Chapter 07: Reaction Kinetics

Definition: Reaction kinetics is the study of the rates of chemical reactions and the factors that affect the rates of chemical reactions. The studies of reaction kinetics involves: • A variety of experimental methods, measuring reaction rates, determining reaction orders, understanding reaction mechanisms. • It is a common observation that the rates of different chemical reactions vary greatly. Types of reaction rates Fast reaction: The reaction of NaC with AgNOs is very fast Moderate reaction: The hydrolysis of an ester occurs at a moderate rate. Slow reaction: The rusting of iron is a slow process. Importance of reaction rates in industry The rate of a chemical reaction and its control are often crucial in industrial processes. These factors can: • Determine the economic feasibility of using a particular reaction on a commercial scale. • Affect production efficiency, safety, and cost-effectiveness. S.Q. Differentiate between rusting and explosion. Ans. An explosion is a swift reaction that happens within a fraction of a second; the rusting of iron is a slow process that may take days or months. The rates of reactions occurring during the explosion are enormous. COLLISION THEORY Collision theory Collision theory explains how and why chemical reactions occur at the molecular level. Basic requirements for a reaction According to the theory: • Reactant particles (atoms, ions, or molecules) must form a homogeneous mixture. • They must collide with each other for a reaction to take place. Collision depends upon energy of colliding molecule. It can be: • Effective - leading to product formation • Ineffective - particles bounce back unchanged Conditions for effective collisions For a collision to be effective, two main conditions must be met: 1. Sufficient Energy: The colliding particles must possess at least a minimum amount of energy known as the activation energy (Ea). 2. Proper Orientation: The particles must approach each other in a specific orientation that allows bond formation or breaking. Activation Energy: The minimum energy required for an effective collision between the reacting specie is called activation energy Most reactions are slow • Not all collisions are effective, many particles may lack the required energy or proper orientation. • This explains why most chemical reactions are slow and require specific conditions to proceed efficiently. Quick Check 7.1 (a) What roles does the activation energy play in chemical reactions? Ans. • Collision theory states that for a chemical reaction to occur, reactant particles must collide. • However, not all collisions lead to a reaction. • Two key conditions must be met: (i) Collision frequency: Molecules must collide frequently for a chance to react. (ii) Proper orientation: Molecules must be aligned correctly during collision to allow bond-breaking and new bond formation. The activation energy required to form activated complex. Without frequent and properly oriented collisions, the reactant molecules will not overcome the activation barrier, and no reaction will occur, if activated complex is not formed. (b) How does the activation energy affect the rate of reaction? Ans. If Ea is low, more molecules have enough energy to overcome the barrier, reaction is faster. If Ea is high, fewer molecules can react, reaction is slower. Conclusion Lower activation energygives Faster reaction rate; Higher activation energy gives slower rate. RATE OF REACTION

Definition: The rate of a chemical reaction is defined as "The change in concentration of a reactant or a product divide by the time taken for the change." This can be mathematically expressed as: Rate =A[Concentration]/ ATime or Rate of reaction = Ax Ax is the change in concentration of a reactant or product. Y At is the time interval over which the change occurs in concentration Change during a reaction During a chemical reaction, the are into converted reactants products: As the reaction proceeds, the concentration of reactants decreases because they are being Concentration consumed Kee The concentration of products increases as they are being formed. Graphical explanation Consider the irreversible reaction Fig: Change in the concentration of reactants and for a reactant A which is products with time for the reaction A → B changing irreversibly to B. A → B • The concentration of reactant A decreases over time. • The concentration of product B increases over time. • The slope is steepest at the beginning, indicating a high reaction rate. • Rapid decrease in concentration of reactants and rapid increase in concentration of products. • As time passes, the slope gradually flattens, showing the rate is decreasing. • Eventually, the curves become almost horizontal parallel, meaning: • No further change in concentrations • The reaction has completed The rate of reaction is not constant - it changes at every moment during the reaction. Units of reaction rate The rate of a chemical reaction has the units of concentration by time. Concentration is expressed in moldm-3 and the time is expressed in seconds so, the units for the rate of reaction are moldm3 sl moles dm " Rate of reaction = - sec onds However, for slower reactions, the time unit may be in minutes or hours, giving units like mol dm3 min*' or mol dm3 h-l At • B A - "B Concentration of product Concentration of reactant Time - = mol.dm3 In gas-phase reactions, pressure is often used instead of concentration, with rate units like atm/s or kPa/min. Reaction rate For a general reaction: A→B The rate of reaction can be written in terms of either: • The rate of disappearance of reactant A • The rate of appearance of product B Mathematically Rate of reaction=-AA=+ AB The negative sign indicates that the concentration of A is decreasing over time. The positive sign for B reflects that its concentration is increasing. Average rate and instantaneous rate of reaction Average rate of reaction The rate of reaction between two specific time intervals or the rate over time period. It gives a general idea of how fast a reaction occurs between two points in time. Instantaneous Rate of Reaction The rate at any one instant during the intervals is called instantaneous rate. Instantaneous Rate= d [Reactant or Product]/dt Relationship between the average and instantaneous rate: The average and instantaneous rates are equal only at one specific point during the interval At the beginning of the reaction, the instantaneous rate is usually greater than the average rate because concentrations are higher, which typically speeds up the reaction. At the end of the interval, the instantaneous rate tends to be lower than the average rate as reactant concentrations drop. Quick Check 7.2 The reaction of hydrogen and iodine to make hydrogen iodide at a particular temperature, H2(g) +12(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 to 0.0080 mol dm. What is the average rate of reaction during this period! Ans. To calculate the average rate of reaction, we use the formula: Change in concentration Average rate = Time interval Given: Initial concentration of 12 = 0.010 mol dm3 Final concentration of I2 = 0.0080 mol dm3 Time interval = 100.0 s Step 1: Change in concentration of I2 4|121=0.010-0.0080=0.0020 mol dm At = 100 - 0 = 100 sec. Step 2: Calculate average rate 0.0020 - = 2.0x10-5 moldm 's! Average rate = 100.0 Answer: 2.0×10 moldm3s! MEASURING THE RATE OF A CHEMICAL REACTION How is the rate of a chemical reaction measured? OR Why is the concentration

concentration of reactants or products change with respect to time at regular intervals as reaction proceeds. Decomposition of HI 1. Taking Samples at Regular Time Intervals Small samples are withdrawn from the reaction mixture at set time intervals. 2. Determining Concentration The concentration of a reactant (which decreases) or a product (which increases) is measured for each sample. 3. Plotting Data A graph is plotted between time on x-axis and concentration on y-axis. Average rate (over a time interval), and Instantaneous rate (at a specific time point) can be calculated. Let us consider the decomposition of hydrogen iodide (HI) into hydrogen (Hz) and iodine (Iz) at 508°C. experimental data for decomposition of HI is given in table 7.1. The following data shows how the concentration of HI decreases over time A graph is plotted by taking X-axis: Time (in seconds) Y-axis: Concentration of HI (in mol dm3) HI is a reactant, its concentration decreases over time, resulting in a falling curve. • The steepness (slope) of the curve indicates the progress of the reaction. • A steeper slope near the beginning implies a faster rate. The slope becomes less steep as the reaction slows down. Determining Instantaneous Rate: To find the reaction rate (e.g., 100 seconds): (i) Draw a tangent to the curve at 100 seconds. monitor how the Table 7.1 Change in concentration of HI with regular intervals Concentration Time (s) of HI (mol, dm3) 0 0.100 50 0.0716 100 0.0558 150 0.0457 200 0.0387 250 0.0336 300 0.0296 0.10 0.09 0.08 A 0.07 [HI] 0.06 /0.04 m01 dm 0.05 0.04 0.03 0.02 0.01 'Y 0 50 100 150 200. 250 300 350 Time (s) Fig: The change in the HI concentration with time to the reaction 2HI(g) - → H2(g) +12(g) at 508°C. (i) Form a right-angled triangle with the tangent line. The slope of the tangent is the rate of reaction after 100 sec. at that point. • A right angled triangle ABC is completed with the tangent as hypotenuse. • Suppose the change in concentration (AC) is 0.04 mol dm3 • The time interval (At) is 100 seconds (iii) Use the formula Rate of reaction=AC/At = 0.04 mol dm3/100 s = 4x10-4 mol dm3 5 1 This means the concentration of HI is decreasing at a rate of 4 × 104 mol dm3 s' at 100 seconds Product Concentration Graph: If we instead plot the concentration of products (H2 or Iz) versus time: • The curve would rise, since product concentrations increase. • The slope of the tangent at 100 seconds would also give the same numerical value for the rate: 4x104 mol dm3s 1 Quick Check 7.3 (a) Plot the data in Table 7.1 for HI in your note book. Ans. Use time (s) on the x-axis and concentration of HI (mol dm ") on the y-axis. Mark each data point and connect them with a smooth curve. This will give you a concentration vs. time graph, which typically increases with time for a product like HI. Table 7.1 Change in concentration of HI with regular intervals Concentration Time (s) of HI (mol dm ) 0 0.100 50 0.0716 0.0558 100 0.0457 150 200 0.0387 250 0.0336 300 0.0296 (b) Calculate the rate after 300 sec (when the concentration is 0.03 mol dm3) by drawing a tangent. Ans. Draw a tangent to the curve at the point where time = s. Choose two points on the tangent line and use them to find the slope: Rate = A HI] X 0.10 0.09 .0.08 0.07 0.04 mol dm 0.05 0.04 B C 0.03 0.02 0.01 0 Y 0 50 100 150 200 250 300 350 Time (s) Let's say the tangent goes from (250 s, 0.026 mol/dm) to (350 s, 0.034 mol/dm?): 0.008 = 80×10- moldm3s! 0.034 - 0.026 - - Rate = 100 350-250 (c) Use the same method to calculate the rate of reaction at HI concentrations of 0.10 mol dm ', , 0.050 mol dm3 and 0.02 mol dm3 Ans. Repeat for other concentrations ([HI] = 0.10, 0.050, 0.020 mol dm3) At each point, draw a tangent from the graph. Measure the slope as above. The rate should decrease as [HI] increases, if the reaction slows with time. (d) What do you deduce about the rate of reaction with time from these calculations? Ans. Deduction: The rate of reaction decreases with time. As the reactant concentrations (H2 and I2) decrease, fewer effective collisions occur. (e) At which concentration is the rate highest and lowest? Ans. The rate is highest when [HI] is lowest (early in the reaction). The rate is lowest when [HI] is highest (later in the reaction). Example (based on typical results): Highest rate at [HI] = 0.020 mol dm Lowest rate at [HI] = 0.10 mol dm3 MEASUREMENT OF CONCENTRATION

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 Hydrolysis of ester This method is particularly suitable for reactions in solution, where it is possible to chemically analyze a reactant or product at different times. A typical example is the acid- catalyzed hydrolysis of an ester, ethyl acetate. In the presence of small amount of an acid. Reaction СН,СООС, Hs() + H,0 (г) -Н → СН,СООН ., + С,Н,ОН () Here, ethyl acetate reacts with water in the presence of an acid catalyst to produce acetic acid and ethanol. Procedure (i) Start the Reaction: Mix ethyl acetate, water, and a small amount of acid as a catalyst. (ii) Sampling: At fixed time intervals, withdraw a small sample using a pipette. (ill) Quenching: Immediately transfer the sample into about four times its volume of ice- cold water • This dilutes and chills the mixture, effectively stopping the reaction at that moment. (iv) Titration: Titrate the quenched sample against a standard NaOH solution using phenolphthalein as an indicator • This neutralizes the acetic acid formed, allowing its concentration to be determined. Purpose: By performing this titration at various time intervals, the concentration of acetic acid formed can be measured the about This provides data concentration of acetic acid 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 Fig: The concentration change for this reaction colorimetry a This method is applicable when reactant or product absorbs light in the ultraviolet (UV), visible, or infrared (IR) radiations. The rate of reaction can be measured by measuring the amount of radiation absorbed. For the reaction shown in figure, the be measured can concentration using the calorimetry. conductivity (ii) Electrical method for Conductometric method measuring reaction rate This method is particularly useful for studying reactions that involve ionic species, either as reactants or products. The electrical conductivity of an aqueous solution depends on the concentration and mobility of ions formed during the reaction. measuring the change in volume of a gas given off As the reaction proceeds, the number and type of ions change, leading to a measurable change in conductivity. The rate of change in conductivity reflects the rate of change in ion concentration, which in turn relates to the rate of the reaction. (iii) Volume change method Gas Volume Measurement Method for Reaction Rates This method is useful for reactions where there is a change in the volume of gas - either due to the production or consumption of a gaseous substance. During a reaction, if gas is produced or consumed, the total gas volume in the system changes. Light red Light red Light red Light red Light red Dark red can be determined using colorimetry. Fig: Rate of reaction can be followed by in a reaction. This change in volume is directly proportional to the extent of reaction and the change in concentration of gaseous reactants or products. By measuring gas volume at regular time intervals, the rate of reaction can be determined. Interesting Information! S.Q. What is stopped-flow spectrophotometry? Ans. 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. FACTORS AFFECTING RATE OF A CHEMICAL REACTION

Factors affecting the rate of 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. The factors affecting the rates of reactions are (i) Concentrations of the reactants (11) Temperature of the system (iii) Surface area (iv) Catalyst free 1m. Did you know? S.Q. What is the effect of pressure on the reaction rates of the gases? Ans. 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. (i) Concentration (a) According to the law of mass action "The rate of a chemical is reaction directly to the proportional of the product concentrations of the reactants, each raised to a power equal to its in the coefficient balanced chemical equation." Fig: The reaction in (box a) will occur faster than that in, (b) due to the higher. concentration. • Higher concentration means more particles per unit volume. • This leads to increased collision frequency between reactant molecules. • More collisions → greater chance of effective collisions → faster reaction. (ii) Temperature Increase in temperature increases, the reaction rate. (Maxwell-Boltzmann distribution curve) Effect of Temperature on Reaction Rate Molecules collide more frequently and more energetically. A larger fraction of molecules has energy equal to or greater than activation energy (Ea). This results in a higher rate of effective collisions, hence a faster reaction rate. The rate roughly doubles or triples with every 10 °C rise in temperature. Higher temperature = more molecules with sufficient energy (Ea) → increased reaction rate. Boltzmann distribution curve The curve shows the distribution of kinetic energies among particles at a given temperature. Most particles have moderate energy, some have low, and some have high energy. Ea (activation energy) is a threshold: only particles with energy equal to Ea can react upon collision. Shaded area under the curve beyond Ea represents particles that can successfully react. As temperature increases, the curve flattens and shifts right, increasing the shaded area. Ea- at and above this energy the molecules have enough energy to collide effectively Number of molecules 0 → Molecular energy Fig: The Boltzmann distribution curve showing molecular energies and activation energy Raising the temperature increases the average kinetic energy of particles. Particles move faster, leading to: • More frequent collisions. • Higher-energy collisions. A greater proportion of particles now have energy equal to activation energy (Ea) On the Boltzmann distribution curve: • The peak flattens and shifts to the right. • The shaded area beyond Ea (successful collisions) approximately doubles for each 10 °C rise in temperature. Conclusion: Increasing the temperature increases the rate of reaction. Tz>T, Fraction of Collisions Collision Energy Fig: The Boltzmann distribution of molecular energies at temperatures Ti and T2 Quick Check 7.4 What is the Boltzmann distribution curve? (a) Ans. The Boltzmann distribution curve is a graph that shows the distribution of kinetic energies among the molecules in a gas at a given temperature. Key Features • X-axis: Kinetic energy of molecules • Y-axis: Number (or fraction) of molecules with that energy • Most molecules have moderate energy, some have low energy, and a few have high energy. • The area under the curve represents the total number of molecules. Important Point • Only the molecules that have energy equal to or greater than the activation energy (Ea) can react. • Increasing temperature shifts the curve to the right, so more molecules have energy > Ea → faster reaction. (b) Explain why a 10°C rise in temperature approximately double the rate of a reaction? Ans. A 10 °C rise in temperature: • Increases the average kinetic energy of molecules. • More molecules have energy equal to activation energy. • This leads to more successful collisions per second. Result • The rate of reaction approximately doubles because the fraction of effective collisions increases significantly, even with a small temperature increase. (iii) Catalyst Definition: A catalyst is a substance which alters the rate of a chemical reaction but remains chemically unchanged at the end of the reaction. Catalysts are usually present in small amounts compared to reactants. Examples (i) Platinum (Pt) speeds up the reaction between H2 and Oz to form water. (ii) Manganese dioxide (MnOz) speeds up the decomposition of KCIO3. (iii) Copper(II) chloride (Culz) catalyzes the oxidation of HCl to Cl2. Ea (Activation Energy) The peak lowers an shifts to the righ Greater fraction with enough energy to react " →2H2O 2H, +02 2KCI +302 4C/ +02 - The process which takes place in the 1S called of catalyst presence catalysis. Catalysis provides a new reaction pathway with a lower activation energy (Ea). This lower activation energy allows a larger proportion of molecules to have sufficient energy to react, leading to a higher reaction rate. Energy profile with catalyst • The presence of a catalyst lowers the energy barrier, as shown in the energy profile diagram. • The overall energy change (AH) of the reaction remains the same, but the activation energy (Ea) is lowered. 5- • This results in faster reactions because more molecules can overcome the new, lower activation energy

Catalysis Definition: The process which take place in the presence of a catalyst is called catalysis. 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. Example The formation of SO3(g) from SO2(g) and 02(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. 2502 (g) + 02(g) - Did you know! S.Q. Give function and factors affecting influence of enzymes. 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. 4 Activation Energy without catalyst 4 Activation Energy with catalyst Reactants Products Progress of reaction Fig: The energy path diagram for an uncatalyzed and a catalyzed reaction NO (8) → 203(8) Esters are hydrolyzed in the presence of H2SO4. Both the reactants and the catalyst are in the solution state. (ii) Heterogeneous Catalysis Definition: The process in which the catalyst and the reactants are in different phases is called heterogeneous catalysis mostly, the catalysts are in the solid phase. The reactants are in the gaseous or liquid phase. Example: Oxidation of ammonia to NO in the presence of platinum gauze to produce HNO3 industrially through the heterogeneous catalysis. 4NH 3(8) +502(g) Hydrogenation of unsaturated organic compounds are catalysed by finely divided Ni, Pd ar Pt. Ni at 150°C. Interesting Information S.Q. How vitamins behave as co-enzymes? Ans. 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 K, 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 (a) Can a catalyst be consumed in a chemical reaction? Why or why not? Ans. No, a catalyst is not consumed in a chemical reaction. Reason: • A catalyst lowers the activation energy of a reaction by providing an alternative pathway. • It takes part in the reaction mechanism but is regenerated at the end of the reaction. (b) Explain whether the reaction below is an example of heterogeneous or homogeneous catalysis: 2502(g) +02(g) - Ans. This is an example of heterogeneous catalysis if solid V2Os (vanadium(V) oxide) is used as the catalyst (as in the contact process). Reason • The reactants (SO2 and O2) are gases. • The catalyst (VOs) is a solid. • Since the catalyst and reactants are in different phases, it is heterogeneous catalysis. P4800°C → 4NO (g) +61,0(8) Ni,150°C →CH 6(8) 130, (sold) → 2503(8) (c) Draw an energy profile diagram to show a typical uncatalyzed reaction and an enzyme-catalysed reaction. On your diagram show the activation energy for: (i) Catalysed reaction (ii) Uncatalysed reaction Ans. Energy profile diagram Here's a description of the energy profile diagram you should draw: Diagram should include (1) Y-axis: Potential Reactants Energy (ii) X-axis: Reaction Progress / Reaction Coordinate (111) Two curves: • One for uncatalysed reaction (higher peak = higher activation energy) • One for catalysed reaction (lower peak) Label the following: • Reactants and Products • Activation energy (Ea) for both catalysed and uncatalysed paths • Show the difference in height of the two energy barriers (indicating how the catalyst lowers Ea RATE LAW, RATE CONSTANT AND ORDER OF REACTION 07. What is a rate law and rate constant? Ans. 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. Definition of rate law: 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. General reaction: The 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]* [B] 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 4 Activation Energy without catalyst A Activation Energy with catalyst LTY Products Progress of reaction ee. 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 dm3 and [B] = 1 mol dm3 Rate of reaction = k × 1× × 1y = k (rate constant) Definition of rate constant: The rate constant can be defined as "The specific rate constant of a chemical reaction is the rate of reaction when the concentration of the reactants are unity". Under the given conditions, k remains constant, but it changes with temperature. REACTION ORDER

Reaction Order Definition: "The order of a reaction with respect to a specific reactant is the exponent applied to that reactant's concentration within the rate equation" Consider a general reaction aA +bB → cC + dD Rate = k [AlIB! • The exponents 'x' and 'y' in the above equation give the order of reaction with respect to the individual reactants. • Thus, the reaction is of order'' with respect to A and of order 'y' with respect to B. The overall order of reaction is (xty). • 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. Measurement of reaction order Take equation for the reaction of nitrogen (II) oxide (NO) with H2: 2H 2(g) + 2N0(8) N2(g) + 2H0(8). The experimentally determined rate expression for this is, Rate = k [H2][NO] The expression shows that this reaction is: (i) First-order with respect to H2 (i) Second-order with respect to NO 111) Third-order overall (1+2=3) Keep in Mind! S.Q. Give relation between rate equation and chemical equation. Ans. 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. Types of Reaction Orders Order Rate Law Example Zero Rate = k First Rate = k[A] Second Rate = k[A] Third Rate = k[A}[B] Quick Check 7.6 How is the order of reaction derived from the rate law? (a) The order of a reaction is derived by examining the exponents of the Ans. concentration terms in the rate law. Example: If the rate law is: Rate = k[A]™[B]" The order with respect to A is m The order with respect to B is n The overall order of the reaction is: m + n The order is not determined from the balanced chemical equation, but from experimental data. (b) Explain what is meant by the specific rate (rate constant) of a reaction and how is it represented in the rate equation? Ans. The specific rate or rate constant (k) is a proportionality constant that relates the reaction rate to the concentrations of reactants in the rate law. In the rate law: Rate = k[A]"[B]" k is the rate constant Its value depends on temperature and the nature of the reaction Units of k vary depending on the overall order of the reaction It shows how fast a reaction proceeds at a given temperature when the reactant concentrations are known. Types of reaction order (i) Zero order reaction Definition: The rate of a zero order reaction is independent of the concentration of the reactants. Rate law Rate=k[A]°=k Since anything raised to the power of 0 is 1, the rate = constant. (k) Chang in the concentration of reactants does not change the reaction rate. Examples of zero order reactions (1) # 2(g) + Cl 2(g) = sumien → 2HCl(g) (in presence of light): • A photochemical reaction, where the rate is controlled by light intensity, not concentration. Effect of Doubling Reactant) No change in rate Rate doubles Rate quadruples More complex, depends on both terms ng → N2(g) + 3H 2(g) (on hot platinum surface): (11) 2NH 3(g) = • Surface-catalyzed reaction; rate depends on the availability of surface sites, not [NH3]. • Photochemical reactions usually zero order. • Often seen in photochemical and enzyme-catalyzed reactions at saturation. (il) First Order Reaction Definition: A reaction for which some of the exponents of the concentrations in the rate equation is 1. Rate law Rate=k[A]=k[A] This means doubling [A] doubles the rate, tripling [A] triples the rate, and so on. In these reactions, there may be multiple reactants present but concentration of only one reactant affects the rate, even if others are present. The reaction follows exponential decay of the reactant over time. Common in decomposition and radioactive decay processes. Example Decomposition of nitrogen pentoxide is an example of first order as it rate of reaction depends upon one molecules. 2N,05(g) →2N04(8) + 02(g) Rate law Rate=k[N2Os] This equation suggests that the reaction is first order. Overall order = 1 All radio-active disintegration reactions are first order reactions. (iii) Second order reactions Definition A reaction for which sum of the exponents of the concentrations in the rate equation is 2. A second order is that reaction whose rate depends upon the concentration of one reactant raised to the second power or on the concentrations of the two different reactants, each raised to the first power. Hypothetically, it can be expressed as: Rate laws There are two common forms: (a) Single reactant Rate=k[A]2 (b) Two different reactants Rate=k[A]BJ • Doubling [A] in Rate=k[A] increases rate by a factor of 4. • In Rate=k[A][B], doubling both. A and B increases rate by a factor of 4. Units of k mol-1 dms! Oxidation of nitric oxide: Oxidization of nitric acid with ozone is first order with respect to NO and first order with respect to O3. The sum of individual orders gives the overall order of reaction. NO (8) + 03(8) → NO 2(g) + 02(g) Rate law Rate=k[NO][O3] First order with respect to NO, first order with respect to O3, overall order = 2. (iv) Third-Order Reactions Definition: A third order reaction is the reaction for which sum of the exponents of the concentration in the rate equation is 3. Forms of Rate Law Single reactant Rate=k[A] Two reactants Rate=k[A] [B] Three reactants Rate=k[A][B][C] Examples: The reaction involves eight reactant molecules but experimentally it has been found to be a third order reaction. 2Fel 2(ag) + 6KC (ag) + 12 2FeCls(ag) +6Kl (ag) Experimental rate law Rate=k[FeCI][KI]? Order with respect to FeCl = 1, order with respect to KI = 2, overall order = 3. 2NO(g) +02(g) →2N02(g) Rate law Rate=k[NO][02] Overall order = 3 (V) Fractional Order Reactions Definition: A reaction in which sum of the exponents of rate equation is in fraction, is called the fractional order reaction. Example: Consider the formation of carbon tetrachloride. CHCl 3(6) + Cl 2(g) CCl 4(0) + HCl (g) Rate law Rate=k[CHCh][C½]* Order with respect to CHC = 1 Order with respect to Cl = ½ Overall order = 1 + ½ = 1.5 Reactions involving free radicals frequently exhibit fractional orders. Fractional orders often indicate that the reaction mechanism is complex, typically involving free radicals. In this reaction Cl dissociates into Cl radicals, which then initiate the reaction with CHCI. Because the formation of Cl is in dynamic equilibrium, the concentration of Cl affects the rate non-linearly, leading to the ½ order. UNITS OF RATE CONSTANT

Units of rate constant The rate constant is specific for a particular reaction at a certain temperature. Reason: Concentrations are in mol dm3 and the reaction rate is in units of The units for k s' the order depend on the order of the reaction and the units of time. when mole mol dm3 of reaction, the number of molecules at which rate actually depends will charge so, on it of rate constant will also change. General equation: Rate = [Reactants]" Rate k= (Reactants)" kn = (mol. dm3)I-" s kn = (concentration) -ng-1 This equation can be used to determine units of any order of reaction. (i) Units of k of zero order For a zero order reaction (n = 0), ko = (mol. dm3)1-05! ko = mol. dm 's! (ii) Units of k for first order For a first order reaction (n = 1), the rate is directly proportional to the concentration of one reactant. ki = (mol. dm3)1- 5' ki = (mol. dm3)0s1 Therefore, the units of k for a first order rate constant are s (iii) Units of k for second order For a second order reaction (n= 2), k2 = (mol. dm3) - s - s' . k2 = (mol-1 k2 = dm mol s The units of k for a second order rate constant are d? mol-1 sil (iv) Units of k for third order For a third order reaction (n = 3), K3 = (mol. dm ) -3 k3 = mol2 dm' sl k3 = dm mol? sl Therefore, the units of k for a third order rate constant are dm mol? sl Quick Check 7.7 (a) Consider the following rate expression rate = k [NO|? [NH3]° rate = k [BrOз] [Br][H+]2 (i) Calculate the overall order of reactions (ii) What are the orders with respect to each reactant in the expression. Ans. (i) Rate = k[NO|[NH3]° • Order with respect to NO = 2 • Order with respect to NH3 = 0 Overall order = 2 + 0 = 2 (ii) Rate = k[BrO ][Br ][H+]2 • Order with respect to BrO: = 1 • Order with respect to Br = 1 • Order with respect to Ht = 2 Overall order = 1 + 1 + 2 = 4 where n = order of reaction (moldm " )s' = (moldm " )" (b) Why is the sum of the coefficients of a balanced chemical equation not necessarily the order of a reaction? The order of a reaction is determined by experiment, not by the stoichiometric coefficients in the balanced equation. Reason • The balanced chemical equation shows what reacts and in what amounts, but not how the reaction actually occurs. • The rate law depends on the slowest (rate-determining) step, which may involve only some of the reactants. DETERMINATION OF RATE CONSTANT

Determination of rate constant The rate constant (k) of a reaction can be calculated using the following two methods: (i) Initial Concentration Method Reaction of H202 and I ion In the presence of hydrogen ions, hydrogen peroxide H202, reacts with iodide ions to form water and iodine. Н, 02(ag) + 21(ag) + 2H (ag) 2H 0(r) + 1 2(ag) The rate equation for this reaction is to rate of reaction =k|H202] F The progress of the reaction can be followed by measuring the initial rate of formation of iodine. Table 7.2 shows the rates of reaction obtained using various initial concentrations of each reactant. Calculation of rate constant The procedure for calculating k is shown below, using the data for experiment 1. Step 1: Write out the rate equation rate of reaction= [020 Step 2: Rearranged the equation in terms of k k= Step 3 Substitute the values k= (0.0200)x (0.0100) k= 1.75 × 102 dm mol-1 51 Table 7.2: Effect of change in concentrations of reactants on the rate of reaction: || ] moldm |H202| moldm Experiment 1 0.0200 0.0100 0.0300 2 3 0.0050 0.0200 The concentration of hydrogen ions is ignored because [Ht] does not appear in the rate equation. The reaction is zero order with respect to [H+]. rate 3.50x10-6 IH T Initial rate of moldm reaction moldm s' 0.0100 3.5 × 10-6 0.0100 5.3 × 10-6 0.0100 0.0200 1.75 × 10-6 (ii) Half-life method Definition: Half-Life (t1/2) is "the time taken for the concentration of a reactant to fall to half of its original value" Measuring reaction order • Rate constant (k) is calculated by measuring half life time it takes for the concentration of reactant to decrease by half. • For first-order reactions, the half-life is independent of concentration and follows a fixed relationship. • For a first-order reaction, the rate constant and half-life are related by: k = 0.693 + ½ Decomposition of hydrogen peroxide Given • The half-life of hydrogen peroxide is 2 hours • Reaction is first-order • To calculate k, we first need to convert the half-life, which is 2 hours, into seconds 2 × 60 × 60 = 7200 s Put this value into the equation 0.693 K = - =9.6x105s 1 7200 Quick Check 7.8 with a half-life of 15 minutes. If the initial Consider a first-order reaction concentrations of the reactant is 0,100 mol dm, calculate the rate constant (k) for the reaction. Ans. For a first-order reaction, the half-life (t1/2) and rate constant (k) are related by the formula: 51/2 =- Given • t1/2 =15 minutes • Initial concentration is given, but not needed for calculating k in first-order reactions. Solution 0.693 0.693 = k = 15 min 4112 k = 0.0462 min If you want the rate constant in seconds, convert minutes to seconds: 15 min = 15x 60 = 900s 0.693 k = - 900 0.693 k - = 0.0462 min -=7.7x1045-1 REACTION MECHANISM

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 products. Role of reaction mechanism • The overall balanced equation shows only: Reactants and products. • The reaction mechanism shows: Actual steps (called elementary steps) through which reactants turn into products • Each elementary step represents a single molecular event, like: Bond breaking, bond forming and most reactions proceed through multiple steps, not just one. Each step is triggered by the collision of: Atoms, ions and molecules. Molecularity: The number of reactant molecules involved in an elementary step. Types of Molecularity (i) Unimolecular Reaction: A unimolecular elementary reaction involves only a single reactant molecule. Example: Decomposition of N2Os. 1= N,05(8) →→ NO 2(g) + NO3(8) (ii) Bimolecular Reaction: An elementary reaction involves two atoms, ions or molecules and is called bimolecular Examples → NO 2(g) + CO2(g) CO (8) + NO3(8) → NO2(g) + 02(g) NO (8) +03(8) - (ill) Termolecular Reaction: A termolecular reaction involves the simultaneous reaction of three molecules. • Involves three reacting molecules simultaneously • Rare due to the low probability of three particles colliding at once Example: formation of ozone in the atmosphere by reaction of oxygen molecule and atomic oxygen in smog formation. 202(g) +0(8) 03(8) +02(g) Intermediates Definition: They are short lived species that are produced mechanism and consumed in a subsequent step. They are not stable so they do not mansion in balanced chemical equation. Example: carbocation (CH3)3 Ct etc. Rate-determining step Definition: In many reactions, one step is significantly slower than all of the others. This step is called rate-determining step. • Controls the overall rate of the reaction. Balanced equation is net result of sum of all individual steps. • Any step after the Rate determining steps does not affect the rate. • Only reactants involved in the Rate determining steps appear in the rate law. • All the reactants appear in rate determining step will also appear in the rate equation. • Determines the order of the reaction. • Even if the overall reaction involves several steps, the rate-determining step acts like a bottleneck • The rate equation is written based only on the rate-determining step. • This helps chemists predict or control the speed of complex reactions. Example: The overall reaction between NO and H2 is given. 2NO(8) + 2H2(g) 211, (g) + 2(g) Then the rate law would be: Rate=k[NO}[H2]' Even though 2NO and H2 are used in the overall reaction, only one of each appears in the rate law because they are involved in the rate determining step. Table 7.3: Effect of change in concentrations of reactants on the rate of reaction [H2] in (mol dm3) [NO] in (mol dm3) 0.006 0.006 0.006 - 0.003 0.001 0.002 0.003 Experimental setup Six experiments were conducted to determine how the rate of reaction depends on the concentration of H2 and NO. The general form of the reaction: Rateo[H2]*[NO] Effect of H2 Concentration In these experiments, [NO] is kept constant, and only [Ha] is varied Doubling [H2] then Rate doubles Tripling [H2] the Rate triples Rateo[H2]! The reaction is first order with respect to H2. Effect of NO Concentration Here, [H2] is constant, and [NO] is varied. Doubling NO] results Rate increases 4 times Tripling [NO] results Rate increases 9 times Conclusion Rate [NO]- The reaction is second order with respect to NO The overall rate equation of reaction is, Rate oc [H2][NO] Initial rate (atm min) 0.025 0.001 0.050 0.002 0.0075 0.0063 0.025 0.009 0.056 0.009 or Rate = k[H½]'[NO] 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: 2N0 (g) + H 2(g) Fast →2H,0(g) Step 2: H,02(g) + H 2(g) Reaction of F2 and NO2 The step 1 is slow and rate determining. The reaction between nitrogen dioxide and fluorine gas: 2NO2(g) + F2(g) →→2NO,F(8) This reaction is first order in NOz, 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: Slow > NO,E(g) + E(8) NO 2(g) + E2(g) Fast →NO, F(g) NO 2(g) + F(g) 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 An acidified solution of hydrogen peroxide reacts with iodide ions. H, 02(a9) + 2H (aq) + 21 (ag) The rate equation for this reaction is rate = k[H202] |7 The mechanism below has been proposed for this reaction. H2O, + I - Slow > H2O +10- Fast → HIO H+ + 10 - HIO+ H* +T - Fast >1, + H2O Explain why this mechanism is consistent with the rate equation. Ans. Explanation: The rate-determining step (slow step) is the first step: H2O,+Г →H2O+I0™ The rate of the overall reaction depends on the slowest step. The rate law from the slow step is: Rate = k[H2O,][r] • This matches exactly with the experimental rate law given. • The other steps are fast, so they do not affect the rate law. • Therefore, the mechanism is consistent with the rate equation because the slow step includes only H202 and I, and the rate depends only on their concentrations. (rate determining step) Rate = kı[NOz][F2] Rate = kz[NOz][F] SAMPLE PROBLEMS Sample problem 7.1 The reaction for the formation of ammonia in Haber process is: N2(ag) + 3H 2(g) → 2NH 3(g) Calculate the instantaneous rate after i. 1.0 min What is the average rate of ii. production of ammonia for the system, between 1.0 and 4.0 minutes? Solution: The instantaneous rate at 1.0 min can be calculated as Instantaneous Rate = AC 2.7mol.dm3 = -2.7 mol dm min 1mın At If the concentration of ammonia is 3.5 moldm after 1.0 min and 6.2 mol:dm " after 4.0 minutes? 4C=4 NH,] = (6.2 - 3.5)mol.dm; AC = 2.7mol.dm3 At = (4.0-1.0);= 3.0 min Rate of formation of NH3 = 0.90 mol dm min' AC_ A[NH,] _ 2.0moldm" 3 min At At The rate of production of NH3 gas over the given time interval is 0.90 mol dm min 1 Sample problem 7.2 The first-order reaction cyclopropane to propene, for which the half-life is 17.0 min, calculate the rate constant of this reaction. Solution: Step 1 convert minutes to seconds Step 2 substitute the half-life into the expression. 0.693 k = 4112 0.693 k = Im. 17× 60s = 6.79×104 di mol-1s 1