Reaction Kinetics

Reaction kinetics is the study of the rates of chemical reactions. Rates differ greatly: the reaction of NaCl with AgNO₃ is very fast, the hydrolysis of an ester proceeds at a moderate rate, and the rusting of iron is a slow process that may take days or months. An explosion, by contrast, is a swift reaction that happens within a fraction of a second. The rates of reactions and their control are often important in industry — they may decide whether a reaction can be used economically, so it is important to discover the conditions under which the reaction will proceed most economically.

Collision theory explains how reactions occur. According to this theory, for a chemical reaction to take place, the particles (atoms, ions or molecules) of reactants must form a homogeneous mixture and collide with one another. These collisions may be effective or ineffective, depending upon the energy of the colliding particles. When collisions are effective, they give rise to the products; otherwise the colliding particles just bounce back.

An effective collision can take place only when the colliding particles possess a certain amount of energy and they approach each other with the proper orientation. The minimum amount of energy required for an effective collision between the reacting species is called activation energy. Most reactions are slow, showing that all collisions are not equally effective.

📌 Quick Check 7.1 (from the book): (a) Activation energy is the energy barrier that colliding particles must cross before they can change into products — without it, collisions only bounce back. (b) The higher the activation energy, the smaller the fraction of collisions that are effective, so the slower the rate of reaction.

During a chemical reaction, reactants are converted into products. So, the concentration of the products increases with the corresponding decrease in the concentration of the reactants as they are being consumed. The rate of a reaction is defined as the change in the concentration of a reactant or a product divided by the time taken for the change.

Rate of reaction = (change in concentration of the substance) ÷ (time taken for the change), i.e. Rate of reaction = Δx ÷ Δt, where Δx is the change in the concentration of a reactant or a product in a time interval Δt.

It is explained graphically in Figure 7.1 of the book. For the reactant A which is changing irreversibly to the product B (A → B): the slope of the graph for the reactant or the product is steepest at the beginning. This shows a rapid decrease in the concentration of the reactant and, consequently, a rapid increase in the concentration of the product. As the reaction proceeds, the slope becomes less steep, indicating that the reaction is slowing down with time; ultimately both the curves become parallel. This is the stage of completion of the reaction.

Units. The rate of reaction has the units of concentration divided by time. Usually the concentration is expressed in mol dm⁻³ and the time in seconds, thus the units of reaction rate are mol dm⁻³ s⁻¹. However, for a slow reaction the units may be mol dm⁻³ min⁻¹ or even mol dm⁻³ h⁻¹. For a gas phase reaction, units of pressure are used in place of molar concentrations.

The rate of a general reaction, A → B, can be expressed in terms of rate of disappearance of the reactant A or the rate of appearance of the product B mathematically, where [A] and [B] are the concentrations of A and B respectively: Rate of reaction = −Δ[A] ÷ Δt = +Δ[B] ÷ Δt. The negative sign with Δ[A] ÷ Δt indicates a decrease in the concentration of the reactant A. Since the concentration of product increases with time, the sign in the rate expression involving the change of concentration of product is positive.

7.2.1 Instantaneous and average rate. The rate at any one instant during the interval is called the instantaneous rate. Whereas the average rate of reaction is defined as: "The rate of reaction between two specific time intervals or the rate over a time period". The average rate and instantaneous rate are equal for only one instant in any time interval. At first, the instantaneous rate is higher than the average rate. At the end of the interval the instantaneous rate becomes lower than the average rate.

Sample Problem 7.1 (book). The reaction for the formation of ammonia in the Haber process is: N_2(aq) + 3H_2(g) → 2NH_3(g). If the concentration of ammonia is 3.5 mol dm⁻³ after 1.0 min and 6.2 mol dm⁻³ after 4.0 minutes: (i) Calculate the instantaneous rate after 1.0 min. (ii) What is the average rate of production of ammonia for the system, between 1.0 and 4.0 minutes? Solution (book): (i) Instantaneous Rate = ΔC ÷ Δt = 3.5 mol dm⁻³ ÷ 1 min = 3.5 mol dm⁻³ min⁻¹. (ii) ΔC = Δ[NH₃] = (6.2 − 3.5) mol dm⁻³ = 2.7 mol dm⁻³; Δt = (4.0 − 1.0) = 3.0 min; Rate of formation of NH₃ = ΔC ÷ Δt = 2.7 mol dm⁻³ ÷ 3 min = 0.90 mol dm⁻³ min⁻¹.

📌 Quick Check 7.2 (book, worked): The reaction of hydrogen and iodine to make hydrogen iodide at a particular temperature, H_2(g) + I_2(g) → 2HI_(g), was studied at various times. At 100.0 s after the start of the reaction, the iodine concentration had fallen from 0.010 mol dm⁻³ to 0.0080 mol dm⁻³. Average rate during this period = change ÷ time = (0.010 − 0.0080) ÷ 100.0 = 0.0020 ÷ 100.0 = 2.0 × 10⁻⁵ mol dm⁻³ s⁻¹.

7.2.2 Measuring the rate. The measurement of rate of a chemical reaction involves the determination of the concentration of reactants or products at regular time intervals as the reaction progresses. To determine the rate of a reaction for a given length of time, a graph is plotted between time on x-axis and concentration of a reactant on y-axis, whereby a curve is obtained. To illustrate it, let us investigate the decomposition of HI to H₂ and I₂ at 50 °C, for which the experimental data is given in Table 7.1 of the book: [HI] falls from 0.100 mol dm⁻³ (0 s) to 0.0716 (50 s), 0.0558 (100 s), 0.0457 (150 s), 0.0387 (200 s), 0.0336 (250 s) and 0.0296 mol dm⁻³ (300 s).

By using the data, a graph is plotted as shown in Figure 7.2 of the book. The graph is plotted between time on x-axis and concentration of HI in mol dm⁻³ on y-axis. Since HI is a reactant, therefore, it is a falling curve. The steepness of the concentration-time curve reflects the progress of the reaction. Greater the slope of the curve near the start of reaction, greater is the rate of reaction.

In order to measure the rate of reaction, draw a tangent say, at 100 seconds, on the curve and measure the slope of that tangent. The slope of the tangent is the rate of reaction at that point i.e., after 100 seconds. A right-angled triangle ABC is completed with a tangent as hypotenuse. Figure 7.2 shows that in 100 sec, the change in concentration is 0.04 mol dm⁻³. The rate is then calculated as: Rate of reaction = ΔC ÷ Δt; rate = 0.04 mol dm⁻³ ÷ 100 sec = 4 × 10⁻⁴ mol dm⁻³ s⁻¹. If we plot a graph between time on x-axis and concentration of any of the products i.e. H₂ or I₂, then a rising curve is obtained. The value of the tangent at 100 seconds will give the same value of rate of reaction.

📌 Quick Check 7.3 (book): (a) Plot the Table 7.1 data for HI in your notebook. (b) The rate after 300 sec (when the concentration is about 0.03 mol dm⁻³) is found by the same tangent method — the tangent there is much flatter than at 100 s, so the rate is much smaller. (c) The same method gives the rate at HI concentrations of 0.10, 0.050, 0.030 and 0.02 mol dm⁻³. (d) From these calculations you deduce that the rate of the reaction falls with time, as the reactant is used up. (e) The rate is highest at the highest concentration (0.10 mol dm⁻³, at the start) and lowest at the lowest concentration (0.02 mol dm⁻³).

The change in concentrations of reactants or products can be determined by both physical and chemical methods depending upon the type of reactants or products involved.

a) Chemical Method. This is particularly suitable for reactions in solution. In this method, we do the chemical analysis of a reactant or a product. The acid hydrolysis of an ester (ethyl acetate) in the presence of a small amount of an acid is one of the best examples: CH₃COOC₂H_5(l) + H₂O_(l) → (H⁺) CH₃COOH_(aq) + C₂H₅OH_(aq). In the case of hydrolysis of an ester, the solution of ester in water and the acid (acting as a catalyst) are allowed to react. After some time, a sample of reaction mixture is withdrawn by a pipette and run into about four times its volume of ice-cold water. The dilution and chilling stop the reaction. The acid formed is titrated against a standard alkali, say NaOH, using phenolphthalein as an indicator. The analysis is repeated at various time intervals after the start of reaction. This would provide information about the change in concentration of acetic acid formed during the reaction at different time intervals.

b) Physical Methods. Some of the methods used for the measurement of concentration are as follows: (i) Spectrophotometry or colorimetry. This method is applicable if a reactant or a product absorbs ultraviolet, visible or infrared radiation. The rate of a reaction can be measured by measuring the amount of radiation absorbed. For the reaction shown in Figure 7.3 of the book, the concentration can be measured using the colorimetry. (ii) Electrical conductivity method. The rate of a reaction involving ions can be studied by electrical conductivity method. The conductivity of such a solution depends upon the rate of change of concentration of the reacting ions or the ions formed during the reaction. The conductivity will be proportional to the rate of change in the concentration of such ions. (iii) Volume change method. This method is useful for those reactions which involve changes in volumes of gases as shown in Figure 7.4 of the book. The volume change is directly proportional to the extent of reaction and changes in concentration.

📌 Interesting Information (book): The rates of some very fast reactions can be monitored using stopped-flow spectrophotometry. In this technique, very small volumes of reactants are driven at high speed into a mixing chamber. From here they go to an observation cell, where the progress of the reaction is monitored (usually by measuring the transmission of ultraviolet radiation through the sample). A graph of rate of reaction against time can be generated automatically.

In general, 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. The factors affecting the rate of a reaction are: i. Concentrations of the reactants, ii. Temperature of the system, iii. Surface area, iv. Catalyst.

7.3.1 Concentration. According to the law of mass action, the greater the concentration of the reactants, the more rapidly the reaction proceeds. When the concentration of one or more reactants increases, the rate of reaction increases. This is because increasing the concentration results in more collisions between the reacting particles, which speeds up the reaction. Figure 7.5 of the book shows this: the reaction in box (a), with more particles crowded together, will occur faster than that in box (b), due to the higher concentration.

📌 Did You Know? (book): In the case of reactions that involve gaseous reactants, an increase in pressure increases the concentration of the gases which leads 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.

📌 Note on surface area: the book lists surface area as the third factor but gives no separate passage explaining it in this chapter. From collision theory (7.1) it follows directly: a larger exposed surface means more reactant particles are available at the surface to collide at the same time, so the reaction goes faster — the classic example is powdered solid reacting faster than a single lump of the same solid.

7.3.2 Temperature (Maxwell–Boltzmann distribution curve). Increase in temperature increases the reaction rate. It has been observed that rate either doubles or triples for every 10 °C rise in temperature. Temperature usually has a major effect on the rate of reaction. Molecules at higher temperatures have more thermal energy. So, they collide more frequently and with greater energy.

The Boltzmann distribution curve is a graph that shows the distribution of energies at a certain temperature. In a reaction, a few particles will have very low energy, a few particles will have very high energy, and many particles will have energy in between. The distribution of energies at a given temperature can be shown on a graph as shown in Figure 7.6 of the book; this is called a Boltzmann distribution curve. As you know that activation energy is the minimum energy required for colliding particles fruitfully to convert into the product. The shaded area under the graph as shown in Figure 7.6 represents the proportion of molecules that have enough energy to cause a chemical change when they collide. The area under the curve represents the number of particles. The shaded area shows the number of particles with energy greater than the activation energy, E_a.

As the temperature of a reaction mixture is raised, the average kinetic energy of the particles increases. The reacting particles move around more quickly at a higher temperature, resulting in more frequent collisions. Therefore, the proportion of successful collisions also increases greatly as shown in Figure 7.7 of the book. The curve showing the Boltzmann distribution at the higher temperature flattens and the peak shifts to the right. For 10 °C rise in temperature, the shaded area under the curve approximately doubles. In conclusion, increasing the temperature increases the rate of a reaction.

📌 Quick Check 7.4 (book): (a) A Boltzmann distribution curve is a graph of the number (or fraction) of molecules against their molecular (kinetic) energy at a given temperature — it shows how many molecules have each energy. (b) A 10 °C rise approximately doubles the rate because the shaded area beyond E_a roughly doubles: many more molecules now have energy equal to or greater than the activation energy, so the proportion of successful, effective collisions rises sharply — even though the average energy itself rises only slightly.

7.3.3 Catalyst. A catalyst is defined as a substance which alters the rate of a chemical reaction, but remains chemically unchanged at the end of the reaction. A catalyst is often present in a very small amount. For example, the reaction between H₂ and O₂ to form water is very slow at ordinary temperature, but proceeds more rapidly in the presence of platinum. Platinum acts as a catalyst. Similarly, KClO₃ decomposes much more rapidly in the presence of a small amount of MnO₂. HCl is oxidized to Cl₂ in the presence of CuCl₂. Equations: 2H₂ + O₂ → (Pt) 2H₂O; 2KClO₃ → (MnO₂) 2KCl + 3O₂; 4HCl + O₂ → (CuCl₂) 2H₂O + 2Cl₂.

The process, which takes place in the presence of a catalyst, is called catalysis. A catalyst provides a new reaction path with a low activation energy barrier, as shown in Figure 7.8 of the book. A greater number of molecules are now able to get over the new energy barrier and reaction rate increases.

Types of Catalysis. (i) Homogeneous Catalysis. In this process, the catalyst and the reactants are in the same phase and the reacting system is homogeneous throughout. The catalyst is distributed uniformly throughout the system. For example: (i) The formation of SO_3(g) from SO_2(g) and O_2(g) in the contact process for the manufacture of sulphuric acid, needs NO_(g) as a catalyst. Both the reactants and the catalyst are gases: 2SO_2(g) + O_2(g) → (NO_(g)) 2SO_3(g). (ii) Esters are hydrolyzed in the presence of H₂SO₄. Both the reactants and the catalyst are in the solution state: CH₃COOCH₂CH_3(l) + H₂O_(l) → (H⁺) CH₃COOH_(aq) + CH₃CH₂OH_(aq) (ethyl ethanoate → ethanoic acid + ethanol).

📌 Did you Know! (book): Biochemical catalysts, commonly known as enzymes (nature's catalyst) are essential molecules in living organisms' functions by lowering the activation energy required for a chemical reaction to proceed, thereby increasing the reaction rate. Enzymes are typically proteins. Factors such as pH, temperature, and the concentration of substrate molecules can influence enzyme activity.

(ii) Heterogeneous Catalysis. In such systems, the catalyst and the reactants are in different phases. Mostly, the catalysts are in the solid phase, while the reactants are in the gaseous or liquid phase. For example, the oxidation of ammonia to NO in the presence of platinum gauze to produce HNO₃ industrially through the heterogeneous catalysis: 4NH_3(g) + 5O_2(g) → (Pt, 800 °C) 4NO_(g) + 6H₂O_(g). Hydrogenation of unsaturated organic compounds is catalysed by finely divided Ni, Pd or Pt: C₂H_4(g) + H_2(g) → (Ni, 150 °C) C₂H_6(g).

📌 Interesting Information (book): Vitamins are organic compounds that act as catalysts in biochemical reactions, especially when they function as coenzymes. Coenzymes are organic molecules that help enzymes catalyze reactions more efficiently. For example, Vitamin K₁ is necessary for blood clotting. Low levels of vitamin K can cause bleeding diathesis. A lack of vitamins can disrupt metabolic balance in cells and organisms. Vitamin deficiency is an example of a cofactor deficiency.

📌 Quick Check 7.5 (book): (a) No — a catalyst cannot be consumed in a chemical reaction: by definition it remains chemically unchanged at the end, even though it takes part in intermediate steps and is regenerated. (b) 2SO_2(g) + O_2(g) → (V₂O₅ (solid)) 2SO_3(g) is heterogeneous catalysis, because the catalyst is a solid while the reactants are gases — different phases. (c) An energy profile diagram for a typical uncatalysed reaction and an enzyme-catalysed reaction has the same reactant and product energy levels, but two peaks: the uncatalysed peak is high (large activation energy) and the enzyme-catalysed peak is much lower (small activation energy).

7.4.1 Rate Law and Rate Constant. The rate of a chemical reaction at a given temperature may depend on the concentration of one or more reactants and products. The representation of rate of a reaction in terms of concentration of the reactants is known as rate law. A rate law is an equation that relates the rate of a reaction to the concentrations of reactants raised to various powers according to the experimental data.

For a general reaction between A and B where 'a' moles of A and 'b' moles of B react to form 'c' moles of C and 'd' moles of D: aA + bB → cC + dD. We can write the rate equation as: Rate = k [A]^x [B]^y. Where x and y are the experimentally determined values that may or may not be equal to the coefficient of reactants in the balanced chemical equation, as 'a' and 'b' in the above equation. This expression is called rate equation. The brackets [ ] represent the molar concentrations and the proportionality constant k is called rate constant for the reaction.

If [A] = 1 mol dm⁻³ and [B] = 1 mol dm⁻³: Rate of reaction = k × 1^x × 1^y = k (rate constant). The rate constant can be defined as "The specific rate constant of a chemical reaction is the rate of reaction when the concentrations of the reactants are unity". Under the given conditions, k remains constant, but it changes with temperature.

7.4.2 Reaction Order. The exponents 'x' or 'y' in the above equation give the order of reaction with respect to the individual reactants. "The order of a reaction with respect to a specific reactant is the exponent applied to that reactant's concentration within the rate equation". Thus, the reaction is of order 'x' with respect to A and of order 'y' with respect to B. The overall order of reaction is (x + y). The order of a reaction defines how the reactant concentration influences its rate. For a single-reactant, the order is simply the concentration's power in the rate equation.

The chemical reactions are classified as zero, first, second and third order reactions. The order of reaction provides valuable information about the mechanism of a reaction. It is crucial to differentiate between the order concerning a single reactant and the overall reaction order.

Take the equation for the reaction of nitrogen (II) oxide (NO) with H₂: 2H_2(g) + 2NO_(g) → N_2(g) + H₂O_(g). The experimentally determined rate expression for this is: rate = k[H₂][NO]². The expression shows that this reaction is: (i) first-order with respect to H₂, (ii) second-order with respect to NO, (iii) third-order overall (1 + 2 = 3).

📌 Keep in Mind (book): The order of a reaction is given by the sum of all the exponents to which the concentrations in the rate equation are raised. It is important to note that the order of a reaction is an experimentally determined quantity and cannot be inferred simply by looking at the reaction equation. The sum of the exponents in the rate equation may or may not be the same as in a balanced chemical equation.

📌 Quick Check 7.6 (book): (a) The order of reaction is derived from the rate law by adding the exponents (powers) of the reactant concentrations in the experimentally determined rate equation — not from the balanced equation's coefficients. (b) The specific rate (rate constant) k is the rate when every reactant concentration is 1 mol dm⁻³ (unity); in the rate equation it appears as the proportionality constant, Rate = k[A]^x[B]^y, and it is constant at a given temperature but changes with temperature.

Zero Order Reaction. The rate of a zero order reaction is independent of the concentration of the reactants. A change in the concentration has no effect on the speed of the reaction. For the general reaction: A → Product, Rate = k [A]⁰ = k. Examples: (i) H_2(g) + Cl_2(g) → (sunlight) 2HCl_(g), Rate = k [H₂]⁰[Cl₂]⁰ = k. (ii) 2NH_3(g) ⇌ (high T, catalyst) N_2(g) + 3H_2(g), Rate = k [NH₃]⁰ = k. Photochemical reactions are usually zero order.

First Order Reaction. A reaction for which sum of the exponents of the concentrations in the rate equations is 1. In these reactions, there may be multiple reactants present, but concentration of only one reactant will change during the reaction. Example: the decomposition of nitrogen pentaoxide, 2N₂O_5(g) → 2N₂O_4(g) + O_2(g). The experimentally determined rate equation for this reaction is: Rate = k[N₂O₅]¹ (n = 1). This equation suggests that the reaction is first order with respect to the concentration of N₂O₅.

Second Order Reaction. A reaction for which sum of the exponents of the concentrations in the rate equation is 2. A second order reaction is a reaction whose rate depends either on the concentration of one reactant raised to the second power or on the concentrations of two different reactants, each raised to the first power. The simpler type of reaction involves one kind of reactant molecule. Hypothetically, it can be expressed as: Rate = k[A]² (n = 2); Rate = k [A]¹[B]¹ (n = 1 + 1 = 2). Oxidation of nitric oxide with ozone is first order with respect to NO and first order with respect to O₃. The sum of the individual orders gives the overall order of reaction as two: NO_(g) + O_3(g) → NO_2(g) + O_2(g), Rate = k[NO]¹[O₃]¹.

Third Order Reaction. A third order reaction is the reaction for which sum of the exponents of the concentrations in the rate equation is three: Rate = k[A]³ (n = 3); or Rate = k [A]²[B]¹ (n = 2 + 1 = 3); or Rate = k [A]¹[B]¹[C]¹ (n = 1 + 1 + 1 = 3). Example: 2FeCl_3(aq) + 6KI_(aq) → 2FeI_2(aq) + 6KCl_(aq) + I₂. This reaction involves eight reactant molecules but experimentally it has been found to be a third order reaction: Rate = k[FeCl₃][KI]². Also, the following reaction is third order: 2NO_(g) + O_2(g) → 2NO_2(g), Rate = k[NO]²[O₂]¹ — the exponents add to 2 + 1 = 3, i.e. third order. (Note: the book prints "order = 1 + 1 = 2" beside this last example, which does not match its own exponents; the correct sum from the printed rate equation is 3.)

Fractional Order Reaction. A reaction in which the sum of exponents of rate equation is in fraction, is called fractional order reaction. For example, consider the formation of carbon tetrachloride from chloroform: CHCl_3(l) + Cl_2(g) → CCl_4(l) + HCl_(g), Rate = k[CHCl₃][Cl₂]^1/2, n = 1 + 1/2 = 1.5. Reactions involving free radicals frequently exhibit fractional orders.

7.4.4 Units of Rate Constant. The rate constant is specific for a particular reaction at a certain temperature. Since concentrations are in mol dm⁻³ and a reaction rate is in units of mol dm⁻³ s⁻¹. The units for k depend on the order of the reaction and the units of time: Rate = k[Reactants]ⁿ, where n = order of reaction; k_n = Rate ÷ (Reactants)ⁿ = (mol dm⁻³) s⁻¹ ÷ (mol dm⁻³)ⁿ; or k_n = (mol dm⁻³)¹⁻ⁿ s⁻¹; or k_n = (concentration)¹⁻ⁿ s⁻¹. This equation can be used to determine units of any order of reaction.

For a zero order reaction (n = 0): k₀ = (mol dm⁻³)¹⁻⁰ s⁻¹ = mol dm⁻³ s⁻¹. For a first order reaction (n = 1), the rate is directly proportional to the concentration of one reactant: k₁ = (mol dm⁻³)¹⁻¹ s⁻¹ = (mol dm⁻³)⁰ s⁻¹ = s⁻¹. Therefore, the units of k for a first order rate constant are s⁻¹. For a second order reaction (n = 2): k₂ = (mol dm⁻³)¹⁻² s⁻¹ = mol⁻¹ dm³ s⁻¹ = dm³ mol⁻¹ s⁻¹. For a third order reaction (n = 3): k₃ = (mol dm⁻³)¹⁻³ s⁻¹ = mol⁻² dm⁶ s⁻¹ = dm⁶ mol⁻² s⁻¹.

📌 Quick Check 7.7 (book): (a) (i) rate = k[NO]²[NH₃]⁰ — overall order = 2 + 0 = 2; orders with respect to each reactant: NO = 2, NH₃ = 0. (ii) rate = k[BrO₃⁻][Br⁻][H⁺]² — overall order = 1 + 1 + 2 = 4; orders: BrO₃⁻ = 1, Br⁻ = 1, H⁺ = 2. (b) Note: the book prints this second expression as "rate = k[BrO₃][Br][H⁺]²" without the charges; the reading above follows the usual form of that reaction. (c) The sum of the coefficients of a balanced chemical equation is not necessarily important to give the order of a reaction because order is experimental: it is the sum of the exponents in the measured rate equation, and those exponents often differ from the balanced equation's coefficients (see the FeCl₃ + KI example above, which is third order although eight reactant molecules are involved).

7.5 Determination of Rate Constant. The rate constant (k) of a reaction can be calculated using the following two methods:

7.5.1 Initial Concentration Method. In the presence of hydrogen ions, hydrogen peroxide, H₂O₂, reacts with iodide ions to form water and iodine: H₂O_2(aq) + 2I⁻_(aq) + 2H⁺_(aq) → 2H₂O_(l) + I_2(aq). The progress of the reaction can be followed by measuring the initial rate of formation of iodine. Table 7.2 of the book shows the rates of the reaction obtained using various initial concentrations of each reactant: Experiment 1: [H₂O₂] = 0.0200, [I⁻] = 0.0100, [H⁺] = 0.0100, initial rate = 3.5 × 10⁻⁶ mol dm⁻³ s⁻¹; Experiment 2: 0.0300, 0.0100, 0.0100, 5.3 × 10⁻⁶; Experiment 3: 0.0050, 0.0200, 0.0200, 1.75 × 10⁻⁶ (all concentrations in mol dm⁻³).

The procedure for calculating k is shown below, using the data for experiment 1 in Table 7.2. Step 1: Write out the rate equation. The rate equation consistent with the book's own computed k is: rate of reaction = k[H₂O₂][I⁻]. Step 2: Rearrange the equation in terms of k: k = rate ÷ ([H₂O₂][I⁻]). Step 3: Substitute the values: k = 3.50 × 10⁻⁶ ÷ (0.0200 × 0.0100) = 3.50 × 10⁻⁶ ÷ 2.00 × 10⁻⁴ = 1.75 × 10⁻² dm³ mol⁻¹ s⁻¹. The concentration of hydrogen ions is ignored because [H⁺] does not appear in the rate equation. The reaction is zero order with respect to [H⁺]. (Note: the book's printed Step 3 fraction, taken literally, does not give its own printed answer; the derivation above is the one that does — rate divided by the product of the two concentrations — and it matches the book's answer 1.75 × 10⁻².)

7.5.2 Half-Life Method. "Half-life, t_1/2, is the time taken for the concentration of a reactant to fall to half of its original value". Calculating the rate constant (k) using the half-life method involves measuring the time it takes for the concentration of a reactant to decrease by half. If the reaction is first-order, then the rate constant and the half-life of the reaction are related in the following way: k = 0.693 ÷ t_1/2.

Here is an example to help understand how the process is done. A sample of hydrogen peroxide has a half-life of 2 hours. It decomposes in a first-order reaction. Calculate rate constant, k, for this reaction. To calculate k, we first need to convert the half-life into seconds: 2 × 60 × 60 = 7200 s. We then simply substitute this value into the equation: k = 0.693 ÷ 7200 s = 9.6 × 10⁻⁵ s⁻¹.

Sample Problem 7.2 (book). The conversion of cyclopropane to propene is a first order reaction. The half-life is 17.0 min. Calculate the rate constant of this reaction. Solution: Step 1: convert minutes to seconds: 17.0 × 60 = 1020 s. Step 2: substitute the half-life into the expression: k = 0.693 ÷ t_1/2 = 0.693 ÷ 1020 s = 6.79 × 10⁻⁴ s⁻¹. (Note: the book prints the unit here as dm³ mol⁻¹ s⁻¹; for a first-order reaction the correct unit from 7.4.4 is s⁻¹, as given here.)

📌 Quick Check 7.8 (book, worked): Consider a first-order reaction with a half-life of 15 minutes. If the initial concentration of the reactant is 0.100 mol dm⁻³, calculate the rate constant (k) for the reaction. Solution: t_1/2 = 15 × 60 = 900 s; k = 0.693 ÷ 900 = 7.70 × 10⁻⁴ s⁻¹. (For a first-order reaction the half-life does not depend on the initial concentration, so the 0.100 mol dm⁻³ figure is not needed for k.)

7.6 Reaction Mechanism. A reaction mechanism is a detailed, step by step description of how a chemical reaction occurs at the molecular level to yield the product(s). Unlike the overall balanced chemical equation, which only shows the reactants and products, the reaction mechanism reveals the actually happening individual steps (called elementary steps) that lead to the formation of products. Each of these steps represents a single molecular event, such as the breaking or forming of bonds.

Many reactions do not occur in a single step, but rather proceed through a series of steps. Each step is called an elementary reaction and is directly caused by the collision of atoms, ions or molecules. The number of reactant molecules involved in an elementary step is called its molecularity. A unimolecular elementary reaction involves only a single reactant molecule. The example of a unimolecular reaction is the decomposition of N₂O₅: N₂O_5(g) → NO_2(g) + NO_3(g). An elementary reaction that involves two atoms, ions or molecules is called bimolecular. For example: CO_(g) + NO_2(g) → NO_2(g) + CO_2(g); NO_(g) + O_3(g) → NO_2(g) + O_2(g). A termolecular reaction step involves the simultaneous reaction of three molecules. Such reactions are rare. An example is the reaction between oxygen molecules and atomic oxygen to form ozone in the stratosphere or during smog formation: 2O_2(g) + O_(g) → O_3(g) + O_2(g).

Intermediates are short lived species (ions or free radicals) that are produced in one step of the mechanism and consumed in a subsequent step. They do not appear in the balanced equation because they are not stable products.

Rate Determining Step. In many reaction mechanisms, one step is significantly slower than all the others. This step is called the rate-determining step. This step controls the overall rate of the reaction because it limits the speed at which the reaction can proceed. The balanced equation for the overall reaction is equal to the net result of all the individual steps. In a chemical reaction, any step that occurs after the rate-determining step will not affect the rate. So the atoms, ions or molecules taking part in the mechanism after the rate-determining step do not appear in the rate expression. All reactants that appear in the rate-determining step will also appear in the rate equation. As the rate-determining step limits the rate of overall reaction, the order of a reaction can be deduced from the rate determining step.

Lets consider the following reaction: 2NO_(g) + 2H_2(g) → 2H₂O_(g) + N_2(g). Table 7.3 of the book shows the results of six experiments. In the first three experiments, the concentration of H₂ is increased by keeping the concentration of NO constant. By doubling the concentration of H₂, the rate is doubled and by tripling the concentration of H₂, the rate is tripled. So, the rate of the reaction is directly proportional to the first power of concentration of H₂: Rate ∝ [H₂]¹. In the next three experiments, the concentration of H₂ is kept constant. By doubling the concentration of NO, the rate increases four times and by tripling the concentration of NO the rate is increased nine times. So, the rate of the reaction is proportional to the square of concentration of NO: Rate ∝ [NO]². The overall rate equation of this reaction is: Rate ∝ [H₂][NO]², or Rate = k[H₂]¹[NO]²; Order = 1 + 2 = 3. Hence, the reaction is a third order one. This final equation is the rate law for this reaction. It should be kept in mind that rate law cannot be predicted from the balanced chemical equation. The possible mechanism consisting of two steps for the reaction is as follows: Step 1: 2NO_(g) + H_2(g) → (Slow) N₂O_(g) + H₂O_2(g) (rate determining step); Step 2: H₂O_2(g) + H_2(g) → (Fast) 2H₂O_(g). The step 1 is slow and rate determining.

Another example is the reaction between nitrogen dioxide and fluorine gas: 2NO_2(g) + F_2(g) → 2NO₂F_(g). This reaction is first order in NO₂, first order in F₂ and second order overall. The experimental rate law is first order in NO₂ and in F₂: Rate = k [NO₂][F₂] (Observed). The accepted mechanism for the reaction is: Step 1: NO_2(g) + F_2(g) → (Slow) NO₂F_(g) + F^•_(g), Rate = k₁[NO₂][F₂]; Step 2: NO_2(g) + F^•_(g) → (Fast) NO₂F_(g), Rate = k₂[NO₂][F^•]. The first step is slow and determines the rate, in agreement with the observed rate expression. The second and fast step does not affect the reaction rate because fluorine atoms react with NO₂ as soon as they are produced.

📌 Quick Check 7.9 (book): An acidified solution of hydrogen peroxide reacts with iodide ions: H₂O_2(aq) + 2H⁺_(aq) + 2I⁻_(aq) → 2H₂O_(l) + I_2(aq). The rate equation for this reaction is: rate = k [H₂O₂] [I⁻]. The mechanism below has been proposed: H₂O₂ + I⁻ → (Slow) H₂O + IO⁻; H⁺ + IO⁻ → (Fast) HIO; HIO + H⁺ + I⁻ → (Fast) I₂ + H₂O. Explain why this mechanism is consistent with the rate equation. Answer: the slow (rate-determining) step is the first step, H₂O₂ + I⁻ → H₂O + IO⁻, so the rate depends only on the reactants in that step, [H₂O₂] and [I⁻], giving rate = k[H₂O₂][I⁻] — exactly the observed equation. H⁺ takes part only in fast steps after the slow step, so it does not appear in the rate equation, and the intermediates IO⁻ and HIO cancel out of the overall equation.

📌 SLO note — Gibbs free energy: the chapter's Student Learning Outcomes list explaining the relationship between Gibbs free energy change, ΔG°, and the feasibility of a reaction — but no passage in printed pages 140–162 develops ΔG° or feasibility. It is not invented here: in short, a reaction is feasible when ΔG° is negative, but feasibility says nothing about speed — that is what this chapter's kinetics (activation energy, rate law) is about. Students should learn ΔG° from the energetics chapter's treatment and not confuse a feasible reaction with a fast one.