The first aspects forms the subject matter of chemical equilibrium. The second aspects forms the subject matter of chemical kinetics.

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Chemical Kinetics Introduction In a chemical reaction two important aspects are: (a) How far the reaction will go? and (b) How fast the reaction will occur? The first aspects forms the subject matter of chemical equilibrium. The second aspects forms the subject matter of chemical kinetics. Different chemical reactions occur with different speeds: Extremely fast reaction: Extremely slow reaction: AgNO 3 (aq) + NaCl (aq) = AgCl (s)+ NaNO 3 (aq) Iron + Moist Air = Rust Reaction with moderate rate: 2H 2 O 2 = 2H 2 O + O 2 CHEMICAL KINETICS is the branch of chemistry that deals with rates of chemical reactions. Concept of rate of a reaction In a chemical reaction some bonds in reactant molecules are broken and some new bonds are formed in the product molecules. H H + Cl Cl 2 H Cl For this process a definite time is required. The time required to break a bond and to form a new bond are not same in all reaction. Hence different reactions occur with different speeds. Rates of reaction During a chemical reaction reactants are consumed and products are formed. So rate of reaction is the change of concentration of any one reactant or product per unit time. Reaction rates are not uniform. They in general decreases with the passage of time. So three types of reaction rates may be identified: a) Initial rates b) Average rates c) Instantaneous rate Initial rate of the reaction Reaction rates measured at the very beginning of the reaction is called the initial rates of the reaction. This rates depends on the initial concentration of the reactants.

Average rate of the reaction Let us take a hypothetical reaction. A B Reactant Product Let the concentrations of the reactant A at time t 1 and t 2 be A 1 and A 2 and that of product be B 1 and B 2 respectively. Then average rate of the reaction, Average rate = - = [A] t = + = + [B] t (1) (2) = (3) Equation 1 represents the rate of reaction in terms of concentration of reactant (note ve sign, since concentration of reactant decreases with time and rate of reaction is positive), and equation 2 represents the rate of reaction in terms of increase in concentration of product. Equation 3 shows the rate of reaction in terms of change in concentration with time. Average rate of a reaction may be defined as the change in concn of the reactant or product in a unit interval of time in the specified time interval t. Concentration of reactant or product is expressed in mol litre -1 and time in second, hence mole litre-1 Unit for the rate of a reaction is = sec = mole litre -1 sec -1 Problem 1: For the reaction: Zn(s) + H 2 SO 4 (aq) = ZnSO 4 (aq) + H 2 (g) Write the rate expression in terms of concentration of H 2 SO 4 and H 2 Instantaneous rate of a reaction The rate of reaction measured at any particular instant during the course of reaction is called instantaneous rate. The average rate of the reaction is the rate of the reaction during the interval of time T. If the time interval, T, decreased and tends to zero then the average rate becomes instantaneous rate at that point. So, Instantaneous rate = (average rate) t 0 The instantaneous rate, in the notation of differential calculus, is given by Experimental Determination of instantaneous rate The concentrations of the reactant or product at different times are determined and concentration is plotted against time as shown in the following diagram.

Suppose we want to determined the reaction at 3.5 th second. Then a straight line is drawn at 3.5 th second to meet the curve at the point, and tangent is drawn at this point. The slope of this tangent gives instantaneous rate of the reaction at that moment. rate at the 3.5 th second = slope of tangent PQ or RS = 0.052 mol L -1 sec -1 The instantaneous rate of a reaction at any particular time is the rate of the reaction at that particular moment. Instantaneous reaction = slope of the tangent = Reaction rate and stoichiometry of a reaction H 2 + I 2 2HI rate of disappearance of H 2 = d[h 2 ] d t Rate of formation of HI will be twice that of the rate of disappearance of hydrogen. Thus the rate of reaction which are equivalent to each other rate = [H 2] = 1 t 2 [HI] t In general, for the reaction aa + bb = cc + dd The rate of the reaction = [A] a t = [B] b t = + [C] = + [D] c t d t The rate of reaction may be defined as the rate of decrease in the concentration of reactant species or the rate of increase in the concentration of product species divided by the stoichiometric coefficient of that species in the balanced chemical equation.

Problem 2: In a reaction H 2 + I 2 = 2HI, the rate of disappearance of I 2 is found to be 10-6 mole per litre per second. What would be the rate of appearance of HI? [Ans 2 10-6 mole litre -1 sec -1 ] Problem 3: Nitrogen pentoxide decomposes according to 2 N 2 O 5 (g) = 4 NO 2 (g) + O 2 (g) Write the rate of reaction in terms of rate of formation of (a) NO 2 (b) rate of formation of O 2 (c) rate of disappearance of of N 2 O 5. Also write the expression for the rate of the reaction which will all be equivalent to each other. Problem 4: For the reaction: N 2 (g) + 3 H 2 (g) = 2 NH 3 (g) Write the expression for the rate of the reaction. Factors influencing the rate of a reaction The following four factors influence the rate of a reaction: 1. Particle size of reactants 2. Concentration of reactants 3. Temperature and 4. Presence of catalyst Particle size of the reactants Smaller the particle size faster will be the rate of reaction. CaCO 3 + 2HCl CaCl 2 + H 2 O + CO 2 The rate of evolution of carbon dioxide will be faster in which powdered marble is used. Effect of concentration Rate of a chemical reaction increases with increase in the concentration of reactants. Effect of temperature Rate of a reaction increases with increase in temperature. In most cases the rate of a reaction increases by 2 to 3 times with every rise in temperature by 10 o C. Effect of catalyst The rate of a chemical reaction increases with the use of catalyst. For example hydrogen peroxide decomposes slowly on itself as : 2H 2 O 2 = 2H 2 O + O 2 If a small quantity of potassium iodide is added, the decomposition of hydrogen peroxide occurs with greater speed. Potassium is the catalyst in the decomposition of hydrogen peroxide. A catalyst is a substance which enhances the rate of a reaction without consuming itself.

Rate laws and rate constant The rate of reaction in general depends on the concentration of reactant. As the concentration decreases the rate of the reaction also decreases. With the passes of time the concentration of reactant decreases. These facts can be expressed mathematically which are called rate laws. Rate laws may be a) Differential rate law b) Integrated rate law (a) Differential rate law Differential rate law (or simply rate law) is a mathematical expression that gives the rate of a reaction in terms of concentration of reactant. Let us consider a general reaction, A Product (1) rate = d[a] dt [A] n (2) or, rate = d[a] dt = k[a] n (3) This equation 3 is called the differential rate equation or simply rate equation or rate law. Rate constant and order of the reaction The constant k in the rate equation 4 is called rate constant and n is the order of the reaction. If [A] = 1, in equation 3 then rate = k. Thus rate constant is numerically equal to the rate of a reaction when the concentration of the reactant is unity. Rate constant of a particular reaction is constantan at a given temperature. (a) The value of the rate constant is independent of concentration of the reactants. (b) The value of the rate constant increases with temperature. (c) Rate constant relates the rate of reaction with concentration of the reactant. The n in the rate equation 3 is the order of the reaction. Order is the power to the concentration in the rate equation. if n = 0 then the reaction is zero order if n = 1 then the reaction is first order if n = 2 then the reaction is second order if n = 3 then the reaction is third order and so on.

Let us consider more general reaction ma + nb Product Rate = d x dt = k[a]x [B] y Here x is the order with respect to A y is the order with respect to B and (x + y) is the overall order of the reaction. Order of a reaction with respect to a given reactant may be defined as the powder to the concentration of that reactant in the rate law. Overall order of the reaction is the sum of all the powers to which the concentration terms in the rate law are raised. Units of rate constant Units of rate constant depends upon the overall order of the reaction. If the general reaction is A Products Rate = k[a] n (a) For zero order reaction, n = 0 rate = k[a] o or, k = rate = mol L 1 sec = mol L 1 s 1 (b) For first order reaction n = 1 rate = k[a] or, k = mol L 1 s 1 mol L 1 = s 1 (c) For second order reaction, n = 2 rate = k[a] 2 or, k = mol L 1 s 1 (mol L 1 ) 2 = mol 1 L s 1 (d) For third order reaction, n = 3 rate = k[a] 3 or, k = mol L 1 s 1 (mol L 1 ) 3 = mol 2 L 2 s 1.

Problem 5 : A reaction involving single reactant is third order with respect to that reactant. What happens to the rate of reaction when its concentration is double? [Ans the rate will be 8 fold] Problem 6: The specific rate constant of a first order reaction is 0.02 sec -1. The initial concentration of the reaction is 2 mol L -1. Calculate the initial rate. Problem 7: For the reaction: 2A + B + C = A 2 B + C The rate law has been determined to be rate = k [A] [B] 2 with k = 2.0 10-6 mol -2 L 2 s -1 : (a) Determine the rate of reaction with [A] = 0.1 mol L -1.[B] = 0.2 mol L -1 and [C] = 0.08 mol L -1. (b) What will be the rate of the reaction when 0.04 mol L -1 of A has been reacted. Problem 8: The decomposition of N 2 O 5 in a carbon tetrachloride solution has been investigated. N 2 O 5 (soln) 2NO 2 (soln) + 1/2O 2 (g) The reaction has been found to be of the first order in N 2 O 5 with first order rate constant 6.2 10-4 s -1.Calculate the rate of the reaction when (a) [N 2 O 5 ]= 1.25 mol L -1 and (b) [N 2 O 5 ] = 0.25 mol L -1. What concentration of N 2 O 5 would give a rate of 2.4 10-3 mol L -1 s -1?[Ans (a) 7.75 10-4 mol L -1 s -1 (b) 1.55 10-4 mol L -1 s -1 (c) 3.8 mol L -1 ] Problem 9: For a hypothetical reaction A + B C, the following data were obtained in three different runs of experiments. Expts Initial conc n Initial conc n Initial rate of reaction of A (mol L -1 ) of B (mol L -1 ) (mol L -1 min -1 ) 1 0.01 0.01 1.0 10-4 2 0.01 0.03 9.0 10-4 3. 0.03 0.03 2.70 10-3 Determine (a) the order of the reaction with respect to A and B (b) overall order of the reaction (c) the rate constant k. Rate determining step and molecularity of a reaction Reactions which occur in a single step are called elementary reactions. Molecularity of a reaction is the number of molecules which simultaneously collide to give the product. For an elementary reaction molecularity and order of the reaction are same. Many reactions occur in a number of steps The detailed stepwise sequence of elementary reactions that convert reactants to the product is called the mechanism of the reaction.

If a reaction occurs through a number of elementary steps then each elementary step occurs with different rate. The slowest elementary step determines the overall rate of the reaction, and is known as the rate determining step. Molecularity of a complex reaction is sometimes defined as the number of species involved in the rate determining step. Molecularity of a reaction must always be positive integer but order of the reaction may be integer or fraction or even may be zero. Order of the reaction is an experimentally determined quantity, and is obtained from the rate law equation. The hydrolysis of methyl acetate in presence of mineral acid occurs as CH3COOCH3 + H2O CH3COOH + CH3OH Here two molecules are involved and is a bimolecular reaction. But in presence of excess of water, the order of reaction is found to be 1 i.e first order. This type of reaction is referred to as pseudo unimolecular reaction. Integrated rate law Differential rate law equation relates the rate of reaction with concentration. The differential rate equation may be integrated to obtain integrated rate law. The integrated rate law shows how the concentration of species in the reaction depends on time and using integrated rate equation rate constant can be calculated. Once order and rate constant is known, rate of reaction at any instant and at any concentration can be calculated. First order reaction In a first order reaction, the rate of the reaction is proportional to the concentration of the reactant. A products rate = d[a] [A] dt = d[a] dt = k[a] If "a" is the initial concentration and x be the amount of "A" reacted in time 't', then the remaining conc n [A] at the end of the time 't' is (a x). Hence the rate at time 't' is or,, dx dt or, [d(a x)] dt = k(a x) dx (a x) = kdt Integrating, = k(a x) dx (a x) = kdt) or, 1n(a x) = kt + c when, t = 0, x = 0

1n a = c 1n(a x) = kt 1n a or, kt = 1n a 1n(a x) or, k = 1 t 1n a a x k = 2.303 t log a a x This equation is the integrated rate equation for first order reaction, rearrange the above equation log (a x) = (k/2.303) t + ln a This equation is similar to the equation of a straight line y = mx + c Therefore, if we plot log (a x) against t we will get a straight line. If the reaction is first order the slope will be equal to -k/2.303 and intercept equal to log a (figure 19.3). Thus from the slope, rate constant k can be evaluated. Half life period t 1/2 The time required for the concentration of a reaction to decrease to half of its initial concentration is known as half life period, t1/2. The integrated rate equation for first order is k = 2.303 t At t1/2, x = a/2 log a (a x) k = 2.303 t1/2 log a (a a/2) = 2.303 t1/2 log 2 = 0.692 t1/2 or, t1/2 = 0.692/k

where k is the rate constant. Thus the half life of first order reaction is independent of initial concentration of the reactant. Rate equations and half life for reactions of different orders Order Differential rate equation Integrated rate equation Half life 0 dx/dt = k k 0 = x/t t 1/2 = a/2k 0 1 dx/dt = k[a] k 1 = 2.303/t{log a/(a x)} t 1/2 = 0.692/k 1 2 dx/dt = k[a][b] k 2 = 1/t {x/a(a x)} t 1/2 = 1/k 2 a 3 dx/dt = k[a][b][c] k 3 = 1/2t[x(2a x)/a 2 (a-x) 2 ] t 1/2 = 3/2 k 3 a 2 Concept of activation energy Chemical reactions speed up when the temperature is increased. Only a small fraction of the collisions produces reaction product. The collisions which are capable of bringing about chemical reactions are called effective collisions. For producing the effective and fruitful collisions, the reacting molecules must overcome some energy barrier and should also approach each other in proper orientations. To illustrate this point, let us consider the reaction between hydrogen and iodine to give hydrogen iodide. H2 + I2 2HI H2 and I2 molecules should collide with sufficient energy and should at the same time come in proper favourable orientations such that they can form a short lived molecular complex (H2I2) in which the H H and I I bonds are in the process of breaking and new bonds between H and I are in the progress of formation as shown in the diagram. This short lived molecular complex is called Activated complex.

The activated complex is an unstable transitory state of the reacting system which is midway between the reactants and products. The activated complex decomposes to give the products. Thus the reactants do not directly pass on to the product but they should first have some extra energy and form activated complex which then decomposes to give the products. Thus the reaction involves some energy barrier which must be overcome before the products are formed. This energy barrier is known as Activation energy (Ea). The sum of the activation energy [Ea] plus the average energy of the reactants (Eav) is called the threshold energy (Eth). So the threshold energy is in fact the energy of the activated complex. Eth = Eav + Ea or, Ea = Eth Eav This fact is graphically shown in the following figure. The activation energy is the excess energy required by the reactants (over and above their average energy) to undergo a chemical reaction and forms the energy barrier. If the activation energy is low then the reaction will be fast. If activation energy is high then the reaction will be slow. If we increase the temperature, the kinetic energy of the molecules will be increased and there will be more effective collisions which will enable more reactants to cross the energy barrier, and more products will be formed in less time. That is, increase of temperature enhances the rate of a reaction. Effect of catalyst A catalyst is a substance which speeds up a reaction without being consumed.the reactants have to cross the activation energy to give the product. When we use the catalyst the reaction

occurs with an alternative path with lower activation energy, and as a result in the increase in the rate of the reaction. References A catalyst is a substance which provides a new path for a reaction with lower activation energy and speeds up the reaction. 1. Foundations Chemistry Part Two Class XII by Moti Kaji Sthapit, Raja Ram Pradhananga and Kiran Bajracharya 2. Elementary Chemical Calculation by Moti Kaji Sthapit, Raja Ram Pradhananga 3. A text book of Physical Crhemistry for B.Sc. by Moti Kaji Sthapit, Raja Ram Pradhananga