions may be precipitated as Ca-oxalate by adding excess but known quantity of oxalic acid. The excess of oxalic acid may be determined by its titration with KMnO4. Knowing excess of oxalic acid, the amount of oxalic acid having reacted with Ca2+ ions and hence the amount of Ca2+ ions may well be determined. This method of titration is called back titration.

1. b) Double titration:
2. i) Same titration using two indicators: Suppose a mixture of Na2CO3 and NaHCO3 is titrated against HCl using phenolphthalein as indicator. HCl will neutralise Na2CO3 in following two steps.
3. Na2CO3 + HCl ⎯→ NaHCO3 + NaCl
4. NaHCO3 + HCl ⎯→ NaCl + H2O + CO2

As step I completes, phenolphthalein changes its colour since its indicator range falls in alkaline medium. This is the 1st end point. Now M.O. is added into the same mixture and titration continued. When NaHCO3  (formed during the reaction and already present in the mixture), gets completely neutralised and a drop of excess of acid is added, the medium becomes acidic and hence M.O. changes its colour. This is the 2nd end point. Let V1 mL HCl be needed for the 1st end point to reach while V2 mL be needed for the 2nd end point after the 1st one. Then

V1 mL HCl ≡ of the Na2CO3

∴ 2V1mL HCl ≡ Total Na2CO3 present in the sample

V2mL HCl ≡ of the Na2CO3 + NaHCO3 already present in the sample.

(V2 – V1) mL HCl ≡ NaHCO3 already present in the sample

If the sample contains NaOH or KOH besides Na2CO3 and NaHCO3, we have again.

Phenolphthalein

M.O. [NaHCO3 + HCl NaCl + H2O + CO2

 

4. ii) Carrying out two different titrations with the sample: Suppose we have a mixture of oxalic acid and H2SO4. Total acid can be determined by alkalimetry while oxalic acid being an O.A. agent, the same mixture can be titrated against KMnO4 The number of equivalent of alkali required for the end point of alkalimetry will give the no. of equivalent of total acid present in the mixture while no. of equivalent of KMnO4 required for the end point in second titration will give the no. of equivalent of oxalic acid present in the mixture. The difference between the two will give no. of equivalents of H2SO4 present in the mixture.

Calculation of strength of a solution when % by wt. (w) and sp. gravity (D) are given:

Gram per litre of solute = 10 wd

∴ Normality =

Molarity =

Molality =

SOLVED EXAMPLES

Example 1: A metal weighing 0.43 g was dissolved in 50 ml of N H2SO4. The unreacted H2SO4 required 14.2 ml of N NaOH for neutralisation. Find out equivalent weight of the metal.

Solution: Eq. of metal = (E is equivalent wt.)

Total meq. of H2SO4 = 50 × 1  = 50

Meq. of H2SO4 consumed by metal  = Total Meq. of H2SO4 – Meq. consumed by NaOH

= 50 – 14.2 = Meq of metal 

Meq. of metal = 35.8 =

or E = = 12.01

Example 2: A solution contains Na2CO3 and NaHCO3. 20 ml of this solution required 5 ml of 0.1 M H2SO4 solution for neutralisation using phenolphthalein as the indicator. Methylorange is then added when a further 5 ml of 0.2 M H2SO4 was required. Calculate the masses of Na2CO3 and NaHCO3 in 1 litre of this solution.

Solution: Using phenolphthalein as indicator 

Meq. of acid  = Meq. of Na2CO3 = 5 × 0.1 × 2 = 1 …(i)

Using methyl orange indicator 

Meq. of acid  = Meq. of Na2CO3 + Meq. of NaHCO3 = 5 × 0.2 × 2 = 2 …(ii)

(ii) – (i)

Meq. of NaHCO3 = 1

Eq. of NaHCO3 = 1 × 10–3

Wt. of NaHCO3 in 1 litre  = 10–3 × 84 × = 4.2g 

Putting this value of Meq. of NaHCO3 = 1 in eq—— (ii)

Meq. of Na2CO3 = 2 – 1 = 1 or Meq. of Na2CO3 = 2

Eq. of Na2CO3 in 20 ml  = 2 × 10–3

Eq. of Na2CO3 in 1000 ml = 2 × 10–3 × = 0.1

Wt. of Na2CO3  in 1 litre  = 0.1 × Eq. wt = 0.1 × = 5.3 g 

Example 3: 2 litres of ammonia at 30°C and 0.2 atm is neutralised by 134 ml of a solution of H2SO4. Calculate normality of H2SO4

Solution:  Moles of NH3 = = = 0.016

Moles of NH3 = Equivalents of NH3

Since each NH3 molecule can take up only one H+ ion on reaction with acids so ‘n’ factor is equal to one for complete neutralisation.

Meq. of H2SO4 = Meq. of NH3

N × 134 = 0.016 × 1000

N = 0.12

Example4: A solution contains a mixture of Na2CO3 and NaOH. Using phenolphthalein as indicator, 25 ml of mixture required 19.5 ml of 0.955 N HCl for the end point. With methyl orange, 25 ml of solution required 25 ml of the same HCl for the end point. Calculate grams per litre of each substance in mixture.

Solution: Here we see that if we use phenolphthalein we get end point till step (2) and if we use methyl orange we get end point after step 3. Thus using phenolphthalein [Here Eq. wt of Na2CO3 = ].

Meq. of NaOH + Meq. of Na2CO3 = Meq of HCl = 19.5 × 0.995 …(i)

Using methyl orange 

Meq. of NaOH + Meq. of Na2CO3 = Meq. of HCl = 25 × 0.995 …(ii)

Subtracting equation (i) from equation (ii)

Meq. of Na2CO3 = 5.4725

5. of Na2CO3 in 25 ml = 2× 5.4725 × 10–3 × = 0.58 g

∴ Wt of Na2CO3 in 1 litre  = 0.58 × = 23.2 g/litre

Putting value of Meq. of Na2CO3 in eq. (i) 

Meq. of NaOH = Meq. of HCl – Meq. of Na2CO3

= 19.5 × 0.995 – 5.4725 = 13.93

wt of NaOH in 25 ml  = 13.93 × 10–3 × 40 = 0.5572 g

1. of NaOH in 1 litre  = 0.5572 × = 22.28 g/litre

Example 5: 50 ml of an aqueous solution of H2O2 was treated with excess of KI solution in dil H2SO4. The liberated iodine required 200 ml of 0.1N Na2S2O3 solution for complete reaction. Calculate the concentration of H2O2 in gm/litre. What is its volume strength.

Solution: eq. of H2O2 = eq.of Na2S2O3 (acc. to law of g.eq.)

50 × N = 200 × 0.1 ⇒ N = 0.4 of H2O2

0.4 N of H2O2 means that 1 litre contains 0.4 equivalents of H2O2.

∴ wt. of H2O2 in 1 litre  = eq. × Eq. wt.  = 0.4 × 17 = 6.8  g/L

Volume strength of H2O2 = 5.6 × N = 5.6 × 0.4 = 2.24

EXERCISES

1. 200 gms. of ‘ marble  chips’  are dropped into one kilogram of solution of HCl  containing one-tenth of its weight of the pure acid. How much of chips will remain undissolved. What weight of anhydrous calcium chloride and what weight of CO2 gas could be  obtained from it ?
2. Igniting MnO2 in air converts it  quantitatively to Mn3O4. A sample of pyrolusite is of the following  composition: MnO2=80%, SiO2 and  other  inert impurities = 15% , rest being water. The sample is ignited to constant weight. What is the percentage of Mn in the ignited residue ?
3. The oxides of sodium and potassium contained in a 0.5 g  sample of feldspar were converted to the  respective chlorides. The weight of the chlorides thus obtained was 0.118g.  Subsequent treatment of the chlorides with silver nitrate gave 0.2451g.  of silver  chloride. What is the % of Na2O and K2O in the sample ?
4. Chloride  samples are prepared for  analysis by using NaCl, KCl, NH4Cl  separately or as mixtures. What minimum volume of a 5% (by weight)  AgNO3 solution (sp.gr=1.04) must be added to a sample weighing  0.3 g. in order to ensure complete  precipitation of  chloride in every possible case ?

 

1. 5g of pyrolysite (impure MnO2) were heated with Conc. HCl and Cl2 evolved was passed through excess of KI solution. The iodine liberated required 40 mL of hypo solution. Find the % of MnO2 in the pyrolysite.

6. One gram of an alloy of aluminium and magnesium when  heated with excess of  dil. HCl forms  magnesium chloride, aluminum chloride and hydrogen. The evolved hydrogen collected over mercury at 0oC has a volume of 1.2 litres at 0.92  atm.  pressure. Calculate the composition of the alloy.
7. A mixture of pure K2Cr2O7 and pure KMnO4 weighing 0.561 g. was treated with excess of KI in acidic medium. Iodine liberated required 100 ml. of 0.15 M of sodium thiosulphate solution for exact oxidation. What is the % of each in the mixture?
8. A solution contains a mixture of sulphuric acid and oxalic acid, 25 ml. of the solution requires 35.54 ml. of 0.1 M NaOH for neutralization and 23.45 ml. of 0.02 M KMnO4 for oxidation. Calculate the molarity of solution with respect to sulphuric acid and oxalic acid.
9. The neutralization of a solution of 1.2 g. of a substance containing a mixture of H2C2O4. 2H2O, KHC2O4. H2O and different impurities of a neutral salt consumed 18.9 ml. of 0.5 N NaOH solution. On titration with KMnO4 solution, 0.4 g. of the same substance needed 21.55 ml. of 0.25 N KMnO4. Calculate the % composition of the substance.

 

10. 25 ml of solution of NaOH and Na2CO3 required (a) 25ml of 0.1 N HCl using phenolphthalein indicator (b) 35 ml of 0.1 N HCl using methyl orange indicator . Calculate the masses of NaOH and Na2CO3 present for litre of the solution.

 

ANSWERS

 

1. 63.01 g, 152.06 g, 60.27 g	6. Al = 0.5485g  ;   Mg = 4515g.
2. 59.36 %	7. KMnO4 = 56.33 %, K2Cr2O7=43.67 %
3. % Na2O  = 3.62  ; % K2O = 10.57	8. H2SO4 = 0.0241 M, H2C2O4 = 0.0469 M
4. 18.3 ml	9. H2C2O4 . 2H2O=14.35 %, KHC2O4 . H2O = 81.71%
5. 0.174 g; 3.48%	10. Na2CO3 = 4.24 g/L NaOH = 2.4 g/L

 

ATOMIC STRUCTURE

 

Distance of closest approach of α-particle to the target nucleus vs. velocity of
α-particle

 

The distance of α-particle from the target nucleus in Rutherford’s α-particle scattering expt. at which P.E. of its interaction with the nucleus is just sufficient to overcome its K.E. resulting into reversal of direction of its motion is called effective radius of the nucleus. Effective radius of the nucleus is thus the distance of the closest approach.

(CGS unit) = 9 × 109 Nm2C–2 (SI unit)

or    r = …(1a) (∈0 = permittivity of free space)

or …(1b)

Where 

Z = Nuclear unit +ve charge i.e. at. no. of the target, 

m = mass of α-particle = 4× 1.66 × 10–24g (C.G.S.), v = velocity of α – particle (Cms–1 in C.G.S.) or (ms–1 in S.I. unit), e = 4.802 × 10–10 esu (CGS unit)
= 1.6 × 10–19 coulomb  (SI unit), and r = effective radius of the nucleus.

v = velcoity of α-particle (CmS–1 m C.G.S.) or ms–1 m S.I. unit.

e = electronic charge = 4.802 × 10–10 esu (CGS unit)

= 1.6 × 10–19 coulomb (SI unit), and

r = effective radius of the nucleus (Cm in C.G.S. unit or m in SI unit)

 

Using Eq. 1, v can be calculated knowing r or r  can be calculated knowledge of r or v, and also Z. It is also noted that greater the velocity, samller the distance of closest approach.

 

Rutherford Nuclear Model Of Atom Vs. Bohr’s Model Of Atom

 

Rutherford model of atom with a positive charged heavy central core situated at the centre of atom called nucleus was considered as a mile stone in the development of atomic structure. However, Rutherford’s idea that electrons in an atom were free to revolve round the nucleus in an orbit of desired radius, i.e. energy spectrum of electron was continuous, received criticism since such an atom could never be stable (it can last only for a time interval of the order of 10–8 s), could not give line spectrum (a discrete spectrum) and could not justify Planck’s Quantum theory. Niels Bohr, therefore, corrected Rutherfords model of atom with the help Panck’s Quantum Theory (E = hν = hC/λ = hcν) and keeping in mind discrete nature of atomic spectra. Bohr proposed his own model opatern the main postulate of which is that electrons can revolve round the nucleus only in some selected orbitals and not just in any orbit and for orbit to be permissible, the angular mometum of the revolving electron must be an integral multiple of unit, i.e. 

mvr =

Where n = an integer excepting zero i.e. n = 1, 2, 3, 4 …. ∞, called Principal Q.N. Each value of n denotes an orbit (shell, stationary state or energy level) the numerical value of which equals to the serial number of the orbit.

 

Another important postulate of Bohr’s model of atom is that an electron can jump up to the higher energy state or jump up to the higher energy state (state of lower n) by the absorptium or emission of energy equal to the energy difference between initial and final state, and this energy is absorbed or emitted in one and only one bundle (quantum or photon) per atom.

 

If E1 and E2 be the energies of the lower and higher energy states, respectively, within which transition is taking place then

hν = E2 – E1

 

or ν =

Where v = frequency of electromagnetic  radiation i.e. photon emitted.

 

Expressions For Radius Of Orbit Velocity Of Electron And Energy Of Electron

 

Using his concept of quantization of angular momentum of electron, Bohr derived following expression for an atom or ion containing only one electron like H, He+, Li++ etc. in which there is no scope for inter-electronic repulsion.

 

Bohr on the basis of his theory of atom derived following expressions for an atom or ion free from inter-electronic repulsion in an atom or ion containing only one electron.

Radius of nth orbit, rn = Å

Where V = atomic number n = 1, 2, 3 …∞

Where a0 = = radius of the 1st orbit of H-atom called Bohr’s radius 

m = mass of e–, π = 3.14, e = electronic charge, h = Planck’s constant and n = an integer excepting zero, i.e. n = 1, 2, 3, …∞.

 

CGS units: m = 9.11 × 10–28g 

e = 4.802 × 10–10 esu, 

h = 6.625 × 10–27 erg.s and
k = 1 

a0 will in Cm.

En in erg

Vn in Cm/s

 

SI Unit: m = 9.11 × 10–31kg, 

e = 1.6 × 10–19 coulomb, 

h = 6.625, × 10–34 J.S. 

K = 9 × 109 Nm2 C–2

a0 in meter

En in Joule

Vn in m/s

Velocity of electron in nth orbit, Vn = =

=

where n = 1, 2, 3 …∞

No. of revolutions per second =

Energy of electron, in nth orbit En=

= J/atom

= ev/atom

Where n = 1, 2, 3…∞

As is evident, when n = ∞, E∞ = 0

I.P. of atom or ion containing only one electron:

I.P. = E∞  – En = – En

Where E∞ = Energy of electron at infinity from the nucleus 

En = Energy of electron in nth orbit, the orbit from where electron is to be removed (n = 1 in the ground stae, n = 2 in the 1st excitation state, n = 3 in the 2nd excitation state and so on)

 

Significance of negative energy: At infinity from the nucleus energy of electron is zero.
Starting from infinity as electron goes nearer to the nucleus, it performs work (= force × displacement) at the cost of its own energy and hence its energy falls below zero assuming there by negative value.

 

Atomic Spectra of Hydrogen, He+ Ion, Li+ Ion Etc.

 

When a macroscopic sample of hydrogen gas is subjected to electric discharge at low pressure, there occurs dissociation of hydrogen molecules. The H-atoms so produced absorb energy and get excited to some higher energy state depending upon the amount of energy absorbed. Excited state being unstable, H-atoms return to the less excitation state and finally ground state with emission of em radiation (photon) of some definite wavelength corresponding to the energy difference.

 

All H-atoms whose number in the sample is of the order of 1020 or even more, do not get excited to the same state of higher energy and as such they do not emit em radiations of the same wavelength. If a photographic plate is exposed to the mixture  of emitted em radiation of varying wavelengths after resolution through a prism, each of these produces a line on the photographic plate called spectral line and the photographic plate containing so many lines is called atomic or line spectrum of hydrogen. Similarly, we can have atomic structure of He+, Li+ or of any other elements. Thus, line spectra are actually emission spectra, and they are characteristics of the elements.

 

The wave number of the spectral line resulting from downward transition from a level of higher energy n = n2 to another level of lower energy with n = n1 is as given below.

…(6)

Where RH = = 109737 Cm–1 (Bohr’s value)

The experimental value of RH is 109667 Cm–1 

 

Various Series:

 

Lyman Series n1 = 1 (fixed), n2 = 2, 3, 4, 5, 6, 7, 8 …

(UV region)

Blamer Series n1 = 1 (fixed), n2 = 3, 4, 5, 6, 7, 8…

(Visible region)

Paschen Series n1 = 3 (fixed), n2 = 4, 5, 6, 7, 8 …

(IR region)

Brackett Series n1 = 4 (fixed), n2 = 5, 6, 7, 8…

(IR region)

Pfund series n1 = 5 (fixed), n2 = 6, 7, 8..

(IR region)

 

No. of possible spectral lines emitted by one electron atom or ion excited to nth level 

= .

PHOTO ELECTRIC EFFECT

 

hν = hν0 + K.E.

or K.E. = h(ν – ν0)

= hc

= ………(7)

Where

ν0 = Threshold frequency (The minimum frequency that a radiation may possess so as to cause just the removal of electron from the outermost shell of a metal atom)

ν  = frequency of the incident radiation.

 

and hν0 = I.P. of gaseous atom also called work function (φ)

Thus, K.E. of photoelectron = hν – I.P. = = – I.P. ………(8)

 

de-BROGLIE’S RELATION (DUAL NATURE OF ELECTRON)

In order to explain the quantisation of angular momentum of electron, de-Broglie (1924) conceived the idea that electron revolves round the nucleus like a wave. Thus, electron is a particle by virtue of having mass and charge, while at the same time it has wave nature as well. The wavelength of electron as wave (λ) and momentum (mν) of electron as a particle is intimately linked by the following equation.

…(9)

This equation and hence particle wave dualism holds good for all the micro (minute) material particle.

The wave nature of electron was accidentally confirmed by Davisson and Germer who noticed that a beam of electron accelerated by a potential difference of 7.5 V gets diffracted upon striking the surface of Nickel metal. Here, Nickel crystal actually behaved as a diffraction grating.

Å …(10)

Where v = p.d. of the field in volt

 

HEISENBERG’S UNCERTAINTY PRINCIPLE

Δx × Δp ≥ …(11)

Where Δx = Uncertainty or error in the measurement of position.

Δp = Uncertainty or error in the measurement of momentum.

 

If Δv = uncertainty w.r.t. velocity

Δp = mΔv, and hence 

Δx × mΔv ≥ …(11b)

 

SCHRODINGER WAVE EQUATION (QUANTUM MECHANICAL MODEL OF ATOM)

 

Classical mechanics treats particle and wave with different equations. So, neither particle nor wave equation can be successfully applied to electron possessing particle – wave dualism. Schrodinger after incorporating de-Broglie’s equation in classical wave equation derived an equation which is considered as gate way to Quantum mechanics. The equation which can be applied to calculate energy of electron accurately, is as follows.

= 0

Where ∇2 (pronounced as del two or twice) is as defined below:

, E = total energy of electron, V = P.E. of electron, m = mass of electron and ψ which is the amplitude function of the electron wave is called wave function.

 

ψ2 has great physical meaning. ψ2 measures the probability of finding the electron at a given point with coordinate x, y, z, and ψ2dτ (where dτ = dxdydz) is the probability of finding the electron in an infinitisimal volume element dτ. ψ also refers to AO which is quantum quality mechanical analogue of orbit. The value of ψ can be obtained by solving the Schrodinger wave equation.

 

AO is the region in three dimensional space round the nucleus where there is high probability of finding electron of given energy. As worded, the energy of electron is certain while its position is uncertain. Thus, orbital satisfies the condition as laid down by uncertainty principle for microsopic  systems. Bohr’s orbit of fixed energy and fixed radius has no validity in view of wave nature of electron coupled with uncertainty principle.

 

FOUR QUANTUM NUMBERS

 

The energy of electron in an atom can calculated with the help of following four Quantum Numbers (Q.N.).

 

1. i) Principal Q.N. (n)
2. ii) Acimuthal Q.N. (l)

iii) Magnetic Q.N. (m)

1. iv) Spin Q.N. (S)

 

The first three Q.N.s have come from quantum mechanics while the fourth has come from Spectroscopy.

 

1. i) Principal Q.N. (n)

Allowed values of n: n = 1, 2, 3, 4, ….. ∞

Significance: It determines size of orbital

As n increases, size of orbital (electron-cloud) increases.

n also denotes the principal shell in which electron belongs. The principal shells with their values are: K (n = 1, i.e. the 1st shell), L (n = 2, i.e. the 2nd shell), M (n = 3, i.e. the 3rd shell), and so on.

1. ii) Azimuthal Q.N. (l)

Allowed values of l: l = 0, 1, 2…. (n – 1)

Total number of l values = n

Significance: a) It determines the shape of the orbital.

1. b) It denotes the sub-shell of electron. Sub-shells are denoted by the symbols s, p, d, f,….

Value of l	Sub-Shell	L	Shape of Orbital
l = 0	S	0
l = 1	p
l = 2	d
l = 3	f		Complicated
l = 4	g		More complicated

 

(From f onward sub-shells are denoted by letters in alphabatical order)

Form = 3, l = 0, 1 and 2, meaning third shell is divided into three sub-shells: 3s, 3p and 3d.

1. c) It determines the orbital angular momentum of electron i.e. L.

L =

iii) Magnetic Q.N. (m)

Allowed values of m: All integral values right from – l to + l including zero

Total number of m values = 2l + 1

Significance:

1. a) It determines orientation of orbital. 

For l = 1, m = –1, O, +1, meaning there by that p-electron can assume three different orientations in an applied magnetic field. Each of these orientation is said to be an orbital. This means p-subshell has three orbitals possible and they are px, py and p2. All the orbitals within a given sub-shell are degenerate.

Fig: Three p-orbitals of identical energies differing in their orientation.

px, py and pz orbitals have one lobe each lying in yz, xz and xy planes, respectively.

 

1. b) It determines the z-compoent of orbital angular momentum which is equal to . It θ be the angle between z axis and the component, then 

 

1. iv) Spin Q.N. (S):

Allowed value: Only two

Significance: It determines the direction of spin of electron i.e. whether clock wise or anti clock wise and introduces quantisation in the spin angular momentum according to the following equation.

Along clock wise and anti clock wise direction the components of Ls are: and , respectively.

It helps to determine spin magnetic moment of an atom or ion as given below.

B.M.

= B.M.

where S = Summation of spin Q.N.

n = Number of unpaired electrons = 2S

 

RADIAL DISTRIBUTION CURVE

 

The volume of a spherical shell of thickness dr at a distance r from the centre of atom is 4 πr2dr. The probability of finding electron in this spherical shell is 4πr2ψ2 dr and it is known as Radial distribution function. As we move away from the nucleus 4πr2dr increases while ψ2 decreases but their product vary with the distance as shown in the following curves called radical probability distribution curves of 1s, 2s and 2p orbital.

 

The maximum corresponds to the point of maximum probability of finding the electron. In the ground state of H-atom where electron exists in 1s orbital the radius of maximum probability occurs of 0.529Å. This corresponds to the Bohr’s radius. Further, wave nature of atom leads to the condition of stable orbit that an orbit to be stable the circumference of the orbit must be an integral multiple of the wavelength of electron otherwise electron wave will be out of phase and will be destroyed by destructive interference. 

 

Thus

2πr = nλ (stable orbit)

where n = 1, 2, 3,….

Putting the value of λ from de-Broglie’s equation

2πr =

mvr =

Which is the Bohr’s postulate

 

Thus, we see that although quantum-mechanical model of atom has superheaded Bohr’s model of atom, yet the latter has some relevance still left.

 

ELECTRONIC CONFIGURATION OF ELEMENTS

 

The lowest possible electronic energy state of an atom or ion is called its ground state. The ground state electronic configuration of an atom or ion may be written recalling the following.

 

1. i) a) Electron prefers to occupy, sub-shell with lower n +  value to higher n + 
2. b) Among the Sub-shells with the same value of n + , the one with lower value of n is filled first. The above statement is called Auf-bau principle. Thus, electrons fill up sub-shells in the following order starting from 1s.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s

 

1. ii) No two electrons in the same atom can have all the four Q.N.s equal i.e. an orbital can contain only a maximum of 2 electrons and that too when their directions of spin are opposite (Pauli’s Exclusion Principle).

Thus,  configurations like that shown below violate Pauli’s Exclusion principle (where box represents an orbital)

 

but the configuration like that shown below obeys Pauli’s Exclusion principle and hence is the ground state of atom.

Since d-subsell contain five d-orbitals so the capacity of d-subshell of accommodating electrons will be 5 ×2 = 10.

 

iii) a) Orbitals of a given sub-shell have all the same energy in absence of any field and are called degenerate orbitals. Spin-paring (11) in any orbital within a given sub-shell will not occur unless each of the orbitals within that sub-shell is not filled singly with similar spin first (Hund’s rule).

 

Thus, for p3 the ground state configuration is III and not I or II given below.

1. b) Completely filled, half-filled or completely empty sub-shell is associated with some extra stability (Hund’s rule).

Thus ground state of Cr(24) is [Ar] 3d5 4s1 (Hund’s rule) and not [Ar] 3d4 4s2 (Auf-bau principle)

 

ATOMIC NUMBER IS MORE FUNDAMENTAL THAN ATOMIC MASS
OR MASS NUMBER

 

The wavelength of given x-ray spectral line depends upon atomic number of the target (anticathode) rather than on its mass number, in Moseley’s experiment. Following linear relationship holds.

where σ = shielding constant a = proprotionality constant, z= atomic number and v = frequency of x-ray line 

A plot of √v = vs z will be a straight line of the slope equal to “a” and intercept equal to –aσ. 

Solved Examples

 

Example-1: What will be the K.E,. of photo electron ejected by a metal upon irradiation with electro magnetic radiation of wavelength equal to that of the series limit in Lyman series of He+ ion? RH = 109677 Cm–1, I.P. of metal is 3.8 ev, h = 6.625 × 10–27 erg.s and c = 3 × 1010 cms–1)

 

Solution:

Z = 1, n1 = 1 and n1 = ∞, λ = ?

= 109677 × 4 Cm–1

λ = 2.279 × 10–6 Cm, ν = = 1.316 × 1016 Hz

hν0 = 3.8 ev = 3.8 × 1.6 × 10–12 erg

or ν0 = = 0.917 × 1015 Hz

K.E. = h(ν – ν0) = 6.625 × 10–27 (1.316 × 1016 – 0.0917 × 1016) 

= 8.11 × 10–11 erg

= 50.68 ev

Example-2: i) Calculate wavelength of a base ball (mass = 100g ) moving with a velocity of 100m/s. Is it possible to observe the wave nature of base ball? Give reason for your answer.

1. ii) Calculate wave length of electron accelerated by a potential difference of 150 volts.

 

Solution: i) = 6.625 × 10–33 cm

= 6.625 × 10–25 Å

The wavelength is too small to observe. To observe the wave nature of base ball we will have to perform diffraction experiment using grating of the dimension of 6.625 × 10–25 Å. We cannot have such a small grating since the dimension of atom or molecule is of the order of Å. de-Broglie’s relation applies only to micro particle. Base ball is a macro particle and is a “hard particle”.

 

1. ii) Å = = 1Å

Example-3: An electron has a speed of 600 ms–1 with an accuracy of 0.004%. Determine the uncertainty with which its position can be located. Comment on the result.

 

Solution: Δp = = 4 × 10–5 × 9.11 × 10–31 × 600 Kg ms–1

According to Heisenberg uncertainty principle

Δx × Δp ≥

∴ Δx ≥

or Δx ≥

or Δx = 0.0024m = 2.4 × 10–3 m

The uncertainty in position is many times more than the radius of atom which is of the order of 10–10 m. It cannot be said whether atom is within the atom.

 

EXERCISES

 

1. An electron has K.E. 2.8 × 10–25J. Calculate de-Brogilie wavelength,
   (Me = 9.11 × 10–31 kg)
2. The minimum energy required to overcome the attractive forces between electron and the surface of Ag – metal is 7.52 × 10–19J. What will be the maximum K.E. of electron ejected out from Ag which is being exposed to UV light of λ = 360 Å?

 

3. The ionisation energy of H-like Bohr atom is 4 Rydberg. (a) Calculate the wavelength radiated when electron enertgy jumps from the first excited state to the ground state (b) what is the radius of the 1st orbit of this atom? (1RH = 2.18 × 10–18 J)

 

4. A sample of hydrogen atoms weighing 1.8g is excited to cause radiations. The study of spectra indicates that 27% of the atoms are in 3rd energy level and 15% of atoms in 2nd energy level and the rest in ground state. If IP of hydrogen is 21.7 × 10–12 erg, calculate
5. a) Number of atoms present in 3rd and 2nd energy levels, and 

(b) Total energy evolved when all the atoms return to the ground state.

5. Find the Q.N. “n” corresponding to the excited state of He+ ion if on transition to the ground state that ion emits two photons in succession with wavelengths 108.5 and 30.4 nm

 

6. The dissociation energy of H2 is 430.53 kJ mole–1. If H2 is exposed to radiation of wavelength 253.7 nm, what percent of radiant energy will be converted into K.E.

 

1. The photo electric emission requires a threshold frequency v0. For a certain metal
   λ1 = 2200Å and λ2 = 1900Å produce electrons with a maximum kinetic energy KE1 and KE2. If KE2 =- 2KE1 calculate v0 and corresponding λo.

8. Two hydrogen atoms collide head on and end up with zero kinetic energy. Each then emits a photon with a wavelength 121.6 nm. Which transition leads to this wavelength? How fast were the hydrogen atoms travelling before the collision? Given RH = 1.097 × 107 ms–1 & mH = 1.67 × 10–27 kg.
9. A bulb emits light of wave length 4500Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by bulb per second?
10. Calculate the frequency, energy and wave length of radiation corresponding to the spectral line of lowest frequency in Lyman series in the spectra of hydrogen atom. Also calculate the energy for the corresponding line in the spectrum of Li2+ ion. (RH = 1.09678×107m–1, c = 3 ×108 m/s, h = 6.626 ×10–34 Js–1)

ANSWERS

1. λ = 9.245 × 10–7 m = 9245 Å = 924.5 nm	6. 8.68%
2. 47.68 × 10–19J	7. v0 = 1.1483 × 1015 S–1,  λ0 = 2612.6 Å
3. (a) 303.89Å, (b) 2.645 × 10–9 cm	8. 10. 4.43 × 104 m/sec
4. (a) 2.928 × 1023 atoms in 3rd shell, 162.6 × 1023 atoms in II shell,  (b) 832.50 kJ	9. 4.65 eV
5. n = 5	10. 4.175 × 10–3 moles of I2

GASEOUS STATE

 

1. i) Ideal gas equation for n mole(s): PV = nRT
2. ii) Compressibility factor, Z =

Z = 1 (ideal behaviour) while Z ≠ 1 (deviation from ideal behaviour)

 

iii) Deviation of real gases from ideal behaviour

[A] Plot of Z vs. P at 0°C (showing the effect of pressure on deviation)

 

[B] Plot of Z vs. P for N2 gas at differnet temps (showing the effect of temperature on deviation)

At 51°C, N2 gas behaves ideally upto appreciable pressure. Above 51°C, it show deviation. Below 51°C, it shows negative deviation. The temperature 51°C is called Boyle’s temperature for N2. Every gas has its own Boyle’s temperature.

 

1. iv) Equation of state for n mole(s) of a real gas (vander Waal’s equation):

= nRT

For 1 mole (n = 1)

= RT

“a” and “b” are vander Waal’s constants of a gas. Both are characteristics of a gas. “a” measures the strength of intermolecular attractive force existing in a gas. It is also called internal pressure of gas.

p = pressure correction factor

or, a = =

The constant b is equal to 4 times the actual volume of molecules per mole i.e.

b = 4 × Nπr3

b has the unit of volume

vander Waal’s equation may be rearranged as

–

This equation is used to calculate pressure of real gas.

 

1. v) Kinetic gas equation: PV =

RMS velocity (c) =

Average velocity =

Most probable velocity (Cmp) =

 

1. vi) Mean K.E. per mole =

Mean K.E. for n mole(s) =

Mean K.E. per molecule = k = Boltzmann’s constant.

K.E. per mole or per molecule depends only on temperature and not upon nature of gas.

 

vii) Partial Pressure, pi = Xi Pmix, Xi =

Pmix = (Dalton’s law of partial pressure)

 

22. a) The numerical value of mass in gram of 22.4 litre vapour of a substance at NTP is equal to its M.W.

 

1. b) Knowing number of moles of various gaseous components in a mixture maintained at a given P and T, the density of the mixture can be calculated. The relevant equation for this purpose are:

 

M.W. of the mixture =

PV = nRT =

or =

1. ix) Graham’s law of diffusion:

Spontaneous mixing of two or more gases is called diffusion.

When a gas compressed at high pressure diffuses into air through an orifice, the process is called effusion.

Rate of effusion (r) = =

r depends upon temperature, pressure difference, size of orifice and MW of the gas.

According to Graham’s law of diffusion.

(at a given P, T and size of orifice)

, the time remaining the same =

D2 = V.D. of gas “i” and Mi = MW of gas “i” ri = rate of diffusion or effusion.

If pressure is not constant then

1. x) Eudiometry (Gas Analysis):

Reacting gases are taken in a Eudiometer tube over mercury and made to react explosively by passing electric spark into it. Volume of the gaseous mixture before and after explosion is measured directly by the graduation in Eudiometer tube while volumes of constituent gases are measured by absorbing them in suitable solvents.

Gases Solvent for absorption

O2 alk. Pyrogallol

CO2 KOH

SO2 KOH

CO Ammonical Cu2Cl2

O3 Terpentine oil

NO FeSO4 solution

NH3 CuSO4 solution or any acid solution

 

The combustion of a hydrocarbon takes place according to the following general equation.

CxHy + O2 ⎯→ xCO2 +

 

SOLVED EXAMPLES

Example-1: At 1200°C, gaseous mixture of Cl2 and Cl atoms effuses 1.16 times as fast as Krypton effuses under similar conditions. Calculate the fraction of chlorine molecules dissociated into atoms.

(MW of Kr = 83.8)

Cl2 2Cl

Solution:

 

1.16 =

 

Mmix = 62.28

Cl2 2Cl

1 0

1-α 2α

Mmix =

or 62.28 =

Upon solving

α = 0.14

∴ dissociation = 14%

 

Example-2: 20% N2O4 molecules are dissociated into NO2 in a sample of gas at 27°C and 760 torr. Calculate the density of equilibrium mixture.

 

Solution: = 0.2; N2O4 2NO2 

1 0

1 – 0.2 2 × 0.2

M.W. of the mixture = = 76.66

d = = = 3.116 gL–1

Example-3: 1 litre mixture of CO and CO2 when passed over red hot coke, volume increased to 1.6 litre. Find the mole fraction of CO and CO2 in the original mixture.

 

Solution: CO2(g) + C(s) ⎯⎯⎯→ 2CO(g)

1 vol 2 vols

x L 2x L

So, volume of CO(g) after reaction = [(1 – x) + 2x] litre

From question

(1 – x) + 2x = 1.6

or x = 0.6

Thus, in the original mixture

Volume of CO2 = 0.6 L

Volume of CO = 0.4 L

Mole fraction of CO2 = = 0.6

Mole fraction of CO = = 0.4

 

Example-4 : A mixture of 1 mole He and 3 moles N2 has a pressure of 16 atm. Due to a hole in the vessel in which mixture is placed, mixture leaks out. What is the composition if mixture effusing out initially?

 

Solution: pHe = = 4 atm

= 12 atm

So, no. of mole of He and N2 effusing out initially are in the ratio 0.88:1

 

Example-5: vander Waal’s constant b of Ar is 3.92 × 10–5 m3 mole–1. Calculate molecular diameter of Ar.

 

Solution: b = 4N × πr3

Putting the values of b, N and π we get

r = 1.47 × 10–10 m

∴ d = 2r = 2.94 × 10–10 m = 0.294 nm

 

EXERCISES

 

1. Bond energy of H2 is 104 kcal mole–1. At what temperature average K.E. of gaseous H2 molecules equals the energy required to dissociate the molecules into atoms.

 

2. A certain volume of H2 effuses from an apparatus in one minute. The same volume of ozonized oxygen (mixture of O3 and O2) took 246 sec. to effuse from the same apparatus and under identical conditions. Find the percentage composition of the ozonized oxygen.

 

3. The partial pressure of O2 respired in air is 158 mm Hg at 25°C while total pressure of air is 760 mm Hg. The air that is respired from your lungs after breathing has a partial pressure of O2 equal to 115 mm Hg under the same conditions. How many mole(s) of O2 is (are) absorbed by your lungs from 1.0 L air.

 

4. 10 mL of a gaseous hydrocarbon was burnt completely in 80mL of O2 at NTP. The remaining gas occupied 70 mL at NTP. This volume reduced to 50 mL on treatment with KOH solution. What is the formula of the hydrocarbon?

5   Using vander Waal’s equation of state calculate the pressure exerted by 88g of CO2 occupying a volume of 8 litres at 27°C (a = 3.6 atm L2 mole–2, b = 0.043 litre mol–1, R = 0.0821L atm K–1 mole–1). Also calculates compressibility factor assuming that observed  pressure is 1% less than the vander Waal’s pressure.

6. At 27°C, H2 is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same T and P as that of H2 is leaked through the same hole for 20 minutes. After the effusion of the gases, the mixture exists a pressure of 6 atm. The H2 content of the mixture is 0.7 mole. If volume of container is 3 litre. What is molecular mass of gas?

 

7. A mixture of CH4(g) and C2H6(g) has a total pressure of 0.53 atm. Just enough O2(g) is introduced in the mixture to bring its complete combustion to CO2(g) and H2O(g). The total pressure of these two gases is found to be 2.2. atm. Assuming constant volume and temperature, determine the fraction of CH4 in the mixture.

 

8. Calculate the relative rates of diffusion of 235UF6 and 238UF6 in gaseous form.

 

9. Pure O2 diffuses through an aperture in 224 seconds, whereas mixture of O2 and another gas containing 80% O2 diffuses from the same in 234 seconds. What is molecular mass of gas?

 

10. 60 ml of a mixture of equal volume of Cl2 and an oxide of chlorine was heated and then cooled back to the original temperature. The resulting gas mixture was found to have volume of 75 ml. On treatment with caustic soda solution, the volume contracted to 15 ml. Assuming that all measurements are made at the same T and P. Deduce the simplest formula for oxide of Cl2. The oxide of Cl2 on heating decomposes quantitatively to O2 & Cl2.

 

ANSWERS

 

1. 34900 K	6. 1033
2. O2 = 89.87%, O3 = 10.13%	7. 0.4245
3. 2.3 × 10–3 mole	8. 1.0043
4. C2H4	9. 46.6 gm
5. Pobs =  5.931 atm                                          Z=   0.963]	10. Cl2O

 

CHEMICAL KINETICS 

1. INTRODUCTION

Chemical Kinetics is the branch of science that deals with the following:

(i) Rate of reaction (definition and determination)

(ii) Factor affecting the rate of reactions like concentration, temperature, surface area of the reactant, catalyst etc.

(iii) Reaction – mechanism (with the help of the rate law i.e., concentration dependence of rate)

Ionic reactions and reactions involving free radicals or odd electron molecules occur at such an enormously high rate that they complete within twinkling of eye.  Their half – lives are of the order of negligible fraction of second. On the other hand some molecular reactions which involve breaking of very strong bond(s) occur extremely slowly such that their half – lives are of the order of several years. An example of extremely slow reaction is “rusting of iron”.  However, most organic reactions occur at measurable rates with their half – lives of the order of days, hours and minutes. In chemical kinetics we actually deal the reactions occurring at measurable rates. Now a days kinetics of very fast reactions are also being studied using lasers:

 

Rate of reaction:

The change in the concentration of any of the reactants or products per unit time is called rate of reaction i.e.,

Rate  =

Where ΔC the change in conc. of reactant or product whichever is convenient to measure, during the time interval Δt, –ve sign signifies decrease in conc. of reactant where +ve sign signifies increase in conc. of product.

The  rate of reaction is  not uniform . It decrease from moment to moment due to decrease in concentration (s) of reactant(s) with the passage of time. So the rate expressed above is actually the average rate of reaction within the time interval considered. The rate of reaction at any given instant of time t also called instantaneous rate is as defined below.

rt  or rinst = 

Where  dc is the infinitesimal change in concentration during infinitesimal time interval dt after time t such that the rate remains constant during the time interval t and  t + dt.

For the reaction

2A ⎯⎯→ 3B + C

The rate of reaction at any time t may be expressed by any one of the following.

–

Since for every 2 moles of A consumed, 3 moles of B and one mole C are produced within the same time so the above three individual rates can be equated as 

Thus, the individual rates in terms of various species involved in any reaction will be equal to one another  when rate is calculated at the rate of per mole i.e., when each of them is divided by its respective stoichiometric coefficient involved in the chemical equation.

Let the reaction mentioned above obeys the following rate below  

.

Rate =  k[A]2, k  = rate constant of reaction 

If kA, kB and kC were the rate constants of reaction as defined low 

and = 

Thus these rate constants will be interrelated as .

The rate of reaction at any time t is determined in the following way 

(i)  Concentration  (either of reactant or product) vs time curve is plotted 

(ii) A tangent is drawn at the point (p) of the curve that corresponds to the time (t) at which rate is to be determined 

(iii) The slope of the tangent gives the rate of reaction at the required time as shown below.

 (CR = conc. of reactant and CP = conc. of product)

Unit of rate  = = conc. time –1 i.e., mole L–1 s–1

Molecularity and order

Single step reaction is called elementary reaction while a reaction occurring in two or more steps is called complicated or consecutive reaction. Each step of a complicated reaction is an elementary reaction.

The minimum number of molecules, atoms or ions of reactants required for an elementary reaction to occur as indicated by the sum of the stoichiometric coefficients of the reactant(s) in the chemical equation, is known molecularity of the reaction. Thus for an elementary reaction represented by the general chemical equation.

aA  + bB ⎯→ products

molecularity = a + b

Examples Reactions Molecularity

PCl5 PCl3 + Cl2 1

H2 + I2  2HI 2

 

A complicated reaction has no molecularity of its own but molecularity of each of the steps involved in its mechanism.

 

For example:

The reaction: 2NO + 2H2 N2 + 2H2O, takes place in the sequence of following three steps.

1. NO + NO N2O2 (fast and reversible)
2. N2O2 + H2 ⎯→ N2O + H2O (slow)

III. N2O + H2 ⎯→ N2 + H2O (fast)

 

The molecularity of each step involved in mechanism is 2 i.e., each step is biomolecular. So as regards the molecularity of the reaction under consideration we simply say that the reaction has mechanism and each step involved in it is biomolecular

 

However, there is another view also –  according to which molecularity of a complicated reaction is taken to be equal to the number of molecules, atoms or ions of reactant(s) and/or intermediates coming into contact and colliding simultaneously in the slowest step i.e. the rate-determining step of the reaction.

For example ROH + HCl RCl + H2O which is actually a nucleophilic substitution reaction

 

1. (fast)

III. (fast)

The mathematical equation showing the relation of rate of reaction with concentration of reactant(s) is  called rate-law of the reaction. For example, the kinetic experiment carried on the reaction.

2NO+ 2H2(g) ⎯→ N2(g) + 2H2O (g)

has shown that (i) rate increase 4 – fold when conc. of NO is doubled keeping that of H2 constant (ii) rate gets just doubled when conc. of H2 is doubled keeping that of NO constant, and (iii) rate increases 8 – fold when concentrations of both NO and H2 are doubled simultaneously. These experimental rate data fit into following equation. 

Rate α [NO]2[H2]

This equation is called experimentally observed “Rate law” of the reaction.

Order of reaction is defined as the sum of the powers of the concentration terms appearing in experimentally observed rate law. Thus for the reaction (reduction of nitricoxide by hydrogen) considered above.

Order w.r.t. NO = 2

Order w.r.t. H2 = 1

Overall order  = 2+ 1 = 3

In general, let a reaction represented by the chemical equation:

aA  + bB ⎯⎯→ Products 

obeys the following rate law.

Rate α [A]m [B]n, or Rate  = k[A]m [B]n

Where K = rate constant of reaction, a constant at constant temp and is actually the rate of reaction when conc. of  each reactant is equal to unity k is also called velocity coefficient or specific reaction rate.

Note that m and n are experimental quantities which may or may not be equal to the respective stoichiometric  coefficients.

Also note that if either of A or  B is taken in large excess as compared to another, the order w.r.t this reactant will be zero so order  = m (when B is in large excess), and order  = n (when A is large excess),i.e.,

Rate = k′[A]m, k′ = k[B]n  another constant, when B is in large excess.

Order w.r.t. A = m

Order w.r.t. B = 0

Rate = k′′[B]n, k′′ = k[A]m = another constant,  when A is in large excess.

Order w.r.t. A = 0

Order w.r.t. B = n

k′ or k′′ is actually known as pseudo rate constant 

Thus, a biomolecular reaction conforms to the first – order when one of the reactants is taken in large excess and the reaction is said to be pseudo unimolecular or pseudo first order. An example of this is the hydrolysis of ester by dil. acid i.e.

CH3COOC2H5 + H2O+  CH3COOH + C2H5OH

The reactions is originally of second – order obeying the kinetics.

Rate = k [CH3COOC2H5] [H2O]

But the reaction is usually carried out taking dilute aqueous solution of ester and acid (HCl)  such that in the reaction mixture water exists in large excess as compared ester. So there is no appreciable change in the concentration of water and the same remains practically constant. k[H2O] becomes another constant (k′) called pseudo rate constant and reactions oebys the following 1st order kinetics 

Rate = k′ [CH3COOC2H5]

Thus, the molecularity of the above reaction is ‘two’ but its order is ‘one;.

In the light of the above concept one may define order of reaction as ” the number of molecules of the reactant(s) whose concentration alters during the chemical change is the order of reaction”.

 

Distinction between molecularity  and order:

(i)  Molecularity is theoretical property of reaction while order is its experimental property.

(ii) Molecularity concerns to mechanism, order concerns to kinetics 

(iii) Molecularity is always a whole number and never zero while order may be any number zero, fractional and integral.

(iv) Molecularity is invariant but order may vary with change in experimental condition. As for example, the isomerisation of cyclo – propane represented by the chemical equation given below is a first order reaction at high cyclo propane pressure  and is a second order reaction at low cyclo propane pressure.

(v) Reactions of higher molecularity is rare since the chance of coming into contact and colliding simultaneously decreases with number of molecules involved in collision.   

 

Rate law and mechanism

As already mentioned above the reaction:

2NO + 2H2⎯⎯→ N2 + 2H2O

Obeys the following third order kinetics 

Rate α [NO]2 [H2]

The kinetics is not in tune with chemical equation. As the law of mass action suggests each concentration term should be raised to power 2. Neither experimental facts nor those coming from Law of mass Action can be unacceptable to us. Under this condition we are led to believe that the reaction does not occur according to the chemical  equation as written. That is the reaction is not elementary but is complicated. In order to explain the observed rate law following mechanism has been proposed.

I NO + NO N2O2 (fast and reversible)

II N2O2 + H2 ⎯→ N2O + H2O (slow)

III N2O + H2  ⎯→ N2 + H2O (fast)

The step II being the slowest i.e., the rate – determining step (r.d.s.). Note that the rate of formation of N2 cannot be faster than the rate of formation of N2O. So, the rate of overall reaction or rate of formation of N2 will be equal to the rate of step II which according to law of mass action may be given as.

Rate of overall reaction = Rate of step II = k[N2O2] [H2]

Where k = rate constant of step II

The conc. of the intermediate (N2O2) may be evaluated by applying law of mass action upon the equilibrium existing in step I as mentioned below.

Kc =

or [N2O2] = Kc [NO]2

Putting this in the rate expression, we get

Rate = k.kc[NO]2[H2] = kobs [NO]2[H2]

Where kobs = k.kc = observed rate constant of the reaction

Rate α [NO]2 [H2]

This is the same rate – law as observed experimentally.

 

Kinetics of first order reaction 

A first order reaction is one whose rate increases as number of times as concentration of reactant is increased i.e.

α C

Let us consider an unimolecular  reaction obeying  the first order kinetics represented by the following equation.

A  ⎯⎯⎯→Product(s)

a x = 0 … t = 0

a – x x = x … t = t

The initial concentration of  reactant be a mole L–1 and its concentration after  the expiry of time be is (a – x) mole L–1. This means during the time interval t, x mole L–1 of A has reacted.

The rate of reaction at any time t is given by the following first – order kinetics.

K = rate constant of the above first order reaction a constant at constant temperature

or = k (a – x),

( = 0 as a has a  definite value in the given expt.)

or 

Integrating both sides between proper limites, i.e.,

we get 

k =

Converting from loge  to log10, we get 

k = ——- (1)

This is the expression for the rate constant of a first order reaction.

 

Characteristics:

(i) Unit of k: 

The numerical value of k for a first – order reaction will not change by any change in concentration unit but will change if time unit is change. If k is 6.0 × 10–3 min–1 then it would be 6.0 × 10–3 / 60 i.e.  1 × 10–4 s–1.

A general unit of k is (mole L–1)1–n time–1

Where n = order of reaction 

When n = 0, k will be in mole L–1 time–1 which is the unit of rate.

When n = 1, k will be in time-1

When n = 2, k will be in mole–1 L time–1.

 

(ii) Amount consumed during the time interval

x = a(1–e–kt)

 

(iii) Expression for half – time (t1/2)

Half time is the time required for 50% completion of the reaction i.e., reducing the concentration of the reactant to half of its initial value. t1/2 = = a constant independent of initial concentration of the reactant 

A general relationship between t1/2 and initial concentration of reactant i.e, a is follows.

t1/2  α , n = order of reaction.

It is to be noted that not only half – time but time for any fraction of change in a first order reaction is a constant independent of initial concentration of the reactant, and in general, time for any fraction of change is inversely proportional to the initial concentration of reactant raised to the power one less than the order of reaction.

No. of halves (n) = =

∴ t1/2 =

Amount  left after  n – halves  =

Amount distinegrated (decayed) or 

Consumed during n – havles  = No –

Graphical representation of a first – order reaction:

1.  , or = Kc

∴ a plot of rate vs. concentration of reactant will be a straight line passing through the origin and having the slope equal to K.

(B) Taking logarithm of both side 

-= log (a–x) + log k

So, a plot of logarithm of rate vs. logarithm of concentration will be a straight line of unit slope and with intercept equal to log k.

(C) The integrated rate expenssion may be rearranged as log(a–x) =

Thus, a plot of logarithm of concentration vs. time will be a straight line of slope equal to
– and intercept equal to logarithm of initial concentration of reactant.

Radioactivity as a First order Nuclear reaction:

(i)  The disintegration or decay constant (λ) of a radioactive element is expressed as 

–   = λN  (Differential rate expression)

λ =   log (Integrated rate expression)

Where No is the number of atoms present in the sample of radioactive element or its compound, N is the number of atoms remaining undecayed at any time t, and λ is a constant for a given radioactive element.

(ii) unit  of λ : Time –1

(iii) Half – life period (t1/2) = = a constant independent of initial  concentration 

(iv) Average life (tav) =

(v) t1/2 = 0.693 tav or tav = 1.44 t1/2

(vi) No. of halves (n) = , T = total time 

∴ t1/2 =

Amount left after n – halves  =

(vii) Unit of radioactivity : Curie denoted as Ci 

1 Ci = 3.7 × 1010 dps = activity of 1g Ra Disintegrations per second (dps) is also called Becquerel . dps  is the SI unit of radio activity.

(viii)  Carbon – dating 

C – 14 is a radioactive isotope of carbon and is a β – emitter  with half – life  of 5760 years.

6C14 ⎯→ 7N14 + –1β°

(unstable) (stable)

In living tissues the β activity due to C  – 14 content remains in equilibrium. The equilibrium value is nearly 14 dpm per gram of total carbon contents. However, when tissues are cut the activity goes on decreasing due to disintegration of C – 14 without at the same time being supplemented from the external source. By measuring the β – activity of any remains of living bodies like piece of wood,  fossils etc. recovered from archeological excavations, it is possible to determine how long before the living tissues of which the remains form a part died or cut. This process is called carbon – dating.

 

Energy of Activation 

(i)  ΔH = Activation energy of FR – Activation energy of BR= Ea1 – Ea2, FR =  Forward reaction

Ea1 < Ea2 for exothermic reaction, and hence ΔH = – ve     BR = Backward reaction 

Ea1 > Ea2 for endothermic reaction, and hence ΔH = +ve 

 

The minimum activation energy for an exothermic reaction is zero. The minimum activation energy for an endothermic reaction is ΔH.

Note that FR exothermic while BR is endothermic in the  above Fig.

 

(ii) Arrhenius equation in different forms:

I

II k =

III log k  =

IV log =

(iii) From II: T ⎯⎯→ ∞, k ⎯⎯→ A. Thus A, the pre exponential factor also called frequency factor and which is constant for a given reaction, is the rate constant of reaction at infinite temperature maximum rate constant.

 

From iii: A plot of log k vs. is a straight line of the slope equal to and intercept 

Form III: A plot  of  log K is   is a straight line of the slope equal to – and intercept equal to log A.

 

From IV: Ea can be calculated with the knowledge of rate constant of a reaction at two different temperature.

(v) A catalyst lowers the energy of activities of a reaction. If E′a  and be the energy of activation of uncatalysed and catalysed reaction, respectively, thus < Ea. If K and Kcat be the rate constants of reaction in absence and in presence of catalyst, respectively at temp. T this Kcat > k and 

=

(vi) The rate of reactions in the presence of catalyst at any temperature T1 may be made equal to the rate of the reaction in absence of the catalyst but for this salt we will have to raise the temp. to Ti i..e,

 

=

∴ =

SOLVED EXAMPLES

Example 1: From the following data show that the decomposition of hydrogen peroxide in aqueous solution is a first – order reaction. What is the value of the rate constant ?

Time in minutes	0	10	20	30	40
Volume V in ml	25.0	20.0	15.7	12.5	9.6

where V is the number of ml of potassium permanganate required to decompose a definite volume of hydrogen peroxide solution.

 

Solution: The equation for a first order reaction is

The volume of KMnO4 used, evidently corresponds to the undecomposed hydrogen peroxide.

Hence the volume of KMnO4 used, at zero time corresponds to the initial concentration a and the volume used after time t, corresponds to (a – x) at that time. Inserting these values in the above equation, we get

when t = 10 min. k1 = log = 0.022318 min-1 = 0.000372 s-1

when t = 20 min. k1 =

when t = 30 min. k1 =

when t = 40 min. k1 =

The constancy of k, shows that the decomposition of H2O2 in aqueous solution is a first order reaction.

The average value of the rate constant is 0.0003879 s-1.

 

Example 2: 5 ml of ethylacetate was added to a flask containing 100 ml of 0.1 N HCl placed in a thermostat maintained at 30°C. 5 ml of the reaction mixture was withdrawn at different intervals of time and after chilling, titrated against a standard alkali. The following data were obtained :

Time (minutes)	0	75	119	183	∞
Volume of alkali used in ml	9.62	12.10	1.10	14.75	21.05

Show that hydrolysis of ethyl acetate is a first order reaction.

 

Solution: The hydrolysis of ethyl acetate will be a first order reaction if the above data confirm to the equation.

Where V0, Vt and V∞ represent the volumes of alkali used at the commencement of the reaction, after time t and at the end of the reaction respectively, Hence

V∞ – V0 = 21.05 – 9.62 = 11.43

 

Time V∞ – Vt k1

75 min 21.05 – 12.10 = 8.95

119 min 21.05 – 13.10 = 7.95

183 min 21.05 – 14.75 = 6.30

A constant value of k shows that hydrolysis of ethyl acetate is a
first order reaction

 

Example 3: The optical rotations of sucrose in 0.5 N HCl at 35°C at various time intervals are given below. Show that the reaction is of first order :

Time (minutes)	0	10	20	30	40	∞
Rotation (degrees)	+32.4	+28.8	+25.5	+22.4	+19.6	-11.1

 

Solution: The inversion of sucrose will be first order reaction if the above data confirm to the equation ,

Where r0, rt and r∞ represent optical rotations initially, at the commencement of the reaction after time t and at the completion of the reaction respectively

In the case a0 = r0 – r∞ = +32.4 – (-11.1) = +43.5

The value of k at different times is calculated as follows :

Time rt rt – r∞ k

10 min +28.8 39.9 = 0.008625 min-1

20 min +25.5 36.6 = 0.008625 min-1

30 min +22.4 33.5 = 0.008694 min-1

40 min +19.6 30.7 = 0.008717 min1

The constancy of k1 indicates that the inversion of sucrose is a
first order reaction.

 

Example 4: The reaction given below, involving the gases is observed to be first order with rate constant 7.48 × 10−3 sec−1.Calculate the time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise to 0.145 atm and also find the total pressure after 100 sec.

2A(g) → 4B(g)   +  C(g)

 

Solution: 2A(g) → 4B(g)   + C(g)

initial Po     0   0

at time t Po − P′ 2P′ P′/2

  Ptotal = Po − P′ + 2P′ + P′/2

= Po +

P′ = (0.145 − 0.1)

= 0.03 atm

k =

t =

t = 47.7 sec

Also, k =

7.48 × 10−3 =

0.1 − P′ = 0.047

P′ = 0.053

Ptotal = 0.1 + (0.053)

≈ 0.180 atm.

 

     Example 5: For the reaction :

C2H5I + OH− → C2H5OH + I−

the rate constant was found to have a value of 5.03 × 10−2 mol−1 dm3 s−1 at 289 K and 6.71 mol−1 dm3 s−1 at 333 K. What is the rate constant at 305 K.

 

Solution: k1 = 5.03 × 10−2 mol−1 dm3 s−1 at T1 = 289 K

k2 = 6.71 mol−1 dm3 s−1 at T2 = 333 K

log =

On solving we get, Ea = 88.914 kJ

The rate constsnt at 305 K may be determined from the relation:

log

log

On solving we get, k2 = 0.35 mol−1 dm3 s−1

 

EXERCISE

 

1. At 380°C, the half life period for the first order decomposition of H2O2 is 360 min. The time required for 75% decomposition of 450°C is 20.40 min. Calculate Ea.

 

2. A small amount of solution containing Na – 24 radio nuclide with activity 2× 103 dps was injected in the blood stream of a man. The activity of 1cc of blood sample taken at the end of 5 hours turned out to be 16 dpm. If the half life of the radio nuclide is 15 hours, find the volume of the blood in the man’s body.

 

3. The half time of first order decomposition of nitramide is 2.1 hour at 15oC.

NH2NO2(aq) → N2O(g) + H2O(l)

If 6.2 g of NH2NO2 is allowed to decompose, calculate (i) time taken for NH2NO2 to decompose 99%; and (ii) volume of dry N2O produced at this point, measured at STP.

4. A drop of solution (volume 0.05 ml) contains 3.0 × 10-6 moles of H+ ions. If the rate constant of disappearance of H+ is 1.0 × 107 mole litre-1 sec-1. How long would it take for H+ ion in the drop to disappear ?

 

5. A sample of uraninite, a uranium containing mineral, was found on analysis to contain 0.214 g of Pb206 for every gram of uranium. Assuming that the lead all resulted from the radioactive disintegration of the uranium since the geological formation of the uraninite and that all isotopes of uranium other than 238U can be neglected, estimate the date when the mineral was formed in the earth’s crust. The half life of 238U is 4.5 × 109 y.
6. In nature a decay chain series starts with 90Th232 and finally terminates at 82Pb208 . A thorium ore sample was found to contain 8 × 10-5 ml of helium at STP and 5 × 10-7 g of Th232. Find the age of the ore sample assuming the source of helium to be only due to the decay of Th232. Also assume complete retention of helium within the ore. (Half life of
   Th232 = 1.39 × 1010Y)
7. The gas phase decomposition of dimethyl ether follows first order kinetics:

CH3—O—CH3(g) → CH4(g) + H2(g) + CO(g)

The reaction is carried out in a constant volume container at 500oC and has a half life of 14.5 minutes. Initially only dimethyl ether is present at a pressure of 0.40 atmosphere. What is the total pressure of the system after 12 minutes? Assume ideal gas behaviour.

8. The half – life for a first – order decomposition reaction at 650 K is 363 min and its energy of activation is 217.58 kJ mol-1.

(a) What fraction of the molecules at 650 K have sufficient energy to react ?

(b) Calculate the time required for the compound to be 75 % decomposed at 723 K.

9. A first order reaction, A → B, requires activation energy of 70 kJ mol-1. When a 20% solution of A was kept at 25oC for 20 minutes, 25% decomposition took place. What will be the percent decomposition in the same time in a 30% solution maintained at 40oC? Assume that activation energy remains constant in this range of temperature.

 

10. An optically active compound A is hydrolysed by dilute acid to give two optically compounds B and C according to the following chemical equation, A + H2O ⎯⎯→ 2B + C. The angle of rotation after 40 minutes was observed to be 26° and that after completion of reaction was 10° at 27°C. Find the half – time of the reaction assuming it to follow pseudo first order kinetics. The observed rotation per mole of A,B and C are 60°, 50° and  –80°. A plot of logarithm of rate constant of the above reaction vs T–1 give straight line with intercept equal 15.046 on log K axis. Calculate at what temperature half – time  of the reaction will be 31.1 min.        

 

ANSWERS

 

1. 200 kJ mol–1 2. 6 litre
2. (i) 13.96 hrs 4. 6 × 10–9 sec

(ii) 2.217 litre

5. 1.4 × 109 years 6. 4.89 × 109 years
6. 0.7488 atm 8. (a) 4 × 10–18 (b) 12.5 minute
7. 67% 10. (a) 124.4 minute (ii) 310.80 K

 

CHEMICAL EQUILIBRIUM

 

Reaction Quotient (Q) and Equilibrium Constant (Kc)

 

For the reaction:

 

aA + bB cC + dD

 

The reaction Quotient (Q) may be difined as

where square bracket terms denote active masses of the species enclosed.

 

Active mass is idealised or effective concentration and is less than molar concentration. The two are inter related as 

a = C × f

Where f = activity coefficient

a = activity or active mass, and

c = molar concentration

 

f < 1, so a < c. As solution is diluted more and more, the interionic or intermolecular interaction decreases and values of f increases. At infinite dilution when interionic or intermolecular interaction ceases to exist completely, f approaches unity and hence activity becomes equal to molar concentration. For very low concn of solution or gas, activity is taken to be equal to the molar concentration as an approximation. The active mass of a pure substance in solid or liquid state is taken to be unity.  Replacing activity terms by molar concentration terms in quotient expression, we get

At the point of commencement of the reaction i.e. zero time Cc = 0 = CD so Q = 0. As reaction proceeds some C and D are formed and some A and B are consumed so Q > 0. With the passage of time concentrations of A and B go on decreasing while those of C and D go on increasing, hence Q goes on increasing with time. At equilibrium, rate of forward reaction and that of backward reaction become equal. There is no net change in the concentration of any substance whether reactant or product. So, Q attains some maximum value which does not increase with time. This maximum value of Q is known as equilibrium constant of the reaction. Thus, at equilibrium.

…(1)

If Q < Kc, then reaction will proceed in the forward direction. If Q = Kc, the reaction will be in equilibrium. If Q > Kc, the reaction will occur in the backward direction.

 

For an irreversible reaction which goes to completion, Q increases with time and finally when the reaction is complete it becomes infinity. It is to be noted that no reaction is irreversible. Reversible reactions with very high values of their equilibrium constants are said to be irreversible. A reversible reaction giving one or more gaseous products may become irreversible if it is carried out in an open vessel.

 

The magnitude of equilibrium constant of a reaction measures the extent to which a reversible reaction will occur before attainment of equilibrium i.e. it measures the position of equilibrium from the initial point. Higher values Kc implies that the concentrations of products at equilibrium are high while those of reactants are low i.e., the equilibrium lies farther away from the initial point. Equilibrium constant is the ratio of rate constant of the forward reaction to the rate constant of the backward reaction. For gaseous equilibria, molar concn terms in Eq.1 may be replaced by respective partial pressure terms since pi ∝ Ci at a given P and T, so as to get the expression for Kp as shown below.

…(2)

(Provided that A, B, C and D are all gases)

 

If A, B and C are gases while D is a solid or liquid then for the reaction mentioned above, we have, since a solid or liquid represents standard state of unit activity.

Kc =

Kp =

It is to be noted that if any of the reactants or products be in the solution form, we cannot write Kp expression for such an equilibrium. An example of such reaction is as follows:

CO2(g) + H2O(l) 2H+ + (aq)

 

A general expression for Kp 

For the reaction

aA(g) + bB(g) cC(g) + dD(g)

 

where Δn = (c + d) – (a + b)

 

Relation between Kp and Kc

Kp = Kc (RT)Δn

Where Δn = Change in number of moles during the reaction

Or, Δn = (c + d) – (a + b), provided A, B, C and D are all gases

Unit of Kc: (mole L–1)Δn … from Eq. 1

Unit of Kp: (atm)Δn or (mm)Δn … from Eq. 2

 

Relation between ΔG and K

ΔG0 = – 2.303 RT log K

Where K = Kc or Kp

 

It is to be noted that ΔG0 calculated using Kc and Kp differ which is due to the difference in the standard states (1 mole L–1 at 25°C in the case of former while 1 atm at 25°C in the case of latter).

 

Le-Chatalier’s Principle

 

The principle which is the principle of dynamic equilibrium states that whenever chemical change in equilibrium is subjected to some constraints i.e., change in the factors like pressure, temperature and concentration, the equilibrium shifts itself in such a way so as to reduce the effect of the change as far as possible.

 

This principle helps to determine the favourable conditions for high yield of the product of a reversible reaction.

1. i) Effect of Pressure

If Δn = difference between number of moles of the gaseous products and number of moles of the gaseous reactants, then

1. a) The yield of the product of a gaseous equilibrium with Δn = –ve, increases with increase of pressure.

Ex: N2(g) + 3H2(g) 2NH3(g)

 

1. b) The yield of the product of a gaseous equilibrium increases by lowering the pressure if Δn for the reaction is +ve, 

Ex: PCl5 PCl3 + Cl2(g)

 

1. c) If Δn = 0, there is no effect of pressure on equilibrium.

Ex: N2(g) + O2(g) 2NO(g)

1. ii) Effect of Temperature
2. a) The yield of the product of an exothermic reaction (ΔH – ve) decreases with increase of temperature,

Ex: N2 + 3H2 2NH3, ΔH = –ve

1. b) The yield of the product of an endothermic reaction (ΔH +ve) increases with increase of temperature

N2 + O2 2NO(g), ΔH = +ve

The equilibrium constant of a reaction varies with temperature as per the following equation called Van’t Hoff equation.

or,

where and are the equilibrium constants of the reaction at temperatures T1 and T2, respectively.

 

iii) Effect of passing inert gas in the reaction system at constant volume in closed volume. No effect since partial pressure of each component remains unaltered.

 

1. iv) Effect of adding inert gas at constant pressure (increasing volume): Equilibrium shifts to the product side if the reaction is accompanied by increase in number of moles, i.e. Δn > Since in this case molar concentration or partial pressure of the reactant increases while that of product decreases.

 

1. v) Adding reactant in the reaction vessel or removing the product formed during the reaction: Equilibrium shifts towards products side and hence the yield of product increases.

 

CALCULATION OF DEGREE OF DISSOCAITON (α) AND DENSITY MEASUREMENT

 

The density of an ideal gas at constant T and P is inversely proportional to the number of moles for a given weight, so

, Final V.D. = V.D. of the equilibrium mixture

or =

PCl5(g) PCl3(g) + Cl2(g)

Initially 1 mole 0 0

At equilibrium (1 – α) moles α mole α mole

 

= = 1 + α

Thus, α can be calculated by V.D. or M.W. ( MW = 2 × V.D.) data and knowing α, Kc or Kp can well be calculated.

 

If dissociation be supposed to occurs as

A2B4 2AB(g) + B2(g)

Initially 1 mole 0 0

At equilibrium (1-α) mole 2α α

 

= 1 + 2α

 

EXERCISE

1. An equilibrium mixture at 300 K contains N2O4 and NO2 0.28 and 1.1 atm respectively.
   If the volume of container is doubled, calculate the new equilibrium pressures of the two
   gases.
2. Calculate the temperature at which equilibrium constant is 1.0 × 106, given equilibrium constant is 9.1 × 102 at 800 K and ΔH is –1.8 × 105 J mol–1.

 

3. The degree of dissociation of HI at a particular temperature is 0.8. Calculate the volume
   of 2M sodium thiosulphate required to react with iodine present at equilibrium when 1.35
   mol each of H2 and I2 are heated at this temperature in a closed vessel of 2 L capacity.
4. At 46°C, KP for the reaction N2O4(g) 2NO2(g) is 0.66. Compute the percent dissociation of N2O4 at 46°C and a total pressure of 0.5 atm. Also calculate partial pressures of N2O4 and NO2 at equilibrium.

 

5. Sulphide ions in alkaline solution react with solid sulphur to form polyvalent sulphide
   ions. The equilibrium constant for the formation of  S2-2 and S3-2 from S and S2- ions
   are 1.7 and 5.3 respectively. Calculate equilibrium constant Kc for the formation of S3-2
   from S22- and S.
6. At 627°C and 1 atm., SO3 is partly dissociated into SO2 and O2. The density of the
   equilibrium mixture is 0.925 g/l. Calculate Kp for the reaction : SO3 SO2 + ½O2.

 

7. 0.15 mole of CO taken in a 5L flask is maintained at 600 K along with a catalyst, so that
   the following reaction takes place.
   CO(g) + 2H2(g) CH3OH(g)
   Hydrogen is introduced until the total pressure of the system is 5.91 atmosphere at
   equilibrium and 0.1 mole of methanol is formed. Calculate
   (i) Kp and Kc
   (ii) the final pressure if the same amount of CO and H2 as before are used, but with no
   catalyst, so that the reaction does not take place.
8. The reaction 2A(g) + B(g) C(g) was allowed to come to equilibrium. The initial amounts of the reactants present in a 1.80 L vessels were 1.18 mole A and 0.78 mol B. The amount of A at equilibrium was found to be 0.92 mol. What is the value of KC for this reaction.

                                                                

9. A mixture of N2 and H2 in the molar ratio 1:3 at 50 atm and 650°C is allowed to react till equilibrium is obtained. The NH3 present at equilibrium is 25% by weight, calculate KP for N2(g) + 3H2(g) 2NH3(g).

 

10. At 300 K, the equilibrium constant Kp for the reaction N2O4(g) 2NO2(g) is 0.174. What would be the average molar mass of an equilibrium mixture of N2O4 and NO2 formed by the dissociation of pure N2O4 at a total pressure of 1 atm at this temperature?

 

ANSWERS

 

1. 0.64 atm,   0.0953 atm 2. 640 K
   3. 1080 ml 4. 50%, 0.167 and 0.333 atm
2. 3.11 6.   0.187
3. Kc = 246.9, Kp = 0.102, 8. 0.70 mol2 L–2, 0.977 atm P = 7.872 atm
       9. 1.434 × 10–4 atm–2 10. 76.5  g mol-1

 

IONIC EQUILIBRIA

 

pH of a weak monobasic acid solution

 

HA  H+ + A–

C 0   0 … initially (mole L–1)

C(1–α) Cα Cα … At equilibrium (mole L–1)

Ka = = α2C ( α <<1)

(For this approximation to apply α ≤ 0.1)

[H+] = Cα = =

pH = –log (Ka⋅C)1/2 = – [log Ka + logc]

= [pKa – logC]

pH + pOH = pKw

 

pOH of a weak mono acidic base solution

BOH B+ + OH–

C(1 – α) Cα Cα … At equilibrium

Kb = ≈  α2C

[OH–] = Cα =

∴ pOH = [pKb – logC]

 

pH of a mixture of two weak mono basic acids

HA H+ + A– HB   H+        +  B–

C1 (1 – α1) (C1α1 + C2 α2) C1α1 C2(1 – α2) (C1α1 + C2α2) C2α2

Using same approximations, α1 and α2 can be calculated and hence H+ ion concentration of the solution also.

 

pH of a solution of strong acid/strong base

Strong acid remains completely ionised in solution (α = 1) and hence.

[H+] = molarity of solution × basicity of acid

= Normality of solution.

Similarly, [OH–] = Molarity of the solution × acidity of strong base

= Normality of solution

 

However, if the solution of acid or base be very dilute (Normality ≤ 10–7), it is imperative to add H+ ion concentration of water to that of the acid itself after considering common ion effect of acid or base on the ionization of water.  

 

Illustration: To calculate pH of 10–8 M HCl solution.

We must proceed as follows:

H2O     H+ + OH–

X + 10–8 x

x(x + 10–8) = 10–14

Upon solving, x = 95 × 10–8

So [H+] = 9.5 + 10–8 + 10–8 = 10.5 × 10–8

pH = 6.9

 

BUFFER MIXTURE

 

Two types:

1. i) Acid Buffer, e.g., CH3COOH + CH3COONa
2. ii) Base Buffer, e.g., NH4OH + NH4Cl

 

Henderson’s equation for calculating pH of buffer mixture

pH = pKa + … For acid buffer

pH = pKb + … For base buffer

 

Example-1: 20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid to give 70 mL of solution. What is the pH of the solution? Calculate the addition  volume of 0.2M NaOH required to make pH of the solution 4.74. (Ka for CH3COOH
= 1.8 × 10–5).

 

Solution: Added NaOH reacts with acetic acid to from equimolecular amount of CH3COONa, salt.

Concentration of acetic acid after addition of NaOH =

Concentration of sodium acetate solution formed =

pKa = – log Ka = – log (1.8 × 10–5) = 4.745

∴ pH = pKa + = 4.745 +|  = 4.568

Alternatively: Millimoles of NaOH = 0.2 × 20 = 4, Millimoles of
CH3COOH = 0.2 × 50 = 10 

Millimoles of salt formed = 4, Millimoles of acid left = 10 – 4 = 6

 

Buffer Capacity

Amount of acid or base added to buffer solution to charge its pH by unit. It is the capacity of a buffer solution to resist the change in pH on addition of acid or alkali.

Buffer capacity (φ) =

Where b is no.of mole(s) of acid or base added to one litre solution and d(pH) is change in pH. Buffer solution registers maximum buffer capacity when [acid] or [base] = [salt]

 

pKa and pKb for a conjugate acid – base pair

HA        H+ + A–

Acid (weak) base conjugate (strong)

A– + H2O HA + OH–

Ka ⋅ Kb = = [H+] [OH–] = Kw

So, pKa + pKb = pKw = 14 (at 25°C)

For example, pKa of CH3COOH is 4.75 so pKb of CH3COO– ion will be 14 – 4.75 = 9.25.

 

SALT HYDROLYSIS

 

1. i) Salt of weak acid with a strong base like CH3COONa, Na2CO3

It is the anion of the salt that interacts partially with solvent water to furnish tree OH– ion in solution thereby rendering the solution faintly alkaline.

A– + H2O + OH–

Initially: C mole L–1, constant 0 0

At equilibrium: C (1 – h), constant Ch Ch

KH = ∴

Other ionic equilibria existing side by side in this solution:

HA H+ + A–, Ka =

H2O H+ + OH–, Kw = [H+] [OH–]

KH = , h = , [OH–] = Ch =

∴ pOH = – log[OH–] = – log

or pOH = [pKw – pKa – logC]

Using this equation pOH may be calculated.

 

1. ii) Salt of strong acid with a weak base like NH4Cl, FeCl3

Hydrolysis if this salt means partial interaction of cation of the salt to furnish free H+ ion in solution thereby rendering the solution faintly acidic.

B+ + H2O + H+

C(l-h) constant Ch Ch … At equilibrium

, KH =

[H+] = Ch =

pH = [pKw – pKb – logC]

With the help of this equation, pH can be calculated

 

iii) Salt of Weak acid with a weak base like CH3COONH4

Both cation and anion of the salt interacts with solvent water to furnish free H+ and OH– ions in solution which combine to form freshly ionised water molecule. The resulting solution may be either faintly acidic, alkaline or neutral depending on [H+] > [OH–], [H+ ≤ [OH–] or [H+] = [OH–] i.e., Ka > Kb, Ka < Kb or Ka = Kb

CH3COO– + + H2O CH3COOH + NH4OH

At equilibrium: C(l-h) C(l-h) Ch Ch

KH = ≈ h2

Other ionic equilibria existing here are:

CH3COOH CH3COO– + H+, Ka =

NH4OH + OH– , Kb =

H2O H+ + OH–, Kw = [H+] [OH–]

KH = , ∴

From acetic acid dissociation equilibrium

[H+] = =

∴ pH = –log [H+] = [pKw + pKa – pKb]

pH of a solution of the salt of a weak acid with a weak base is independent of the concentration of salt solution. Knowing Kw, Ka and Kb, pH can be calculated.

When Ka = Kb, pH = = 7

When Ka > Kb, pH < 7

When Ka > Kb, pH > 7

 

Example-2: Calculate pH of 0.5 M solution of ammonium acetate at 25°C. Given

Ka for = 5.88 × 10–10

Kb for CH3COO– = 5.55 × 10–10

Kw = 1 × 10–14

 

Solution: Since, Ka ⋅ Kb = Kw

So, Ka for CH3COOOH = = 1.8 × 10–5

Kb for NH4OH = = 1.7 × 10–5

pKw = 14

pKa = –log (1.8 × 10–5) = 4.74

pKb = –log (1.7 × 10–5) = 4.76

pH = [14 + 4.74 – 4.76] = 6.99

 

SOLUBILITY PRODUCT

 

In a saturated solution of AxBy (sparingly soluble salt), following equilibrium exists.

 

AxBy(s) xA+ + yB–

Ksp = [A+]x [B–]y, Ksp = Solubility product of AxBy

Thus, for satd solution of AgCl

Ksp = [Ag+] [Cl–]

In satd solution of Ag2SO4

Ksp = [Ag+]2 []

 

Relation between solubility(s) and solubility product Ksp

 

Let S be the solubility of AxBy in mole L–1. If AxBy be sparingly soluble, whole of AxBy passing into solution will  remain completely ionised so.

[A+]x [B–]y = Ksp

(xs)x (ys)y = Ksp

or xx⋅yy⋅sx+y = Ksp

or S =

 

Solubility Product Principle

 

For Unsaturated solution: Ionic product < Ksp

Saturated solution: Ionic product = Ksp

Super saturated solution: Ionic product > Ksp

Ionic product means [A+]x [B–]y

 

Since super saturation is the condition of precipitation so for precipitation of any salt, the ionic product must exceed the solubility product.

 

Example-3: Ksp BaSO4 is 1.5 × 10–9. Calculate its solubility in (a) pure water and (b) 0.10 M Na2SO4 assuming complete ionisation of 0.10 M Na2SO4 solution.

 

Solution: Let the solubility of BaSO4 in 0.1 M

Na2SO4 sbe mole L–1

Thus, in pure water

[Ba++] = S g – ion [L–1 = []

So, [Ba++] = Ksp

S2 = Ksp

S = = = 3.87 × 10–5 mole L–1

In 0.1 M Na2SO4 solution

[Ba++] = S gram-ion L–1

= s + 0.1 ( Na2SO4 → )

S × 0.1 = 1.5 × 10–9

S = 1.5 × 10–8 mole L–1

 

Solubility of a salt when Acid – Base Reaction occur simultaneously or the anion or cation of the salt gets hydrlysed.

 

Let us consider CH3COOAg, a salt of weak acid with a strong base. The solubility of CH3COOAg be S mole L–1

CH3COOAg(S) ⎯→ CH3COO– + Ag+

s s

The acetate ion will get hydrolysed as

CH3COO– + H2O CH3COOH + OH–

Initially: S 0 0

At equilibrium: S (l – h) Sh Sh

S – x x x

(Where Sh = x)

 

So, [CH3COO–] [Ag+] = Ksp

 

or (S – x) S = Ksp

KH = , or

Knowing Ksp, Kw and Ka, S and x can be calculated.

 

Solubility of a salt in a Buffer

Solubility of a solute in acidic or basic buffer can be determined if value of Ksp is given. For example solubility of Mg(OH)2 basic buffer of pH = 12 i.e. [OH–] = 10–2g – ion L–1, is S mole L–1 (say)

Ksp = [Mg2+] [OH–]2 = S × (10–2)2

∴ S = 104 Ksp

 

Example-4: Calculate solubility of CaF2(s) in a buffer solution with pH = 3.00

Given: Ksp (CaF2) = 5.3 × 10–9 

Ka(HF) = 6.7 × 10–4

Solution: CaF2(s) Ca2+ + 2F–, K1 = Ksp

2H+ + 2F– 2HF, K2 =

––––––––––––––––––––––––––––––––

CaF2(S) + 2H+        Ca2+ + 2HF

Initial Concn.: 10–3 mole L–1   0 0

At equilibrium: 10–3 mole L–1   x        2x mole L–1

K =

or =

or x = 1.4 × 10–3 mole L–1

 

Theory of acid-base indicator

An acid-base indicator is a weak organic acid or base which possesses different colour in its ionised and unionised form due to change in structure from benzenoid to quinonoid upon ionisation in solution.

 

H+ +

= Kin

or [H+] =

or pH = pKin – log

Let for colour A to predominate: [HIn] ≥ 10 [In–] then pH ≤ pKin + 1. Similarly, for collur B to prevail pH ≥ pKin + 1Thus, an indicator will change its colour when pH of solution changes abruptly from pKin – 1 to pKin + 1, called indicator range or pH – range of the indicator (Kin = indicator constant, a constant at constant temperature)

 

EXERCISE 

 

1. What is pH of 0.1M H2S solution. Given that;

Ka1 (H2S) = 1.0 × 10–7

Ka2 (H2S) = 1.3 × 10–14 

 

2. 3ml of 0.1M NaOH solution is added into 100 ml of 0.1M CH3COOH solution. What is the pH of resulting solution? What volume of 0.1M NaOH solution be further added so that pH of resulting solution be 4.74 (Given than Ka(CH3COOH) = 1.8 × 10–5

 

3. A 50 ml aliquot of 0.01 M solution of HCOOH was titrated with  0.1M NaOH. Predict the pH of solution

(A) at the beginning of the reaction 

(B) at the half  equivalence point 

(C) at the equivalence point

(D) after 10 ml of the base has been added 

 

4. Find the simultaneous solubility of AgCN and AgBr if their solubility products are 1.0 × 10–12 and 5 × 10–13 respectively.

 

5. Calculate the conc. of fluoroacetic acid which is required to get [H+] = 1.5 × 10–3. Ka (acid) = 2.6 × 10–3.

 

6. Determine the pOH of a solution after 0.1  moles of NaOH is added to 1L of a solution containing 0.15 M CH3COOH and 0.2M CH3COONa. Assume no change in volume.

 

7. Bromo phenol blue is an indicator with Ka 5.84 × 10–5 what percentage of this indicator is in its basic form at pH = 4.84.

 

8. What is [H+] in a 0.0060M H2SO4 solution? The first ionisation of H2SO4 is complete and the second ionisation has the K2 of 1.02 × 10–2. What is in the same solution.?

 

9. The  following solution were mixed : 500 ml of 0.01M AgNO3 and 500 ml. of solution that was both 0.01M in Nacl and 0.01M NaBr. Calculate [Ag+], [Cl–] and [Br–]

Ksp(AgCl) = 1.0 × 10–10

Ksp (AgBr) = 5 × 10–13

 

10. A weak base BOH was titrated against a strong acid. The pH at 1/4th equivalence point was 9.24. Enough strong base (6 m.eq.) was now added to completely convert the salt. The total volume was 50 ml. Find the pH at this point 

 

ANSWERS 

 

1. 6.8617 2. 4.37, 20 ml
2. 2.88, 3.75, 7.86, 11.92 4. 4.1 × 10–7, 8.16 × 10–7 M
3. 2.37 × 10–3 M 6. 8.5
4. 80% 8. 0.0092
5. [Cl–] = 0.005 M 10. 11.2

[Ag+] = 2× 10–8 M

[Br–] = 2.5 × 10–5 M