UNIT 4 CHEMICAL KINETICS PDF

UNIT FOUR: CHEMICAL KINETICS 4.1 INTRODUCTION Every chemical reaction proceeds at a different rate (speed). Some reactions proceed very slowly and may take a number of days to complete; while others are very rapid, requiring only a few seconds. For example, rusting of iron could start quickly, whil...

UNIT FOUR: CHEMICAL KINETICS 4.1 INTRODUCTION Every chemical reaction proceeds at a different rate (speed). Some reactions proceed very slowly and may take a number of days to complete; while others are very rapid, requiring only a few seconds. For example, rusting of iron could start quickly, while ripening of fruits may be completed in a few days. On the other hand, weathering of stone may take more than a decade and the breakdown of plastics in the environment takes more than hundred years. However, other reactions, like the combustion of gasoline or the explosion of gunpowder occur in a few seconds. The area of chemistry that is concerned with reaction rates is called chemical kinetics. The word “kinetic”suggestsmovement or change. Chemical kinetics refers to the rate of reaction, which is the changeover times in the concentration of a reactant or a product. 4.2 THE RATE OF A REACTION The rate of a chemical reaction measures the change in concentration of a reactant or a product per unit time. This means that the rate of a reaction determines how fast the concentration of a reactant or product changes with time. For example, for a general reaction: Reactants → Products This equation tells us that, during the course of a reaction, reactant molecules are consumed while product molecules are formed. As a result, we can follow the progress of a reaction by monitoring either the decrease in concentration of the reactants or the increase in concentration of the products. Consider the progress of a simple reaction in which A molecules are converted to B molecules: A → B The decrease in the number of A molecules and the increase in the number of B molecules with time are shown in the figure below. In general, it is more convenient to express the rate in terms of change in concentration with time. Rate of reaction = ∆c/ ∆t, where∆c= Change in concentration of substance, &∆t =change in time. Note that ∆denotes the difference between the final and initial state. Thus, for the preceding reaction we can express the rate as: Rate =-1/2∆[A]/∆ t or Rate =∆[B]/ ∆t in which, ∆[A] and ∆[B] are the changes in concentration (mol L−1) over a period ∆t. Because the concentration of A decreases during the time interval, ∆[A] is a negative quantity. The rate of a reaction is a positive quantity, so a minus sign is needed in the rate expression to make the rate positive. On the other hand, the rate of product formation does not require a minus sign because ∆[B] is a positive quantity (the concentration of B increases with time). For more complex reactions, we must be careful in writing the rate expression. Consider, for example, the reaction: 2A → B Two moles of A disappear for each mole of B that forms, that is, the rate at which B forms is one half the rate at which A disappears. We write the rate as either: Rate = -1/2∆[A]/∆ t or Rate = ∆[B]/ ∆t Similarly, for the reaction: aA + bB → cC + dD the rate is given by:Rate = 1/a ∆[A]/∆ t = 1/b ∆[B]/ ∆t= 1/c ∆[C]/ ∆t= 1/d ∆[D]/ ∆t Examples 1. Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products: a. I−(aq) + OCl− (aq) → Cl− (aq) + OI− (aq) b. 3O2(g) → 2O3(g) c. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) Solution: a. Because each of the stoichiometric coefficients equals 1, Rate = −∆ [I−]/ ∆ t =− ∆ [OCl− ]/ ∆ t = ∆ [Cl− ]/ ∆ t = ∆ [OI−]/ ∆ t b. Here the coefficients are 3 and 2, so Rate =− 1/3 ∆ [O2 ]/ ∆ t = 1/2 ∆ [O3 ]/ ∆ t c. In this reaction: Rate = −1/4 ∆ [NH3]/ ∆ t =−1/5 ∆ [O2]/ ∆ t =1/4∆ [NO ]/ ∆ t = 1/6 ∆ [H2O]/ ∆ t 2. In the reaction of nitric oxide with hydrogen, 2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g) If the rate of disappearance of NO is 5.0 × 10–5mol L–1s–1, what is the rate of reaction for the formation of N2? Solution: The rate of reaction for the formation of N2 Rate = ∆ [N2]/ ∆t =− 1/2 ∆ [NO ]/∆ t =−1/2 (-5.0 × 10−5mol L−1s−1) =2.5 × 10−5mol L−1s−1 Solve exercise 4.1 on page 198 Determination of Rate of Reaction Reaction rate is change in an observable property over time. The observable property should be selected based upon what can be measured in the laboratory. This could be a colour change, a temperature change, a pressure change, a mass change, or the appearance of a new substance (for example, amount of precipitate formed). The observable (measurable) properties can be used to determine the change in concentration over time. Consider, for example, the thermal decomposition of gaseous dinitrogenpentoxide, N2O5, to give the brown gas nitrogen dioxide: 2N2O5 (g) → 4NO2 (g) + O2 (g) Changes in concentration as a function of time can be determined by measuring the increase in pressure as 2 gas molecules are converted to 5 gas molecules. Alternatively, concentration changes can be monitored by measuring the intensity of the brown color due to NO2 formation. Reactant and product concentrations as a function of time at 55 °C are listed in Table 4.1. Using this data, we can calculate the rate: Rate = −1/2 ∆ [N2O5]/ ∆ t = 1/4 ∆ [NO2]/ ∆ t = ∆ [O2 ]/ ∆ t Table 4.1: Concentrations (mol L−1s−1) as a function of time at 55 °C for the thermal decomposition of gaseous dinitrogenpentoxide( N2O5 ) Time (s) N2O5 NO2 O2 0 0.0200 0 0 100 0.0169 0.0063 0.0016 200 0.0142 0.0115 0.0029 300 0.0120 0.0160 0.0040 400 0.0101 0.0197 0.0049 500 0.0086 0.0229 0.0057 600 0.0072 0.0256 0.0064 700 0.0061 0.0278 0.0070 Note: The concentrations of NO2 and O2 increase as the concentration of N2O5 decreases. In the decomposition of N2O5, the rate of formation of O2 is given by the equation: Rate of formation of O2= ∆[O2]/ ∆t = (Conc. of O2 at time t2 −Conc. of O at time t1)/(t2− t1) During the time period 300 to 400 s, for example, the average rate of formation of O2 is 9 × 10−6mol L−1s−1: Rate of formation of O2= ∆[O2]/ ∆t = (0.0049 M − 0.0040 M)/ (400 s −300 s) =9 x 1 0-6mol L−1s−1 The average rate of formation of NO2 during the time period 300-400 s is 3.7 × 10-5molL−1s−1, which is four times the rate of formation of O2, as with the 4:1 ratio of the coefficients of NO2 and O2 in the chemical equation for the decomposition of N2O5. Rate of formation of NO2 = ∆ [NO2]/ ∆ t = (0.0197 M − 0.0160 M)/ (400 s−300 s) =3.7 x 1 0-5molL−1 s−1 Similarly, during the time period 300–400 s, the average rate of decomposition of N2O5 is 1.9 × 10-5mol L−1 s−1; Rate of formation of N2O5 = −∆ [N2O5]/ ∆ t = − (0.0101 M − 0.0120 M)/ (400 s−300 s) =1.9 x 1 0-5mol L−1 s−1 It’s also important to specify the time when quoting a rate because the rate changes as the reaction proceeds. For example, the average rate of formation of NO2 is 3.7 × 10-5mol L−1 s−1 during the time period 300 – 400 s, but it is only 2.2 × 10-5mol L−1s−1 during the period 600 –700 s.In general, the rate of reaction at the beginning is fast and it decreases as the reaction proceeds. What could be the possible reason for this decrease in the rate of reaction with time? Plotting the data of Table 4.1 gives the three curves in Figure 4.2. Looking at the time period 300 – 400s on the O2 curve, ∆[O2] and ∆t is represented, respectively, by the vertical and horizontal sides of a right triangle. The slope of the third side, the hypotenuse of the triangle, is ∆[O2]/∆t, the average rate of O2 formation during that period. The steeper the slope of the hypotenuse, the faster the rate. For example, compare the hypotenuse of the triangle defined by ∆[NO2] and ∆t during the time period 300 – 400 s and 600 –700 s. Figure 4.2: Concentration as a function of time when N2O5 decomposes to NO2& O2 It is important to realize that, given the initial concentrations in Table 4.1 the concentrations of NO2(g) and O2(g) can be calculated from the concentration of N2O5(g) at any time. The following example illustrates such a calculation. Example Using the data in Table 4.1, if [N2O5]= 1.20 ×10–2mol L−1 at 300 s, calculate [NO2] and [O2] at t = 300 s. Solution: The chemical equation is: N2O5 (g) → 2NO2 (g) + ½ O2 (g). The number of moles per liter of N2O5 (g) that have reacted in 300 s is the difference between the initial concentration and the concentration at 300 s [N2O5]. From Table 4.1.[N2O5]0 = 2.00 × 10–2mol L−1. Molarity of N2O5(g) reacted = [N2O5]0 − [N2O5] = (2.00 × 10–2mol L−1 − 1.20 × 10–2mol L−1) = 0.8 × 10–2mol L−1. According to the chemical equation, 2 moles of NO2 (g) are provided for every mole of N2O5 (g) that reacts. So, per liter, we have: Molarity of NO2 (g) produced = (0.8 × 10–2mol L−1 N2O5) x(2 molL−1NO2/1 mol L−1 N2O5) =1.6 x10–2mol L−1 Similarly, for O2 (g) we have: Molarity of NO2 (g) produced = (0.8 × 10–2mol L−1 N2O5) x (0.5mol L−1 NO2/1 mol L−1 N2O5) =0.4 x10–2mol L−1 Often, chemists want to know the rate of a reaction at a specific time t rather than the rate averaged over a time interval ∆t. For example, what is the rate of formation of NO2 at time t = 350 s? Such instantaneous rate is calculated from the slope of a tangent drawn at any points on the graph of concentrations versus time - (Figure 4.2) The slope of tangent taken at the initial point of the graph is assumed to be equal to its initialrate. In the initialratethe change in concentration of a reactant or product as a function of time is measured within minutes (or seconds) the reaction starts. Example Consider the gas-phase reaction that occurs when we mix 1.000 moles of hydrogen and 2.000 moles of iodine chloride at 230 °C in a closed 1.000-liter container. H2(g) + 2ICl(g) → I2(g) + 2HCl(g) From the experimental data given in the table below, determine the instantaneous rate of reaction at time, t = 2 s. Time (s) 0 1 2 3 4 5 6 7 8 [ICl] (mol L−1) 2.000 1.348 1.052 0.872 0.748 0.656 0.586 0.530 0.484 [H2] (mol L−1) 1.000 0.674 0.526 0.436 0.374 0.328 0.293 0.265 0.242 Solution:First, plot H2 concentration versus time for the reaction of 1.000 M H2 with 2.000 M ICl. Then, draw a tangent line to the curve. The instantaneous rate of reaction at any time, t, equals the negative of the slope of the tangent to this curve at time t. Figure 4.3 shows how to find the instantaneous rate at t = 2 seconds. Look Figure 4.3: A plot of the Hydrogen concentration versus time, using data of the above table. Page 203 Try to solve Exercise 4.2 on page 204 Conditions needed for a chemical reaction Chemical reactions are usually explained by the collision theory. The assumption of the collision theory is that chemical reactions take place due to collisions between molecules. Collisions between reactants The collision theory of rate of reaction is that, in order for a reaction to occur between reacting species (atoms, ions or molecules), they must first collide (come in contact). The rate of reaction is directly proportional to the number of collisions per second (the frequency of collision). Rate α number of collisions/Second 1. Proper Orientation According to collision theory, the more collisions there are the faster the rate of reaction would be. However, not all collisions between reacting species result in a reaction. This is because collisions between reactants can be either effective or ineffective. Effective collisions are collisions that result in a reaction to form the desired products. Ineffective collisions are collisions that do not result in a reaction to form the desired products. Thus, the collision between molecules should have the proper orientation. The effect of molecular orientation on the reaction of NO and O3. 2. Activation energy If the collisions between the reactant molecules do not have sufficient energy, then no reaction will occur. Therefore, for the reaction to take place collision must always occur with sufficient energy to break the bonds in the reactants and form new bonds in the product. Thus, minimum amount of energy needed for the reaction is known as activationenergy, Ea. According to the kinetic molecular theory, the average kinetic energy of the particles of a substance is directly proportional to the absolute temperature. As the temperature of the reacting species is raised, the average kinetic energy of the reacting particles increases considerably. This causes the particles of the reactants to collide more frequently and with greater energy. This results in increase in reaction rate. 4.3 Factors Affecting the Rate of a Chemical Reaction The rates at which reactants are consumed and products are formed during chemical reactions vary greatly. Even a chemical reaction involving the same reactants may have different rates under different conditions. Change in temperature, concentration, nature of reactant, surface area and presence of a catalyst, result in changes in rate of reaction. i. Nature of the reactants The rate of a reaction depends on the chemical nature of the substances in the reaction. The combination of two oppositely charged ions usually occurs very rapidly. For example, the reaction of an acid with a base is: H3O+ + OH– → 2H2O The acid-base reaction of HCl and NaOH is much faster than the decomposition of hydrogen peroxide, which involves the reorganization of molecules. 2H2O2 → 2H2O + O2 Even similar reactions may have different rates under the same conditions. For example, if small pieces of the metal iron and sodium are left in air, the sodium reacts completely overnight, whereas the iron is barely affected. The active metals sodium and calcium both react with water to form hydrogen gas and the corresponding metal hydroxide. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive. ii. Surface Area of Reactants Reactions can be classified as homogeneous (same phase) and heterogeneous (different phases). For example, the combustion of graphite is a heterogeneous reaction because it involves a solid with gas. In this case, the reaction occurs only at the interface (boundary) between two phases. A homogeneous reaction occurs only in one phase. For example: NO (g) → ½N2(g) + ½O2(g) Except for substances in the gaseous state or in solution, reactions occur at the boundary, or interface, between two phases. Hence, the rate of reaction between two phases depends to a great extent on the surface area of contact between them. Finely divided solids, react more rapidly than the same amount of the substance in a large body because of the greater surface area available. iii. Concentration of Reactants At a fixed temperature, the rate of a given reaction depends on the concentration of the reactants. Reaction rates often increases when the concentration of one or more of the reactants increases. This is because increasing the concentration produces more contacts between the reacting particles, which results in increasing the rate of reaction. In the case of reactions that involve gaseous reactants, an increase in pressure can increase the concentration of the gases which may lead to an increase in the rate of reaction. However, pressure change has no effect on the rate of reaction if the reactants are either solids or liquids. iv. Temperature of Reactants Temperature usually has a major effect on the rate of reaction. Molecules at higher temperatures have more thermal energy. Generally, an increase in the temperature of a reaction mixture increases the rate of chemical reactions. This is because, as the temperature of the reaction mixture raises, the average kinetic energy of the reacting particles increases. So, they collide more frequently and with greater energy. The effect of temperature on rate of reaction can be experienced in our daily life. For example, foods cook faster at higher temperature than at lower ones. We use a burner or a hot plate in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. In many cases, the rate of a reaction in a homogeneous system is approximately doubled by an increase in temperature of only 10 °C. v. Presence of a Catalyst A catalyst is a substance that changes reaction rate by providing a different reaction mechanism one with a lower activation energy, Ea. An activation energy is the minimum energy required to start a chemical reaction. Catalysts are not used up by the reactions, rather they are recovered at the end of the reaction. Although a catalyst speeds up the reaction, it does not alter the position of equilibrium. Figure 4.9: A catalyst provides an alternative pathway with a lower Eabarrier fot the reaction. Chemical catalysts can be either positive or negative. Positive catalysts increase the rate of reaction by lowering the Ea. Example 1:2SO2 (g) + O2 (g) → 2SO3 (g); V2O5 (s) as catalyst. Negative catalysts or inhibitors decrease the rate of reaction by increasing the value of Ea. Example 2:OCl– (aq) + I– (aq) → OI– (aq) +Cl– (aq); OH– (aq) as inhibitor. Example 1 represents heterogeneous catalysts as state of reactants and catalyst is different, while Example 2 represents homogeneous catalyst as both reactants and catalyst are in same state. A substance that catalyzes one reaction may have no effect on another reaction, even if that reaction is very similar. Many of the most highly specific catalysts are those designed by nature. The chemical reactions in living things are controlled by biochemical catalysts called enzymes. Compiled by: Elias HailemichaelAyele, Ph.D.