1 / 76

Reaction Rates and Equilibrium

Reaction Rates and Equilibrium. What is meant by the rate of a chemical reaction?. Can also be explained as the speed of he reaction, it is the amount of time required for a chemical reaction to come to completion. Different reactions take different times Burning Aging Ripening Rusting.

olathe
Download Presentation

Reaction Rates and Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Reaction Rates and Equilibrium

  2. What is meant by the rate of a chemical reaction? • Can also be explained as the speed of he reaction, it is the amount of time required for a chemical reaction to come to completion. • Different reactions take different times • Burning • Aging • Ripening • Rusting

  3. Collision Theory • Atoms, ions and molecules can react to form products when they collide provided that the particles are orientated correctly and have enough kinetic energy. • The relative orientations of the molecules during their collisions determine whether the atoms are suitably positioned to form new bonds. • Particles lacking necessary kinetic energy to react bounce apart when they collide. • Imagine clay balls… • The minimum amount of energy that particles must have in order to react is called the activation energy

  4. Activated Complex

  5. Activation Energy • Swedish Chemist Svante Arrhenius explored activation energy, Ea • He found that most reaction rate data obeyed an equation based on three factors: • The fraction of molecules possessing the necessary activation energy or greater • The number of collisions occurring per second • The fraction of collisions that have the appropriate orientation • Arrhenius equation k = Ae-Ea/RT

  6. Factors that Affect Reaction Rates • Temperature (Alkaseltzer activity) • Raise temperature, faster reaction rate • Lower temperature, slower reaction rate • Higher temperatures make molecules move faster because they have more kinetic energy so reaction is more likely • Particle Size/Surface Area (lycopodium demo) • The smaller the particle size the larger the surface area. • An increase in surface area, increases the amount of reactant exposed, which increases the collision frequency. • Can be dangerous… coal powder or gas particles reacting to our lungs.

  7. Catalyst (demo) • A substance that increases the rate of a reaction without being used up itself during the reaction. • They help reactions to proceed at a lower energy than is normally required. • Enzymes catalyze reactions in our body • An inhibitor interferes with the action of a catalyst

  8. Concentration • The number of reacting particles in a given volume also affects the rate at which reactions occur. • Cramming more particles into a fixed volume increases the concentration of reactants, the collision frequency and therefore, reaction rate.

  9. Rate Law: Effect of Concentration on Rate • One way of studying the effect of concentration on reaction rate is to determine the way in which the rate at the beginning of a reaction depends on the starting concentrations. a A + b B  c C + d D Rate = k[A]m[B]n • The exponents m and n are called reaction orders and depend on how the concentration of that reactant affects the rate of reaction.

  10. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH4+] doubles, the initial rate doubles.

  11. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO2−] doubles, the initial rate doubles.

  12. Concentration and Rate • This means Rate  [NH4+] Rate  [NO2−] Rate  [NH4+] [NO2−] which, when written as an equation, becomes Rate = k [NH4+] [NO2−] • This equation is called the rate law, and k is the rate constant. Therefore,

  13. Rate Laws • A rate law shows the relationship between the reaction rate and the concentrations of reactants. • The exponents tell the order of the reaction with respect to each reactant. • Since the rate law is Rate = k [NH4+] [NO2−] the reaction is First-order in [NH4+] and First-order in [NO2−].

  14. Rate Laws Rate = k [NH4+] [NO2−] • The overall reaction order can be found by adding the exponents on the reactants in the rate law. • This reaction is second-order overall.

  15. Multistep Mechanisms • In a multistep process, one of the steps will be slower than all others. • The overall reaction cannot occur faster than this slowest, rate-determining step.

  16. Slow Initial Step • The rate law for this reaction is found experimentally to be Rate = k [NO2]2 • CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. • This suggests the reaction occurs in two steps. NO2(g) + CO (g) NO (g) + CO2(g)

  17. Slow Initial Step • A proposed mechanism for this reaction is Step 1: NO2 + NO2 NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast) • The NO3 intermediate is consumed in the second step. • As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

  18. Fast Initial Step • The rate law for this reaction is found to be Rate = k [NO]2 [Br2] • Because termolecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br2(g) 2 NOBr (g)

  19. Step 1: NO + Br2 NOBr2 (fast) Fast Initial Step • A proposed mechanism is Step 2: NOBr2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions.

  20. Fast Initial Step • The rate of the overall reaction depends upon the rate of the slow step. • The rate law for that step would be Rate = k2 [NOBr2] [NO] • But how can we find [NOBr2]?

  21. Fast Initial Step • NOBr2 can react two ways: • With NO to form NOBr • By decomposition to reform NO and Br2 • The reactants and products of the first step are in equilibrium with each other. • Therefore, Ratef = Rater

  22. k1 k−1 [NO] [Br2] = [NOBr2] Fast Initial Step • Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2] • Solving for [NOBr2] gives us

  23. Rate = [NO] [Br2] [NO] k2k1 k−1 Fast Initial Step Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives = k [NO]2 [Br2]

  24. = −kt ln [A]t [A]0 Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us Where [A]0 is the initial concentration of A, and [A]t is the concentration of A at some time, t, during the course of the reaction.

  25. First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. ln [A]t = -kt + ln [A]0

  26. CH3NC CH3CN First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile.

  27. CH3NC CH3CN First-Order Processes This data was collected for this reaction at 198.9 °C.

  28. First-Order Processes • When ln P is plotted as a function of time, a straight line results. • Therefore, • The process is first-order. • k is the negative of the slope: 5.1  10-5 s−1.

  29. 1 [A]0 1 [A]t = kt + Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get also in the form y = mx + b

  30. 1 [A]0 1 [A]t = kt + 1 [A] Second-Order Processes So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k.

  31. NO2(g) NO (g) + O2(g) 1 2 Second-Order Processes The decomposition of NO2 at 300°C is described by the equation and yields data comparable to this:

  32. Second-Order Processes • Plotting ln [NO2] vs.t yields the graph at the right. • The plot is not a straight line, so the process is not first-order in [A].

  33. 1 [NO2] Second-Order Processes • Graphing ln vs. t, however, gives this plot. • Because this is a straight line, the process is second-order in [A].

  34. Half-Life • Half-life is defined as the time required for one-half of a reactant to react. • Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0.

  35. 0.5 [A]0 [A]0 ln = −kt1/2 0.693 k = t1/2 Half-Life For a first-order process, this becomes ln 0.5 = −kt1/2 −0.693 = −kt1/2 NOTE: For a first-order process, then, the half-life does not depend on [A]0.

  36. 1 k[A]0 1 [A]0 1 [A]0 1 [A]0 2 [A]0 1 0.5 [A]0 2 − 1 [A]0 = kt1/2 + = kt1/2 + = = kt1/2 = t1/2 Half-Life For a second-order process,

  37. Temperature and Rate • Generally, as temperature increases, so does the reaction rate. • This is because k is temperature dependent.

  38. Reversible Reactions • Until now, most of the reactions we have examined have gone completely to products. Some reactions are reversible meaning the reaction from reactants to products also goes from products to reactants at the same time. • Consider the following reaction: 2SO2 + O2 2SO3 In this reaction, SO2 and O2 are placed in a container. Initially, the forward reaction proceeds and SO3 is produced. The rate of the forward reaction is much greater than the rate of the reverse reaction. As SO3 builds up, it starts to decompose into SO2 and O2. The rate of the forward reaction is decreasing and the rate of the reverse reaction is increasing. Eventually, SO3 decomposes to SO2 and O2 as fast as SO2 and O2 combine to form SO3 (see Fig. 19.10 from text). When this happens, the reaction has achieved chemical equilibrium. • Chemical equilibrium is when the rate of the forward reaction = the rate of the reverse reaction. It says nothing about the amounts of reactants and products at equilibrium. Simulation

  39. Equilibrium Constants • When a system reaches equilibrium there is a mathematical relationship between the concentrations of the products and the concentrations of the reactants. • Pure solids and pure liquids (including water) do not appear in the equilibrium expression • When a reactant or product is preceded by a coefficient, its concentration is raised to the power of that coefficient in the equilibrium expression. • aA + bB cC + dD • Kc = [C]c * [D]d [A]a * [B]b

  40. Write the equilibrium expression for Kcfor the following reactions: Solution Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each. Plan: Using the law of mass action, we write each expression as a quotient having the product concentration terms in the numerator and the reactant concentration terms in the denominator. Each concentration term is raised to the power of its coefficient in the balanced chemical equation. Practice Exercise Sample Exercise 15.1 Writing Equilibrium-Constant Expressions

  41. (PCc) (PDd) (PAa) (PBb) Kp = The Equilibrium Constant Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

  42. Relationship Between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes Kp = Kc (RT)n where n = (moles of gaseous product) - (moles of gaseous reactant)

  43. In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300 °C. Calculate Kpfor this reaction at this temperature. Solution Analyze: We are given Kcfor a reaction and asked to calculate Kp. Plan: The relationship between Kcand Kpis given by Equation 15.14. To apply that equation, we must determine Δn by comparing the number of moles of product with the number of moles of reactants (Equation 15.15). Solve: There are two moles of gaseous products (2 NH3) and four moles of gaseous reactants (1 N2 + 3 H2). Therefore, Δn = 2 – 4 = –2. (Remember that Δ functions are always based on products minus reactants.) The temperature, T, is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.0821 L-atm/mol-K. Using Kc = 9.60, we therefore have Practice Exercise Sample Exercise 15.2 Converting between Kc and Kp

  44. Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

  45. Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

  46. Equilibrium Can Be Reached from Either Direction For the equation, N2 + H2  NH3 It doesn’t matter whether we start with N2 and H2 or whether we start with NH3: we will have the same proportions of all three substances at equilibrium.

  47. What Does the Value of K Mean? • If K>>1, the reaction is product-favored; since products are on top of the expression when they are bigger than the reactants the constant will be greater than 1.

  48. What Does the Value of K Mean? • If K<<1, the reaction is reactant-favored; since reactants are on the bottom of the expression when they are the constant will be less than 1. . If K= 1, then the reactants and products are in equal amounts.

  49. Kc = = 0.212 at 100 C Kc = = 4.72 at 100 C N2O4(g) 2 NO2(g) 2 NO2(g) N2O4(g) [NO2]2 [N2O4] [N2O4] [NO2]2 Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

  50. The equilibrium constant for the reaction of N2 with O2 to form NO equals Kc = 1 × 10–30 at 25 °C: Using this information, write the equilibrium constant expression and calculate the equilibrium constant for the following reaction: Solution Analyze: We are asked to write the equilibrium-constant expression for a reaction and to determine the value of Kcgiven the chemical equation and equilibrium constant for the reverse reaction. Plan: The equilibrium-constant expression is a quotient of products over reactants, each raised to a power equal to its coefficient in the balanced equation. The value of the equilibrium constant is the reciprocal of that for the reverse reaction. Solve: Writing products over reactants, we have Both the equilibrium-constant expression and the numerical value of the equilibrium constant are the reciprocals of those for the formation of NO from N2 and O2: Comment: Regardless of the way we express the equilibrium among NO, N2, and O2, at 25 °C it lies on the side that favors N2 and O2. Thus, the equilibrium mixture will contain mostly N2 and O2, with very little NO present. Sample Exercise 15.4 Evaluating an Equilibrium Constant When an Equation is Reversed

More Related