INTRODUCTORY CHEMICAL KINETICS Chemical kinetics also known as reaction rate. It is the change in the concentration of a reactant or a product with time (M/s). Reactant → Products A→B Average rate = Average rate =

𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑡𝑖𝑚𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑡𝑖𝑚𝑒

= − =

∆[𝐴] ∆𝑡

∆[𝐵] ∆𝑡

aA → bB Average rate = Average rate =

𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑡𝑖𝑚𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑡𝑖𝑚𝑒

1 𝑑[𝐴]

= −𝑎

𝑑𝑡

1 𝑑[𝐵]

= −𝑏

𝑑𝑡

The rate of change of molar concentration of CH3 radicals in the reaction 2 CH3 (g)  CH3CH3 (g) was reported as d[CH3]dt = -1.2 mol dm-3 s-1 under particular conditions. What is (a) the rate of reaction and (b) the rate of formation of CH3CH3? aA + cC → bB + dD 1 𝑑[𝐴]

Rate = − 𝑎

𝑑𝑡

1 𝑑[𝐶]

= −𝑐

𝑑𝑡

=

1 𝑑[𝐵] 𝑏 𝑑𝑡

=

1 𝑑[𝐷] 𝑑 𝑑𝑡

Molecularity of a reaction is the number of molecules that are altered in a reaction. A reaction A  P is unimolecular (sometimes called monomolecular). A reaction A + B  P is bimolecular. Higher molecularity is extremely rare in one-step reactions, probably, but a reaction A + B + C  P would be trimolecular (or termolecular). Order of a reaction defines how many concentration terms must be multiplied together to get an expression for the rate of reaction. Hence, in a first-order reaction the rate is proportional to one concentration; in a second-order reaction it is proportional to the product of two concentrations or to the square of one concentration; and so on.

The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants. Each concentration is expressed with an order (exponent). aA + bB  cC + dD Rate = k [A]a [C]c , where k = Rate constant a and b are the reactant orders determined from experiment. The rate constant, k, converts the concentration expression into the correct units of rate (Ms−1). The Overall Order of a reaction is the sum of the individual orders: a + b. These orders are not necessarily equal to the stoichiometric coefficients a and b. The order usually is the same as the molecularity for a simple reaction that consists of a single step, or for each step in a complex reaction. Though this may not be apparent if one concentration, for example that of the solvent if it is also a reactant, is so large that it is effectively constant. The molecularity of a complete reaction need not be the same as its order. Indeed, a complex reaction often has no meaningful order, as the overall rate often cannot be expressed as a product of concentration terms.

Advantages of the rate law are: 1. helping in predicting the reaction rates for new values of the concentrations without doing additional experiments. 2. The form of the rate law can usually provide information about the sequence of molecular steps that constitute the reaction (the mechanism of the reaction). Exercise: 1. Ethanoic acid reacts with ethanol to form the pungent ethyl ethanoate and water: CH3COOH + CH3CH2OH  CH3COOC2H5 + H2O Write an expression for the rate of this reaction in a similar form, assuming the reaction proceeds in a single step as written. 2. The following chemical equations are balanced differently: 2NO + Cl2  2NOCl and NO + ½ Cl2 → NOCl a. If the rate of the reaction according to the first version is 7.1 × 10−5 mol L−1s−1, what is the rate of the reaction according to the second version? b. Write an expression for the rate of change of the concentration of each substance. Does this quantity differ for the two versions?

First-order kinetics The rate v of a first-order reaction A → B can be expressed as: 𝑑[𝐵] 𝑑[𝐴] 𝑣= = − = 𝑘[𝐴] = 𝑘([𝐴]𝑜 − [𝐵] 𝑑𝑡 𝑑𝑡 [A] and [B] are the concentrations of A and B respectively at any time t, k is a first-order rate constant and [A]o is the initial concentration of A. Every molecule of A that is consumed becomes a molecule of B. It makes no difference to the mathematics whether the rate is defined in terms of the appearance of product or disappearance of reactant. The difference may be seen experimentally, however, because experiments are not done without some errors, and in the early stages of a reaction the relative changes in [B] are much larger than those in [B] . For this reason it will usually be more accurate to measure increases in [B] than decreases in [A]. At time zero [A] = [A]0 , and [B]= 0 when t = 0, the stoichiometry allows the values of [A] and [B] at any time to be related according to the equation [A] + [B] = [A]0, thereby allowing the last equality in the equation. Using integration method: ln(

[𝐴]0 −[𝐵] [𝐴]0

) = −𝑘𝑡

[A] = [A]0e-kt , [A] is the concentration at time t, [A]0 is the initial concentration k is the rate constant, t is the time.

At t = t½, [B] = ½ [A]0 ln 2 = -kt½ 𝑡½ =

0.693 𝑘

Exercises: 1. The rate of decomposition of azomethane (C2H6N2) was studied by monitoring the partial pressure of the reactant as a function of time. Determine if the data below support a first order reaction. Calculate the rate constant for the reaction.

2. A certain reaction proceeds through t first order kinetics. The half-life of the reaction is 180s. What percent of the initial concentration remains after 900s? 3. N2O5  2NO2 + ½ O2 The above reaction is found to be first order. The rate constant at 337.6K is equal to 5.12 × 10−3 s−1. If the partial pressure of N2O5 is 0.500 atm at time t = 0, find the partial pressure of N2O5 at t = 60.0 s, neglecting any reverse reaction and assuming that the system is at constant volume. 4. The half-life of relaxation time.

235

U is equal to 7.1 × 108 years. Find the first-order rate constant and the

5. Find the time required for a sample of 235U to decay to 10.0% of its original amount. Second-order kinetics A + B  C +D The rate law for a second-order reaction with a single reactant is

Rate = −

𝑑[𝐴] 𝑑𝑡

= k[A]2

Using integration method, 1 [𝐴]𝑡

= 𝑘𝑡 +

1 [𝐴]0

Exercise: 1. NO3(g)  NO2(g) + ½ O2(g) The reaction above obeys second-order kinetics. At a temperature of 20oC and an initial concentration of NO3 equal to 0.0500 mol L−1, the concentration after 60.0 minutes is equal to 0.0358 mol L−1. a. Assuming the reverse reaction to be negligible, find the value of the forward rate constant. b. Find the concentration after 145 minutes if the initial concentration is 0.100 mol L−1. Getting the half-life for second order reaction: 2 [𝐴]0

1

− [𝐴] = 𝑘𝑡1 where [A]t = ½ [A]0 0

2

1

𝑡1 = 𝑘[𝐴] 2

0

2. Find the half-life of the above reaction with the given initial concentration. For a bimolecular second order reaction: A + B  Product

Rate =

𝑑𝑃 𝑑𝑡

= k[A][B] = k([A]0-[P]) ([B]0-[P])

Using the integration method, [𝐴]0 ([𝐵]0 − [𝑃]) = 𝑒 ([𝐵]0 −[𝐴]0 )𝑘𝑡 [𝐵]0 ([𝐴]0 − [𝑃]) For a pseudo first-order reaction: [𝑃] = [𝐴]0 (1 − 𝑒 −𝑘[𝐵]0 𝑡 Zero-order kinetics Reactions with zero order kinetics are with a constant rate, independent of the concentration of reactant. If a reaction is zero order with respect to only one reactant, this may simply mean that the reactant enters the reaction after the rate-limiting step. Reactions with overall zero-order are independent of all reactant concentrations. Such reactions are invariably catalyzed reactions and occur as every reactant present in such large excess that the full potential of the catalyst is realized. Enzyme-catalyzed reactions commonly approach zero-order kinetics at very high reactant concentrations. Rate = −

𝑑[𝐴] 𝑑𝑡

= k[A]0 = k

[A] = [A]0 – kt

Exercises: 1. A zero-order reaction with the rate constant of 0.0150 mol L−1 s−1, initial concentration of the single reactant is 1.000 mol L−1 was carried out in 5.00s at a given temperature. Find the concentration after the reaction time. 2. Find an expression for the half-life of a zero-order reaction and the value of the half-life of the reaction of the above question (1). nth-Order Reactions The rate law for an nth-order reaction with a single reactant Rate = −

𝑑[𝐴] 𝑑𝑡

= k[A]n

Using integration method, 1 1 1 [ 𝑛−1 − ] = 𝑘𝑡 𝑡 𝑛 − 1 [𝐴]𝑡 [𝐴]𝑛−1 0

2𝑛−1 − 1 𝑡1 = (𝑛 − 1)𝑘𝑡 [𝐴]𝑛−1 2 0 Note: These formulas are not valid for n =1 or n = 0 Exercises: 1. For a third-order reaction with a single reactant and negligible reverse reaction, find an expression for the time required for 80% of the reactant to react.

2. In terms of t1/2, how long will it take for 7/8 (87.5%) of the reactant of the above question (1) to react? METHODS OF DETERMINING THE RATE LAW The method for determining the rate law for a reaction is to mix the reactants and then to determine the concentration of one of the reactants or products as a function of time. A variety of methods have been used to determine concentrations, including measurement of the following: 1. The absorbance of radiation at some wavelength at which a given product or reactant absorbs. 2. The intensity of the emission spectrum of the system at a wavelength at which a given product or reactant emits. 3. The volume of a solution required to titrate an aliquot removed from the system. 4. The pressure of the system (for a reaction at constant volume). 5. The volume of the system (for a reaction at constant pressure). 6. The electrical conductance of the system. 7. The mass spectrum of the system. 8. The ESR or NMR spectrum of the system. 9. The dielectric constant or index of refraction of the system. 10. The mass loss if a gas is evolved. Exercise: The reaction of nitric oxide with hydrogen at 1280°C is: 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) From the following data, determine the rate law and rate constant.

THE TEMPERATURE DEPENDENCE OF REACTION RATES Temperature increase of most reactions results in the increase of their rate constants. For many chemical reactions, when the experiments are the plot of ln k against 1/T gives a straight line. Mathematically, this behaviour could be expressed by introducing two parameters, one at the intercept and the other at the slope of the straight line, and writing the Arrhenius equation.

ln 𝑘 = 𝑙𝑛𝐴 −

𝐸𝑎 𝑅𝑇

k = Ae-Ea/RT k is the rate constant; A is Arrhenius constant; 1/T = Temperature; Ea = Activation energy; (-Ea/R) = slope; ln A = Intercept. The fact that Ea is given by the slope of the plot of ln k against 1/T means that, the higher the activation energy, the stronger the temperature dependence of the rate constant. A higher activation energy signifies that the rate constant depends strongly on temperature. The activation energy is the minimum kinetic energy that is required of the reactants in order to form products. Exercise: The rate constant for the decomposition of a certain substance is 2.80 x 10-3 dm3 mol-1 s-1 at 30oC and 1.38 x 10-2 dm3 mol-1 s-1 at 50oC. Evaluate the Arrhenius parameters of the reaction.