Chemical Kinetics PDF

# CHAPTER-4 ## CHEMICAL KINETICS Date-July-25 → It is a branch of physical chemistry which deals with the study of rates of chemical reactions. ## Types of chemical reaction:- On the basics of rates, there are 3 types of chemical reactions, such as: - Instantaneous Reactions.. - Moderate Reactions...

# CHAPTER-4 ## CHEMICAL KINETICS Date-July-25 → It is a branch of physical chemistry which deals with the study of rates of chemical reactions. ## Types of chemical reaction:- On the basics of rates, there are 3 types of chemical reactions, such as: - Instantaneous Reactions.. - Moderate Reactions. - Slow Reactions. ## Instantaneous Reaction → These are the reactions which are at once. _**eg:**_ AgNO₃ + NaCl -> AgCl + NaNO₃ (Silver nitrate) (Sodium chloride) (white ppt) (Sodium nitrate) (of silver chloride) ## Moderate Reaction → Moderate Reactions are the reactions which are completed at measurable rates, or → These reactions are neither be slow nor be fast. _**eg:**_ CH₃COOC₂H₅ + NaOH -> CH₃COONa + C₂H₅OH (ethyl acetate) (sodium hydroxide) (sodium acetate) (ethanol) or (alkals) ## Slow Reactions → These are the reactions which take a long time to complete. _**eg:**_ 3Fe + 4H₂O -> Fe₃O₄ + 4H₂ (Iron) (water) (Iron oxide) (hydrogen) ## Rate of Reaction:- → The quantity of a reactant species consumed or the quantity of a product species formed in unit time in a chemical reactions. Consider a General reaction A -> B r = -∆[A]/∆t = +∆[B]/∆t _**eg:**_ CH₃COOH + CH₃OH -> CH₃COOCH₃ + H₂O (actic acid) (methanol) (methyl acetate) (water) r = -∆[CH₃COOH]/∆t = -∆[CH₃OH]/∆t or r = +∆[CH₃COOCH₃]/∆t = +∆[H₂O]/∆t ## Significance of -ve sign- - As the reaction proceeds the molecule concentration of the reactant decrease.. - Thus the change in concentration of ∆[A]Final - ∆[B]Initial, will have a -ve value. - The rate of reaction depends upon the concentration of reactant species. - Hence the rate of reaction: r = -∆[A]/∆t = +∆[B]/∆t ## Units of Rate of reaction:- We know, concentration = mol/Liter but r = Concentration/time = mol/l/sec = mol.l⁻¹.sec⁻¹ or mol.l⁻¹.s⁻¹ ## Mathmetical Expression for Rate of Reaction There are 2 sets of chemical reactions are considered. 1. Reaction involving same stoichiometric co-efficient of all reactants and products. _**eg:**_ NO₂(g) + CO(g) -> NO(g) + CO₂(g) (nitrogen dioxide) (carbon monoxide) (carbon dioxide) r = - ∆[NO₂]/∆t = - ∆[CO]/∆t or r = + ∆[NO]/∆t = + ∆[CO₂]/∆t 2. Reaction involving different stoichiometric co-efficient of reactants and products. _**eg:**_ H₂(g) + I₂(g) -> 2HI(g) (hydrogen) (iodine) (hydrogen iodide) r = - ∆[H₂]/∆t = - ∆[I₂]/∆t or r = + 1/2 ∆[HI]/∆t Consider a general reaction aA + bB -> cC + dD r= -1/a ∆[A]/∆t = -1/b ∆[B]/∆t or r = +1/c ∆[C]/∆t = +1/d ∆[D]/∆t ## Numericals - 01. Given the Reaction C₂ + 2I -> 2CI + I₂ The initial conc. of Iodide ions was 0.62mol/l and conc. after 12 minutes. was 0.56 mol/l. Calculate the rate of disappearance of I⁻ and rate of appearance of Iodine. *Ans:* Given [I⁻]initial = 0.62 mol/l [I⁻]final = 0.56 mol/l ∆t = 12 minutes ∆[I⁻] = [I⁻]Final - [I⁻]initial = 0.56 mol/l - 0.62 mol/l = -0.06 mol/l Rate of disappearance of I⁻ ions r = -∆[I⁻]/∆t = -(-0.06) mol/l/12 min = 0.06 mol/l/12 min * 60 s / 1 min = 0.06 mol/l/200 s = 0.005 mol/l.sec⁻¹ Rate of appearance of I₂ ie: ∆[I₂]/∆t = ∆[I⁻]/∆t = +1/2 ∆[I⁻]/∆t = 1/2 * +0.005 = 0.0025 mol/l.sec⁻¹ ## Numerical - 2 In a Reaction 4NH₃(g) + 5O₂(g) -> 4NO(g) + 6H₂O(g) the rate of formation of NO is 3.6 x 10⁻³ mol/l.s⁻¹ Find the rate of disappearance of oxygen. _**Ans:**_ Given: Rate of appearance of NO = 3.6 x 10⁻³ mol/l.s⁻¹ Rate of disappearance of O₂ = ? r = 1/4 ∆[NO]/∆t = +1/5 ∆[O₂]/∆t Rate of disappearance of O₂ = 5/4 x Rate of appearance of NO = 5/4 x 3.6 x 10⁻³ mol/l.s⁻¹ = 14.5 x 10⁻³ mol/l.s⁻¹ ## Average Rate:- (Expected value) (<r>) → It is defined as the change in concentration of a reactant or product of a chemical reaction per unit time. Considere a general reaction A -> B Average rate = change of conc. in a reactant/time taken = -∆[A]/∆t = [A₂-A₁]/t₂-t₁ where A₁ and A₂ are the conc. of a reactant at time t₁ and t₂ respectively. ## Instantaneous Rate:- The instantaneous rate of reaction is defined as the rate of change of conc. of any one of the reactant or the product species at particular time. for a rea A -> B instantaneous rate of reaction = -d[A]/dt = +d[B]/dt ## Date-30-07-2020 ## Factors Influencing the rate of reaction:- The rate of reaction depends upon the following factors. - Concentration of reactants - Temperature - Nature of reactants - Catalyst - Radiation ## Concentration of reactants:- → As a chemical reaction progresses, the concentration of the reactants decrease and those of the product increases. → When the concentration of reactant decreases the rate of reaction is also decreases. → When the conc of reactants are high then the rate of reactants should increases. ## Nature of Reactants:- → It has been observed that if the conc. and temperature of the reactants are constant then the reaction rates varies from reaction to reaction. _**eg:**_ - NO + 1/2 O₂(25°) -> NO₂ (instantaneous reaction) - CO + 1/2 O₂(25°) -> CO₂ (slow reaction) → The rate of reactant is fundamentally determined by its inherent instentati constitution and the strength of the chemical bonds contained in it.. ## Temperature:- → Temperature influence the rate of reaction to a great extent. _**eg:**_ N₂ + 3H₂ -> 2NH₃ (∆) ## Catalyst: → Catalyst is defined as a substance whose presence increased the rate of chemical reaction, but itself, doesn't undergo any change in the chemical composition at the end of the reaction. or a catalyst is a substance that speeds up a chemical reaction, but it is not consumed by the reaction. _**eg:**_ 2H₂ + O₂ -> 2H₂O ## Radiation:- → It undergoes all chemical reactions are photo-chemical reactions. _**eg:**_ - H₂ + Cl₂ -> 2HCl - H₂ + I₂ -> 2HI ## Rate Constant:- → A co-efficient of proportionality relating the rate of a chemical reaction at a given temperature to the conc. of reactant. → The rate of reaction depends upon the molar concentration of reactant species.. Considere a general reaction: aA + bB -> cC + dD r = k[A]ᵃ[B]ᵇ where 'k' is a Rate constant & [A] = [B] are the molar conc. of reactants. ## Characteristic of Rate constant:- → Different reactions have different values of rate constant at a given temperature. → The value of 'k' doesn't depends upon the concentration of reactant species.. → The unit of rate constant depends upon the order of reaction. → The unit of rate constant is s⁻¹ on time⁻¹ → Thus, the rate constant of a particular reaction is defined as the rate of reaction when molecular concentration of each reactant is unity. That's why the rate constant also called specific-rate constant. ## Rate Law Date-01-08-2020 → The rate law is experimentally determined and can be used to predict the relationship between the rate of reaction and the concentration of reactant. → Rate law gives the actual observed rate of reaction. → Consider a general reaction. aA + bB -> cC + dD According to the law of mass action r α [A]ᵃ[B]ᵇ or r = K[A]ᵃ[B]ᵇ where 'K' is Rate constant and a b are the stoichiometric co-efficient. _**eg:**_ 2NO(g) + O₂(g) -> 2NO₂(g) r α [NO]²[O₂] or r = k[NO]²[O₂] ## Molecularity of a reaction → Molecularity of a reaction is defined as the no. of molecules, atom or ions which called simultaneously to bring about a chemical reaction. ## Molecularity of reaction - Unimolecular - Bimolecular - Trimolecular ## Unimolecular Reaction → It’s a reaction involves a single reacting species are called unimolecular reaction _**eg:**_ N₂O₅ → N₂O₄ + 1/2 O₂ (Nitrogen pentoxide) (nitrogen tetroxide) (oxygen) ## Bimolecular Reaction → Bimolecular Reaction are involves two reactant species. _**eg:**_ CH₃COOC₂H₅ + NaOH -> CH₃COONa + C₂H₅OH (ethyl acetate) (alkali) (so acetate) (ethanol) ## Trimolecular Reaction:- → It contains 3 reactant species. _**eg:**_ 2SO₂ + O₂ -> 2SO₃ (sulphers dioxide) (oxygen) (sulphur trioxide) ## Characteristics of molecularity:- - Molecularity of a reaction, a single step of a complex reaction, which involves a no. of steps for completion - Molecularity is a whole no. and it is never in fraction or zero - The molecularity of a reaction doesn’t exceed three. ## Order of a Reaction:- → The order of a reaction is defined as the sum of the poweres to which the concentration terms of reactants for a given chemical reaction. _**eg:**_ mA + nB -> Product r α [A]m[B]n Order = m+n → Order of reaction may be - 1st order - 2nd order - fractional order - zero order ## 1st Order Reaction:- → These are the reactions only one reactant species involves a chemical reactions. _**eg:**_ P₂O₅ -> P₂O₄ + 1/2 O₂ (Phosphorous pentoxide) (Phosphorous tetroxide) (oxygen) r α [P₂O₅]¹ So, it is a 1st order reaction. ## 2nd Order Reaction:- → These are the reaction which involves two reactant molecule. _**eg:**_ CH₃COOC₂H₅ + NaOH -> CH₃COONa + C₂H₅OH (ethyl acetate) (sodium hydroxide) (so-d acetate) (ethanol) r α [CH₃COOC₂H₅]¹[NaOH]¹ Order = 1+1 =2 So it is a 2nd order reaction. ## 3rd order Reaction:- → Thus reaction contains 3 reactant species. _**eg:**_ 2A +B -> Product r α [A]²[B]¹ Order = 2+1 =3 Another example of 3rd order reaction:- 2NO + O₂ -> 2NO₂ r α [NO]²[O₂]¹ Order = 2+1 = 3 So, It is a 3rd order reaction. ## Zero Order Reactions:- → In these reactions the conc. of the reactant do not influence the rate of reaction because they don’t change appreciably during the reactions.. → All zero order reaction are photochemical reaction. _**eg:**_ - H₂+ Cl₂ -> 2HCI - H₂ + I₂ -> 2HI - Na + 3H₂ -> 2NH₃ ## Integrated Rate equation:- The concentration depence of rate is called differential rate equation. → We integrate the differential rate equation to get relation between directly molar conc of different times and rate constant.. ## Integreated Rate expression for Zero order reaction:- → In zero order reaction, the rate of reaction is proportional to zero power of the concentration of reactants. → Consider a zero order reaction. A -> product r = -d[A]/dt α [A]⁰ or -d[A]/dt = k[A]⁰ or -d[A]/dt = k[A]⁰ * [A]⁰=1 where 'k' is a Rate constant. → -d[A]/dt = k.1 → -d[A]/dt = k → -d[A] = k.dt → d[A] = -kdt (1) Both side integrating in eq(1) ∫d[A] = -k∫dt or ∫d[A] = -k∫dt = [A] = -kt + C (2) where 'c' is a integration constant. when, t = 0, the concentration of reactant [A] = [A]₀ Now, Put this value in eq(2) [A] = -kt + C [A]₀ = -k0 + C [A]₀= 0 + C [A]₀ = C C = [A]₀ (3) where [A]₀ is the initial conc. of reactant A. Now, the ‘c’ value putting in eq(2) [A] = -kt + C [A] = -kt + [A]₀ [A] = [A]₀ -kt k = [A]₀ - [A]/t This is the rate expression for zero order reactions. ## Half-like period (t₁/₂): → Half-like period as the time in which half of the substance has reacted. When [A] = [A]₀/2 t = t₁/₂ Now, substituting this value in rate expression for zero order reaction. k = [A]₀ - [A]/t k = [A]₀ - [A]₀/2 / t₁/₂ k = [A]₀/2/t₁/₂ k = [A]₀/2t₁/₂ k = [A]₀ / 2t₁/₂ t₁/₂ = [A]₀ / 2k This is the half-life period zero order reactions ## Integrated Rate expression for the 1st order reaction In the 1st order reaction, the rate of reaction is directly proportional to the molar concentration of single reacting substance. Consider a 1st order reaction: A -> B + C r = -d[A]/dt α [A] or r = -d[A]/dt = k₁[A] -d[A]/dt = k₁[A] where ‘k₁' is a rate constant. -d[A]/[A] = k₁dt (1) Let the initial concentration of A be ‘a’ molecules at the initial time ‘n’ moles of A decomposes to give ‘n’ moles of B and ‘n’ moles of C. Thus molar conc. of A after time ‘t ‘ is (a-n) . Thus the rate of formation of B and C is dn/dt. r α (a-n) or d(a-n)/dt = k₁(a-n) dn/dt = k₁(a-n) dn/(a-n) = k₁dt (2) Both side integration in eq(2) ∫dn/(a-n) = k₁∫dt ∫(a-n)⁻¹dn = k₁∫dt -ln(a-n) = k₁t + C (3) when t = 0 , n = 0 Now, putting this value in eq(3) -ln(a-n) = k₁t + C -ln(a-0) = k₁.0 + C -lna = C C = -lna (4) Now, substituting the ‘c’ value in eq(3) -ln(a-x) = k₁t + C -ln(a-x) = k₁t - lna -ln(a-x) = k₁t + ln a = 0 -k₁t = [-lna + ln( a-n)] -k₁t = [-lna - In(a-n)] k₁t = lna – ln(a-n) k₁t = ln [a/(a-n)] k₁ = 1/t ln [a/(a-n)] then put ln a/a-n = 2.303log a/a-n k₁ = 1/t * 2.303log a/a-n k₁ = 2.303/t log a/a-n So, this is the integrated rate expression for 1st order reactions. ## Half life period:- → Half-life period is the time during which the concentration of reactant falls down to half of its initial value. According to 1st order rate eq: k₁ = 2.303/t ln a/a-n → t = 2.303/k₁ log a/a-n According to definition: x = a/2 , t = t₁/₂ Now putting this value in eq (2) k₁ = 2.303/t log a/a-n k₁ = 2.303/t₁/₂ log a/a-x k₁ = 2.303/t₁/₂ log a/a/2 k₁ = 2.303/t₁/₂ log 2a/a k₁ = 2.303/t₁/₂ log 2 t₁/₂ = 2.303/k₁ log 2 t₁/₂ = 0.693/k (Heart log 2 = 0.3010) t₁/₂ = 0.693/k t₁/₂ = 0.693/k ## Date-August-6/2020 ## Numericals on the based of Rate expression for 1st order reaction and Half-lite Periods ## Numericals - 1 A 1st order reaction has rate constant of 1.15 x 10⁻³ s⁻¹. How long will 5g, this reactant take to reduce to 3g. _**Ans:**_ Given: a = 5g (a-x) = 3g rate constant = k = 1.15 x 10⁻³ s⁻¹ t = ? We know that expression for the 1st order reaction is: t= 2.303/k log a/a-x t = 2.303/1.15 x 10⁻³ s⁻¹ * log 5/3 t = 2.303/1.15 x 10⁻³ s⁻¹ * (log 5 - log 3) t = 2.303/1.15 x 10⁻³ s⁻¹ * (0.6990 - 0.4771) t = 1.15 x 10⁻³ s⁻¹ * 0.2219 t = 4445 s ## Numericals No-2:- A 1st order reaction takes 40 minutes for 30% decomposition. Calculate the it’s half-life period _**Ans:**_ Given: a = 100% (a-n) = (100-30)% = 70% t = 40 minutes t₁/₂ = ? We know: t = 2.303/k log a/a-x k = 2.303/t log a/a-n k = 2.303/40 minutes log 100/70 k = 2.303/40 minutes * log10- log7 k = 2.303/40 minutes *(1- 0.8451) k = 0.00892 min⁻¹ Hence, Half-life period is: t₁/₂ = 0.693/k = 0.693/(0.00892 min⁻¹) = 77.69 min t₁/₂ = 77.69 min ## Collision Theory of Reaction Rate According to collision theory. → Reactants are made up of molecules. → Molecules are always in a state of Random motion and hence go on colliding with one another. → Collision frequency (z) is the no. of intermolecular collision taking place per unit volume per second at a given temperature. → A Chemical reaction takes place due to inter-molecular collision of reactants. → All intermolecular collisions do not bring about the reaction, only effective collisions bring about the reaction. → The collisions which are bringing about a chemical reaction are called effective collision. → An effective collision is that in which the colliding molecules. → The minimum energy that is required to bring about a chemical reaction is called Threshold energy. → It is properly oriented. → Thus, for an effective collision, there are two barriers are there. i.e (a) Energy barrier (b) Orientation barrier → Applying for collision theory to bimolecule gases reactions, the expression for the rate constant is written as: k = P.Ze⁻Ea/RT where k = Rate constant P = Preobability or steric factor Z = Pre-exponential factor e⁻Ea/RT = fraction of colliding molecules which have the necessary activation energy. → It is important to note that greater the threshold energy, lessore will be rate of reactions. → Because only a small fraction of the molecules possess enough energy and react. → On the other hand, if threshold energy is small, than larger no. of molecules can bring about effective collisions. → The no. of collisions per unit volume per second between the reactant molecules the fraction of effective collisions. → Rate of reaction fαZ f α e⁻Ea/RT →K = P.Ze⁻Ea/RT ## Failure of Collision Theory: → For some reaction, the calculated and the experimental value of ‘k’ differ widely. → For such molecules the eq. become K = P.Ze⁻Ea/RT → It bacess to explains the abnormally high values of rate constant for some reactions. → It bacess to explain the reaction rate of irreversible reactions. → It is not possible to predict the correct orientation before the reactant molecules collide effectively. ## Activation energy :- (Ea) → Activation energy is the difference bet. the threshold energy (Eth) and average energy (Er) possessed by reacting molecule. Activation Energy (Ea). = Threshold Energy – Average energy (Eth (Er) :: Ea = E’th - E’r → It may be noted that each reaction at a particular temp has a definite value of activation energy.. → The value of Activation energy decides the fraction of the total no. of collisions which are effective. → Thus, - (i) for first reaction → Low Activation energy. - (ii) for slow reaction → High Activation energy. ## Application of Activation Energy (Ea):- → Energy barrier may be very high for some reaction and very small for some othere. → Depending upon the value of Ea the fraction of the total no. of collisions (f) which are may be very small moderate or high. → Activation Energy for explosive mixture as that of methane and oxygen, hydrogen and any gen. etc. is very high. ## Arrenhius Equation:- → Arrenhius (1889) developed a quantitative relationship between rate constant and temperature known as Arrenhenius eq. → The Arrhenius equation is: K = A.e⁻Ea/RT where: k = Rate constant A = frequency factor Ea = Activation energy T= Temperature R= Gas constant → e-Ea/RT is called Boltzmann factor. So, The equation is: K α A.e⁻Ea/RT Both side logarithim in eq(1) lnK = ln A.e⁻Ea/RT lnK = lnA - Ea/RT → ln k = lnA + ln e⁻Ea/RT → ln k = ln A - Ea/RT put lnA = 2.303 log A → ln k = ln A- Ea/RT → 2.303 logk = 2.303log A - Ea/RT → 2.303 logk = 2.303 log A – Ea/RT → log k = 2.303 log A – E/RT) / 2.303 → log k = 2.303 log A - E/RT / 2.303 → log k = log A – E/2.303RT → for a reaction of k₁ is the rate constant at temperature T₁ and k₂ is the rate constant at temperature T₂, then: log k₁ = log A - E/2.303RT₁ (1) log k₂ = log A - E/2.303RT₂ (2) Subtract the eq(1) from (2) log k₂ - log k₁ = log A - E/2.303RT₂ - ( log A - E/2.303RT₁) log k₂ – log k₁ = log A - E/2.303RT₂ - log A + E/2.303RT₁ log k₂/k₁ = E/2.303R (1/T₁ - 1/T₂) log k₂/k₁ = E/2.303R (T₂- T₁/T₁T₂) log k₂/k₁ = E/2.303R (T₂- T₁/T₁T₂)