Chemical Kinetics - Measuring the Rate of Reaction PDF

# Chemical Kinetics - Measuring the Rate of Reaction The speed, or rate of a chemical reaction can be measured by one of two ways: - the disappearance of reactants, or - the appearance of products Either will yield a quantitative measure of the rate of a reaction. There are many methodologies for measuring the appearance or disappearance of reactants/products. The state of the involved chemicals is what determines how to measure the rate. i.e. gas, liquid, solid, or aqueous. A homogeneous system means that all reactants are present in the same state. A heterogeneous system means that reactants are present in different phases. Reactions that produce gases can be measured using gas collection techniques. Reactions that involve ions can be measured with conductivity meters, or pH meters if acids/bases are involved. Reactions that involve colour changes can be measured using a spectrophotometer. | Method | Description | | -------- | -------- | | Gas collection | Measures the volume of gas produced | | Conductivity meter | Measures the conductivity of the solution, which changes as ions are produced or consumed | | pH meter | Measures the pH of the solution, which changes as acids or bases are produced or consumed | | Spectrophotometer | Measures the absorbance or transmittance of light through the solution, which changes as the concentration of colored species changes | Recall that a rate of anything is the change in X in a given period of time: $Rate = \frac{\Delta x}{\Delta t}$ Chemicals can be measured with a variety of quantitative measures: - Mass (grams) - Volume (Litres) - Moles - Concentration (mol/L) Concentration will be used most often in this course. Therefore: $average\ rate\ of\ reaction = \frac{change\ in\ concentration}{change\ in\ time}$ or: $r= \frac{\Delta c}{\Delta t}= \frac{moles\ per\ liter}{seconds} = \frac{mol}{L \cdot s}$ If we measure the disappearance of reactant, Δc will yield a negative number (decreasing concentration). Naturally, measuring the appearance of product will yield a positive Δc (increasing concentration). # Chemical Kinetics: Reaction Rates The rate of reaction refers to how quickly or slowly reactants are consumed or products are formed in a reaction. $Average\ Reaction\ Rate = \frac{Amount\ of\ Reactant\ Consumed}{Time}$ $Average\ Reaction\ Rate = \frac{Amount\ of\ Product\ Produced}{Time}$ For example, for the decomposition of water: $2H_2O (l) \to 2H_2 (g) + O_2 (g)$ The rate of reaction can be expressed in several ways: - 2.0 mol of water is consumed per hour (r = 2.0 mol/h or 2.0 mol h⁻¹) - 2.0 mol of hydrogen is produced per hour (r = 2.0 mol/h) - 1.0 mol of oxygen is produced per hour (r = 1.0 mol/h) - 32.0 grams of oxygen is produced per hour (r = 32.0 g/h) - 44.8 L of hydrogen is produced per hour at STP (r = 44.8 L/h) It is common to calculate the rates as the concentration change per unit time: $Average\ Reaction\ Rate = \frac{Concentration\ Change}{Time} = r = \frac{\Delta c}{\Delta t}$ # Analyzing the Progress of a Reaction ## 2A --> B | The Change in Concentration of Substance B Over Time <start_of_image> scale | Time (s) | [B] (mol/L) | |---|---|---| | 0 | 0.0 | | 2 | 0.1 | | 4 | 0.2 | | 6 | 0.3 | | 8 | 0.4 | | 10 | 0.5 | | 12 | 0.6 | | 14 | 0.7 | | 16 | 0.8 | | 18 | 0.9 | <start_of_image> scale | Time (s) | [B] (mol/L) | |---|---|---| | 0 | 0.0 | | 2 | 0.1 | | 4 | 0.2 | | 6 | 0.3 | | 8 | 0.4 | | 10 | 0.5 | | 12 | 0.6 | | 14 | 0.7 | | 16 | 0.8 | | 18 | 0.9 | 1. What happens to the rate of reaction over time? 2. What is the average rate? 3. What is the rate at 6 s? 10 s? 4. What would the rate of this reaction be in terms of [A]? ## X --> Y | The Change in Concentration of X Over Time | scale | Time (s) | [X] (mol/L) | |---|---|---| | 0 | 4.0 | | 2 | 3.8 | | 4 | 3.5 | | 6 | 3.0 | | 8 | 2.5 | | 10 | 2.0 | | 12 | 1.5 | | 14 | 1.0 | | 16 | 0.5 | 1. What happens to the rate of reaction over time? Explain. 2. What is the average rate over 15 s? 3. What is the rate at A? At B? # Rate Laws and Order of Reaction -a mathematical relationship between reaction rate and the various factors that affect the reaction rate ## Rate Law ... the rate of a reaction is exponentially proportional to the product of the initial concentrations of the reactants For aX + bY --> product(s) r c [X]m[Y]n and therefore, r = k[X]m[Y]n **k = a rate constant (p 372) Note: the "m" and "n" describe the mathematical dependence of rate on initial concentrations... they are the "ORDERS OF REACTIONS" ## Overall Order of Reaction ... the sum of the exponents in the rate law Equation ## Example If you are given..... r = k[X]m[Y]n[Z]o Because r ∝ [X]m - if the initial concentration of X is doubled, the rate will double (multiply by 2m) - if the initial concentration of X is tripled, the rate will multiply by 3 (i.e. 3m) Because r ∝ [Y]n - If the initial concentration of Y is doubled, the rate will multiply by 4 (2n) - If the initial concentration of Y is tripled, the rate will multiply by 9 (i.e. 3n) Because r ∝ [Z]o - If the initial concentration of Z is doubled, the rate will multiply by 1 (2o) - If the initial concentration of Z is tripled, the rate will multiply by 1 (i.e. 3o) ***i.e. unchanged... because the rate does not depend on the initial concentration of Z, the rate for this reaction will be r = k[X]m[Y]n HWK: read p 372-377 practice #2-6(p 377) and 7-8(p 381) section 6.3 (p 381) # 2-4 # Rate Law Equation Problem Set ## Part 1: Rate Law Equation Practice 1. For a reaction where the rate equation is r = k[NH4+(aq)][NO2(aq)], a) calculate k at temperature T₁, if the rate, r, is 2.40x10⁷ mol/L.s) when [NH4+(aq)] is 0.200 mol/L and [NO2(aq)] is 0.00500 mol/L. b) calculate r at temperature T2, if the rate constant, k, is 3.20x10⁴ L/(mol·s) when [NH4+(aq)] is 0.100 mol/L and [NO2(aq)] is 0.0150 mol/L. 2. A series of experiments is performed for the system 2A + 3B + C --> D + 2E When the initial concentration of A is doubled, the rate increases by a factor of 4. - When the initial concentration of B is doubled, the rate is doubled. - When the initial concentration of C is doubled, there is no effect on rate. a) What is the order of reaction with respect to each of the reactants? b) Write an expression for the rate equation. 3. The experimental observations in Table 4 are obtained for the reaction 2A + B + 2C --> 3X | Trial | Initial [A] (mol/L) | Initial [B] (mol/L) | Initial [C] (mol/L) | Rate of production of X (mol/L·s) | |---|---|---|---|---| | 1 | 0.10 | 0.10 | 0.10 | 3.0 x 10⁻⁴ | | 2 | 0.20 | 0.10 | 0.10 | 1.2 x 10⁻³ | | 3 | 0.10 | 0.30 | 0.10 | 3.0 x 10⁻³ | | 4 | 0.20 | 0.10 | 0.20 | 2.4 x 10⁻³ | (a) What is the order of reaction with respect to each of the reactants? (b) Write an expression for the rate equation. (c) Calculate a value for the rate constant. (d) Calculate the rate of production of X when [A] = [B] = [C] = 0.40 mol/L. ## Part 2: More Rate Law Expressions 1. Consider the data for hydrogen concentration [H2], iodine concentration [I2] and rate of reaction (moles per litre per second or mol·L⁻¹/s) for this reaction: H2(g) + I2(g) --> 2 HI(g) | Trial | [H2] (mol/L) | [I2] (mol/L) | Rate (mol·L⁻¹/s) | |---|---|---|---| | 1 | 0.01 | 0.05 | 0.04 | | 2 | 0.02 | 0.05 | 0.08 | | 3 | 0.03 | 0.05 | 0.12 | | 4 | 0.05 | 0.01 | 0.02 | | 5 | 0.05 | 0.02 | 0.16 | | 6 | 0.05 | 0.03 | 0.54 | a) Determine the rate law expression for this reaction. b) What is the overall reaction order? c) What would happen to the rate of the reaction if the concentration of both reactants was doubled? 2. Consider a hypothetical reaction: A + B --> C + D Doubling the concentration of A causes the reaction rate to increase by a factor of four. This is done while the concentration of B is held constant. Tripling the concentration of B, while the concentration of A is held constant, causes the reaction rate to increase by a factor of nine. a) What is the rate law expression for this reaction? b) What would happen to the reaction rate if the concentration of A was tripled and the concentration of B was doubled simultaneously? 3. Consider the reaction: NH4+ (aq) + NO2 (aq) --> N2(g) + 2 H2O (l) | Experiment | [NO2] (mol/L) | [NH4+] (mol/L) | Rate (mol.L⁻¹/s x 10⁻⁷) | |---|---|---|---| | 1 | 0.0100 | 0.200 | 5.4 | | 2 | 0.0200 | 0.200 | 10.8 | | 3 | 0.0400 | 0.200 | 21.6 | | 4 | 0.200 | 0.0202 | 10.8 | | 5 | 0.200 | 0.0404 | 21.6 | | 6 | 0.200 | 0.0606 | 32.4 | Determine the rate law expression for this reaction. 4. Consider a hypothetical reaction: A + B --> C | Experiment | [A] (mol/L) | [B] (mol/L) | Rate (mol L⁻¹/s x 10⁻⁵) | |---|---|---|---| | 1 | 0.100 | 0.100 | 4.0 | | 2 | 0.100 | 0.200 | 4.0 | | 3 | 0.200 | 0.100 | 16.0 | a) Determine the rate law expression for this reaction. b) What is the order of reaction with respect to A? to B? c) What is the overall reaction order? 5. In a reaction involving only one reactant, A, the rate of the reaction increases by a factor of 27 when the concentration of A is tripled. What is the rate law expression for this reaction? # The Rate Law and Reaction Mechanisms The rate law equation provides a quantitative description about how the concentration of reactants (in the gas or aqueous state) affects the rate of reaction. These relationships are usually determined by experiment. In some reactions, the reactant concentration does not appear to affect the rate at all. These "zero order" reactants are difficult to rationalize by the collision theory, which tells us that higher reactant concentrations should lead to more collisions and hence a faster rate. The existence of zero order reactants can be understood by examining reaction mechanisms. ## One-Step (Elementary Step) Reactions One step reactions cannot be explained in terms of simpler steps since they involve the direct collision of reactant. In these reactions, the order of each reactant is determined by the coefficient of each reactant. Since a coefficient cannot be zero, there cannot be a zero order reactant in an elementary step. Ag+ (aq) + Cl- (aq) --> AgCl (s) r = k[Ag+][Cl-] ## Multi-Step Reactions Many chemical reactions we study actually occur as a series of simpler steps. A reaction mechanism is simply the series of elementary steps by which the overall reaction occurs. Consider the experimentally determined rate law equation for this reaction: NO2(g) + CO(g) --> NO(g) + CO2(g) r = k[NO2]²[CO] .. r = k[NO2]² Since the rate law equation does not include [CO], this reactant must be zero order. This means that it must be a multi-step reaction. This reaction has been studied and is thought to involve two elementary steps: Step 1: NO2(g) + NO2(g) --> NO3(g) + NO(g) slow Step 2: NO3(g) + CO(g) --> NO2(g) + CO2(g) fast Net: NO2(g) + CO(g) --> NO(g) + CO2(g) Note that the reaction intermediate NO3 has been cancelled out when the equations are added. Since the slowest elementary step (called the rate determining step) is the only one that affects the overall reaction rate, only the reactants in this step will appear in the rate law equation for the overall reaction. # Reaction Mechanism Questions 1. In the ozone layer, UV radiation is absorbed, converting ozone to oxygen gas. The proposed mechanism is shown below: O3 + UV light --> O3 + O O3 + O --> 2O2 (fast) (slow) a) Write the overall balanced equation. b) What is the reaction intermediate? 2. A proposed mechanism for the reaction between iodine chloride gas and hydrogen gas is shown below. ICI(g) + H2(g) --> HI(g) + ICl(g) HI(g) + ICl(g) --> HCl(g) + I2(g) (slow) (fast) a) Write the balanced equation for the overall reaction. b) What are the reactions intermediates (if any)? c) The rate law equation was determined experimentally to be r = k[ICI][H2]⁰. Does the proposed mechanism above agree with the experimental results? If it does not agree, explain why not. 3. A two-step reaction is shown below. I2(g) --> 2I(g) H2(g) + 2I(g) --> 2HI(g) (slow) (fast) a) What is the overall reaction? b) What are the reaction intermediates (if any)? c) What step would you expect to be the rate-determining step? d) Write the rate law equation if the proposed mechanism is correct. 4. Consider the following reaction that occurs in the atmopshere on a smoggy day: NO(g) + O3(g) --> NO2(g) + O2 (g) r = k[NO] 03 Which of the following mechanisms is consistent with this rate law? Explain. a) NO + O3 --> NO2 + O2 b) NO + O3 --> N + O + O2 N + O3 --> NO2 + O O + O --> O2 (slow) (slow) (fast) (fast) NO + O3 --> NO2 + O2 (overall) c) O3 --> O2 + O O + NO --> NO2 NO + O3 --> NO2 + O2 (slow) (fast) (overall) # The Collision Theory If a reaction is going to take place: 1. The reactants must collide. 2. The reactants must collide with sufficient energy for a reaction to occur (the activation energy threshold must be overcome). 3. The reactants must collide with the correct orientation. ## Collision Geometry Even if molecules with sufficient energy collide, a reaction may still not take place. Even high-speed collisions may not be effective if the particles involved are not orientated properly. e.g. CO2 + H2O --> H2CO3 --> HCO3- + H+ **Diagram 1** Collision occurs because the molecules have the correct collision geometry and sufficient energy. After forming the activated complex (a) (H2CO3), the reaction can be completed. **Diagram 2** The molecules collide but are not oriented properly, therefore no reaction takes place. # Rates of Reaction ## Reactants | Products | Succeessful Collisions # Temperature and Reaction Rate If the temperature of a reaction system is increased, the molecules will move more quickly and each molecule will possess more energy. It therefore follows that i) more collisions will take place and ii) the collisions which do take place will be more effective. Therefore increasing the temperature of a reaction system will increase the rate of the reaction and vice versa. ## Kinetic Energy Distribution Curves - Maxwell Boltzmann Curves At a specific temperature, not all the molecules of a reaction system will have the same kinetic energy. A graph of these energies is called a Kinetic Energy Distribution Curve or Maxwell Boltzmann Curve. As a sample of matter is heated, the average kinetic energy of the molecules increases, and the the distribution therefore moves to the right: # Potential Energy Diagrams ## Endothermic Reactions - convert kinetic energy into potential energy (Ep) (decrease temperature) - Ep increases as molecules approach; if they molecules have less than the required threshold (activation) energy, no reaction will occur - if the molecules have enough kinetic energy, the activated complex can form (highest Ep); the activated complex breaks down to form the either the stable reactants or the stable products H2 (g) + I2 (g) + kinetic energy --> 2 HI (g) ## Exothermic Reactions - convert potential energy into kinetic energy (increase temperature) CO (g) + NO2 (g) --> CO2 (g) + NO (g) + kinetic energy # The Effect of a Catalyst ## K.E Distribution Curve | Potential Energy Diagram -a catalyst lowers the Ea.... therefore, more molecules have enough K.E. to react.... therefore, there is a faster reaction ***there is NO effect on ΔΗ -a catalyst provides an alternative pathway with an activated complex of lower P.E. -this lowers the Ea, for both the forward and reverse reactions # Effect of a Temperature Change ## K.E Distribution Curve | Potential Energy Diagram -regardless of the temperature, the Ea, is the same -at higher temperatures, more molecules have the required K.E. for the reaction.... therefore, a faster reaction ***the number of molecules under the curses are equal ## Interpreting Potential Energy Diagrams Practice **Graph 1** 1. What type of reaction (exothermic or endothermic) does the graph represent? 2. Calculate the following: i) ΔΗ = ii) Ea = 3. Would you predict this reaction to happen at room temperature? List 2 reasons for your answer. 4. Calculate the ΔΗ and Ea for the reverse reaction. **Graph 2** 1. What type of reaction (exothermic or endothermic) does the graph represent? 2. Calculate the following: i) ΔΗ = ii) Ea = 3. Would you predict this reaction to happen at room temperature? List 2 reasons for your answer. 4. Calculate the ΔΗ and Ea for the reverse reaction. **Graph 3** Consider the following reaction: A --> B --> C --> F i) What does B represent in the reaction? ii) Is the overall reaction exothermic or endothermic? iii) On the graph, draw a line to represent the net enthalpy change (heat of reaction). iv) What does [X] represent? v) Which step would you expect to be the rate-determining step? Why? # Rates, Temperature and Potential Energy Diagrams Problem Set Read p. 383-390 of Nelson Chemistry 12. Do Q. 1 (p. 387) and 3 (p. 391). ## Problems: 1. Which diagram is described by each of these statements? .. Consider the potential energy diagrams below: (1) Ep Reaction Path (2) Ep Reaction Path (3) Ep Reaction Path .. a) an exothermic reaction that is unlikely to occur at room temperature. b) an endothermic reaction. c) the activation energy (Ea) is greater than the energy released (ΔH) d) a spontaneous exothermic reaction. 2. Draw a potential energy (Ep) diagram for a reaction in which ΔH = -80 kJ/mol and Ea = +28kJ/mol. Label the axes, activation energy, ΔH, site of the activated complex, reactants and products. 3. Using the potential energy diagrams for an endothermic and exothermic reaction shown, choose the letter that best fits each statement. .. Reaction I | Reaction 2 ..Ep | ..Ep | a) Ep of the reactants b) Ep of the products c) AH of the reaction d) activation energy of the forward reaction e) site of activated complex f) Ep of the activated complex g) activation energy of the reverse reaction 4. In the following reaction, the enthalpy change of the forward reaction is ∆H = -36 kJ/mol and the activation energy for the forward reaction is 73 kJ/mol. A + B --> C + 36 kJ a) Draw a potential energy diagram for the the reaction. b) What is the activation energy of the reverse reaction? 5. The activation energy of a forward and reverse reaction are as follows: i) C2H4 (g) + H2 (g) --> C2H6 (g) Ea = 180 kJ/mol ii) C2H6 (g) --> C2H4 (g) + H2 (g) Ea = 317 kJ/mol a) Draw a potential energy diagram for this reversible reaction. b) Calculate the enthalpy change (ΔH) for each reaction. 6. Compare these reactions: i) C2H5Cl (1) --> C2H4 (g) + HCI (g) Ea = 254 kJ/mol ii) C2H5Br (1) --> C2H4 (g) + HBr (g) Ea = 219 kJ/mol Which of these two reactants would decompose more rapidly under the same reaction conditions and temperature? Explain your response. 7. A reaction consists of two steps as follows: Step 1: A + B --> AB Ea (kJ/mol) | ΔΗ (kJ/mol) + 45| - 72 Step 2: A + E --> C + E + 80| +28 a) Write the overall reaction equation. b) What is meant by the "rate-determining step"? c) Which of these steps is the rate-determining step? Why? d) What is the effect on the overall rate of increasing the concentration of A? What is the effect on the overall rate of increasing the concentration of B? Explain. 8. Examine the two Boltzmann distributions showing the distribution of kinetic energy possessed by the reactant molecules in two different reactions and the activation energy (Ea) for each reaction. a) Which reaction will be fastest at room temperature? Explain b) When the temperature eis increased to 60°C, what will happen to the rate of (i) Reaction 1 and (ii) Reaction 2? Why? c) In which case does the activation energy requirement change when the temperature is increased? 9. For the reaction: H2 + Cl2 --> 2 HCI + 184 kJ the activation energy for the process is 156 kJ/mol. What is the activation energy for the decomposition of HCI to produce H2 and Cl2? 10. Using the concepts of kinetic energy and surface area, explain why sugar cubes can be used in hot coffee but granulated sugar is preferable for making iced tea. 11.Given the kinetic energy distribution curves and threshold energy (Ea) for reactions A and B: ... a) Which reaction will be faster at room temperature? b) Which reaction will show the greatest increase in the rate of reaction if the temperature is increased? * Additional problems 12 - 14 available on-line (filename: more potential energy problems.doc) # Part 2: Rate Law Problems 1. Given the reaction: 2 NO(g) + O2(g) --> 2 NO2(g) the following data was collected. | | Initial Concentration | Initial Rate | |---|---|---| | | [NO] (M) | [02](M) | (M/s) | | I | 0.0010 | 0.0010 | 7.10 | | II | 0.0010 | 0.0040 | 28.4 | | III | 0.0030 | 0.0010 | 63.9 | | IV | 0.0020 | 0.0030 | ? | a) Determine the rate law expression. b) What would happen to the rate if the concentrations were changed to those in experiment IV? c) Calculate the initial rate of the reaction for experiment IV. d) Identify the order of the reaction with respect to i) O2 ii) NO iii) overall. 2. Given the reaction (CH3)3CBr + OH --> (CH3)3COH + Br the following data was collected. | | Initial Concentration | Initial Rate | |---|---|---| | | [(CH3)3CBr](M) | [OH](M) | (M/s) | | I | 0.10 | 0.10 | 1.0x10⁻³ | | II | 0.10 | 0.20 | 1.0x10⁻³ | | III | 0.20 | 0.10 | 2.0x10⁻³ | | IV | 0.30 | 0.30 | ? | a) Determine the rate law expression for the reaction. b) By what factor has the rate changed in experiment IV? c) Calculate the rate of the reaction in experiment IV. d) What is the order of the reaction in terms of i) (CH3)3CBr? ii) OH-? iii) the overall reaction? # Part 3: Factors Affecting the Rate of Reaction 1) Which of the following three reactions is likely to be most rapid? Why? Which is likely to be the slowest? Why? a) Cr²+(aq) + Fe³+(aq) --> Cr³+(aq) + Fe²+(aq) b) Cu(s) + 2Ag+(aq) --> 2Ag(s) + Cu²+(aq) c) CO(g) + NO2(g) --> CO2(g) + NO(g) 2) Consider the following reaction:C25H52 + 38O2 --> 25CO2 + 26H2O a) 55g of C25H52 reacts in 20 minutes. Calculate the rate of, the reaction in terms of moles of C25H52 reacted per hour. b) Calculate the rate of the reaction in terms of moles of CO2 produced per minute. 3) For each of the following, state whether-they are directly or inversely proportional to each other: a) concentration and reaction rate b) surface area and reaction rate c) temperature and time 4) Explain what a catalyst is and provide one example of a useful catalyst. 5) A+ B --> C+ D For the above hypothetical reaction, doubling the concentration of A, while the concentration of B is held constant, causes the reaction rate to increase by a factor of four. Tripling the concentration of B, while the concentration of A is held constant, causes the reaction rate to increase by a factor of twenty-seven. a) What is the rate law expression for this reaction? b) What would happen to the reaction rate if the concentration of A was tripled and the concentration of B was doubled simultaneously? c) What is the overall reaction order? 6) A cube takes 15 minutes to dissolve in a liquid. Each side of the cube is 10 cm long. An identical cube is divided into 1000 smaller cubes. They take 1.5 minutes to dissolve. What would be the dimensions of each of the small cubes? 7) A + B --> C rate = k[A]² The rate of the' reaction is 1.6 x 10⁵ M/sec when the concentration of A is 0.200 M and the concentration of B is 0.100 M. Calculate the rate of the reaction when the concentration of A is 0.490 M and the concentration of B is 0.575 M. # Answers: 1. a) Fastest (e- tranfer only) b) or d) Slowest (bonds are broken in solid copper) 2. a) 0.48 mol/h 5. a) r = k[A][B]³ b) 0.20 mol/min 3 a) directly n) directly c) inversely b) Rate increases 72-fold 6. 1 cm x 1cm x 1cm 7. 9.6 x 10⁵ M/s c) 5

Chemical Kinetics - Measuring the Rate of Reaction PDF

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