Pharmaceutical Analytical Chemistry I Lecture Notes PDF

Pharm. D Pharmaceutical Analytical Chemistry I PC_101 Lecture No. 10 Assistant Prof. Ashraf Mohamed Taha 8. Temperature and Rate 9. Effect of Catalyst 29 8. Temperature and Rate How do chemical reactions o...

Pharm. D Pharmaceutical Analytical Chemistry I PC_101 Lecture No. 10 Assistant Prof. Ashraf Mohamed Taha 8. Temperature and Rate 9. Effect of Catalyst 29 8. Temperature and Rate How do chemical reactions occur? We have already given some indications. For example, we have seen that the rates of chemical reactions depend on the concentrations of the reacting species. The initial rate for the reaction. 𝒂𝑨 + 𝒃𝑩 → 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 can be described by the rate law: 𝑹𝒂𝒕𝒆 = 𝒌 [𝑨]𝒙 [𝑩]𝒚 (6) Where the order of each reactant depends on the detailed reaction mechanism. This explains why reaction rates depend on concentration. But what about some of the other factors affecting reaction rates? For example, how does temperature affect the speed of a reaction? With very few exceptions, reaction rates increase with increasing temperature. For example, the time required to hard-boil an egg in water is much shorter if the “reaction” is carried out at 100°C (about 10 min) than at 80°C (about 30 min). Conversely, an effective way to preserve foods is to store them at subzero temperatures, thereby slowing the rate of bacterial decay. Figure 14 shows the rate constant for this reaction as a function of temperature. The rate constant and, hence, the rate of the reaction increases rapidly with temperature, approximately doubling for each 10 °C rise. Fig. 14: Temperature dependence of the rate constant for methyl isonitrile conversion to acetonitrile. To explain this behavior, we must ask how reactions get started in the first place. 30 8.1. The Collision Theory of Chemical Kinetics The collision theory is intuitively appealing, but the relationship between rate and molecular collisions is more complicated than expected. The collision theory implies that a reaction always occurs when an A and a B molecule collide. However, not all collisions lead to reactions. Calculations based on the kinetic molecular theory show that at ordinary pressures (say, 1 atm) and temperatures (say, 298 K), there are about 1 x 10 27 binary collisions (collisions between two molecules) in 1 mL of volume every second in the gas phase. Even more collisions per second occur in liquids. If every binary collision led to a product, then most reactions would be complete almost instantaneously. In practice, we find that the rates of reactions differ greatly. This means that, in many cases, collisions alone do not guarantee that a reaction will take place. For a reaction to occur, though, more is required than simply a collision—it must be the right kind of collision. For most reactions, only a tiny fraction of collisions leads to a reaction. For example, in a mixture of H2 and I2 at ordinary temperatures and pressures, each molecule undergoes about 1010 collisions per second. If every collision between H2 and I2 resulted in the formation of HI, the reaction would be over in much less than a second. Instead, at room temperature the reaction proceeds very slowly because only about one in every 1013 collisions produces a reaction. What keeps the reaction from occurring more rapidly? 8.2. The Orientation Factor In most reactions, collisions between molecules result in a chemical reaction only if the molecules are oriented in a certain way when they collide. The relative orientations of the molecules during collision determine whether the atoms are suitably positioned to form new bonds. For example, consider the reaction: 31  which takes place if the collision brings Cl atoms together to form Cl2, as shown in the top panel of Figure 15. In contrast, in the collision shown in the lower panel, the two Cl atoms are not colliding directly with one another, and no products are formed. Fig. 15: Molecular collisions may or may not lead to a chemical reaction between Cl and NOCl. 8.3. Activation Energy Molecular orientation is not the only factor influencing whether a molecular collision will produce a reaction. Any molecule in motion possesses kinetic energy; the faster it moves, the greater the kinetic energy. But a fast-moving molecule will not break up into fragments on its own. To react, it must collide with another molecule. When molecules collide, part of their kinetic energy is converted to vibrational energy. If the initial kinetic energies are large, then the colliding molecules will vibrate so strongly as to break some of the chemical bonds. This bond fracture is the first step toward product formation. If the initial kinetic energies are small, the molecules will merely bounce off each other intact. 32  We postulate that to react, the colliding molecules must have a total kinetic energy equal to or greater than the activation energy (Ea), which is the minimum amount of energy required to initiate a chemical reaction. When molecules collide, they form an activated complex (also called the transition state AB‡) in Figure 16, a temporary species formed by the reactant molecules as a result of the collision before they form the product. If the products are more stable than the reactants, then the reaction will be accompanied by a release of heat; that is, the reaction is exothermic [Figure 16(a)]. On the other hand, if the products are less stable than the reactants, then heat will be absorbed by the reacting mixture from the surroundings, and we have an endothermic reaction [Figure 16 (b)]. In both cases we plot the potential energy of the reacting system versus the progress of the reaction. Fig. 16: Potential energy profiles for (a) exothermic and (b) endothermic reactions. These plots show the change in potential energy as reactants A and B are converted to products C and D. The activated complex (AB‡) is a highly unstable species with a high potential energy. The activation energy is defined for the forward reaction in both (a) and (b). Note that the products C and D are more stable than the reactants in (a) and less stable than those in (b). The situation during reactions is analogous to that shown in Figure 17. The golfer hits the ball to make it move over the hill in the direction of the cup. The hill is a barrier between the ball and the cup. To reach the cup, the player must impart enough kinetic energy with the putter to move the ball to the top of the barrier. If he does not impart enough energy, the ball will roll partway up the hill and then back down toward him. In the same way, molecules require a certain 33 minimum energy to break existing bonds during a chemical reaction. We can think of this minimum energy as an energy barrier. Fig. 17: Energy is needed to overcome a barrier between initial and final states. 8.4. The Arrhenius Equation The dependence of the rate constant of a reaction on temperature can be expressed by the following equation, known as the Arrhenius equation (Equation 22): −𝐸𝑎⁄ 𝑘= 𝐴𝑒 𝑅𝑇 Eq. (22) where - Ea is the activation energy of the reaction (in kJ/mol), - R the gas constant (8.314 J/K. mol), - T the absolute temperature, and - e the base of the natural logarithm scale. - The quantity A represents the collision frequency and is called the frequency factor. It can be treated as a constant for a given reacting system over a fairly wide temperature range. Equation (22) shows that the rate constant is directly proportional to A and, therefore, to the collision frequency. In addition, because of the minus sign associated with the exponent Ea/RT, the rate constant decreases with increasing activation energy and increases with increasing temperature. This equation can be expressed in a more useful form by taking the natural logarithm of both sides: 34  −𝐸𝑎⁄ ln 𝑘 = ln 𝐴 𝑒 𝑅𝑇 𝐸𝑎 ln 𝑘 = ln 𝐴 − 𝑅𝑇 Or Equation (23) Equation 21 can be rearranged to a linear equation: 𝐸 1 ln 𝑘 = − ( 𝑎 ) + ln 𝐴 Eq.23 𝑅 𝑇 ↨ ↨ ↨ ↨ 𝑦 =𝑏 𝑥+𝑎 Thus, a plot of ln k versus 1/T gives a straight line whose slope m is equal to Ea/R and whose intercept b with the y-axis is ln A. Example 9 demonstrates a graphical method for determining the activation energy of a reaction. Example 9: The rate constants for the decomposition of acetaldehyde: CH3CHO(g) → CH4(g) + CO(g) were measured at five different temperatures. The data are shown in the table. Determine the reaction's activation energy (in kJ/mol). Note that the reaction is “3/2” order in CH, so k has the units of 1/M1/2.S K (1/M1/2). S T (k) 0.011 700 0.035 730 0.105 760 0.343 790 0.789 810 35 Solution First, we convert the data to the following table: Ln k 1/T(K-1) -4.51 1.43 x 10-3 -3.35 1.37 x 10-3 -2.254 1.32 x 10-3 -1.070 1.27 x 10-3 -0.237 1.23 x 10-3 A plot of these data (Plot ln k versus 1/T) yields the graph in Figure 18. The slope of the line is Ea/R. Fig. 18: Plot of ln k versus 1/T. The slope of the line is equal to Ea/R. Slope = -2.09 x104 k From the linear Equation 23 An equation relating the rate constants k1 and k2 at temperatures T1 and T2 Can be used to calculate the activation energy or to find the rate constant at another temperature if the activation energy is known. To derive such an Equation 24. 36  Eq. 24 Example 10 The rate constant of a first-order reaction is 3.46 x 10-2 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? Solution The data are According to Equation 24 37 9. Effect of Catalyst 9.1. Effect of Catalysis Alternatively, reaction rates can be increased by using a catalyst, a substance that increases the rate of a chemical reaction but is not consumed by the reaction. A catalyst works by providing an alternative mechanism for the reaction— one in which the rate-determining step has a lower activation energy (Figure 19). For example, consider the noncatalytic destruction of ozone in the upper atmosphere, which happens according to this reaction: For the decomposition of hydrogen peroxide we saw that the reaction rate depends on the concentration of iodide ions even though I- does not appear in the overall equation. We noted that I- acts as a catalyst for that reaction. A catalyst is a substance that increases the rate of a reaction by lowering the activation energy. It does so by providing an alternative reaction pathway. The catalyst may react to form an intermediate with the reactant, but it is regenerated in a subsequent step, so it is not consumed in the reaction. Fig. 19: Catalyzed and Uncatalyzed Decomposition of Ozone In the catalytic destruction of ozone (red), the activation barrier for the rate-limiting step is much lower than in the uncatalyzed process (blue). 38 9.2. Homogeneous and Heterogeneous Catalysis Fig. 20: Homogeneous and Heterogeneous Catalysis A homogeneous catalyst exists in the same phase as the reactants. A heterogeneous catalyst exists in a different phase than the reactants. Often a heterogeneous catalyst provides a solid surface on which the reaction can take place. We categorize catalysis into two types: homogeneous and heterogeneous (Figure 20). In homogeneous catalysis, the catalyst exists in the same phase (or state) as the reactants. The catalytic destruction of ozone by Cl is an example of homogeneous catalysis—the chlorine atoms exist in the gas phase with the gas-phase reactants. In heterogeneous catalysis, the catalyst exists in a different phase than the reactants. The catalysts used in catalytic converters are examples of heterogeneous catalysts—they are solids while the reactants Are gases. The use of solid catalysts with gas-phase or solution-phase reactants is the most common type of heterogeneous catalysis. 9.2. Enzymes: Biological Catalysis We find perhaps the best example of chemical catalysis in living organisms. Most of the thousands of reactions that must occur for an organism to survive are too slow at normal temperatures. So, living organisms rely on enzymes, biological catalysts that increase the rates of biochemical reactions. Enzymes are usually large protein molecules with complex three-dimensional structures. Within each enzyme’s structure is a specific area called the active site. The properties and shape of the active site are just right to bind the reactant molecule, usually called the substrate. The substrate fits into the active site in a manner that is analogous to a key fitting into a lock (Figure 21). When the substrate binds to the active site of the enzyme—through intermolecular 39 Forces such as hydrogen bonding and dispersion forces, or even covalent bonds—the activation energy of the reaction is greatly lowered, allowing the reaction to occur at a much faster rate. The general mechanism by which an enzyme (E) binds a substrate (S) and then reacts to form the products (P) is: Fig. 21: Enzyme–Substrate Binding A substrate (or reactant) fits into the active site of an enzyme much as a key fits into a lock. It is held in place by intermolecular forces and forms an enzyme-substrate complex. (Sometimes temporary covalent bonding may also be involved.) After the reaction occurs, the products are released from the active site. Sucrase is an enzyme that catalyzes the breaking up of sucrose (table sugar) into glucose and fructose within the body. At body temperature, sucrose does not break into glucose and fructose because the activation energy is high, resulting in a slow reaction rate. However, when a sucrose molecule binds to the active site within sucrase, the bond between the glucose and fructose units weakens because glucose is forced into a geometry that stresses the bond (Figure 22 a, b). Weakening of this bond lowers the activation energy for the reaction, increasing the reaction rate. The reaction can then proceed toward equilibrium—which favors the products—at a much lower temperature. 40 Fig. 22 a: An Enzyme-Catalyzed Reaction Sucrase catalyzes the conversion of sucrose into glucose and fructose by weakening the bond that joins the two rings. Fig. 22 b: Sucrose breaks up into glucose and fructose during digestion. 10. Elementary Reactions and Their Rate Law 41