Chapter 24: Chemical KineticsProblems1. In the reaction 3A + 2B → C + D, reactant A is found to disappear at the rate of -8.2 × 10-4 M s-1.
3. In the reaction B → products, [B] is found to be 0.567M at t=31.6s and 0.356M at t=50.3s. During this time interval, what is the average rate of the reaction? 11. The initial rate of the reaction 2A+B→3C+D can be determined by the following table. Using the given information to (1) find the order of reaction with respect to A and to B. (2) solve the overall reaction order. (3). Solve for the rate constant,k Expt 1 [A],M 0.150 [B],M 0.123 Initial rate,Ms-1 3.52x10-3 13. These rate were obtained in three experiments with the reaction 2CO(g)+F2(g) → >2COF2(g) intial CO 1. .0155M initial F2 1. .0265 initial Rate of Reaction 1. 2.27*10-5 2. .0155M 2. .053 2. 4.55*10-5 3. .0310M 3. .0265 3. 9.08*10-5 What is the rate law? 17. The first order reaction A → Products has t1/2 = 180 s:
19. The reaction A → products is first order in A.
27. Which is a) zero order, b) first order, c) second-order reaction? Time: 0 25 50 75 100 125 150 1. 1 .8 .6 .4 .2 0 0 2. 1 .7 .65 .585 .448 .342 .262 3. 1 .656 .488 .388 .323 .276 .241 29. What is the approximate half-life of the first-order reaction? 1 2 3 25 .80 25 .77 25 .82 33. The reaction A + B → C + D is in second order in A and first order in B. The value of k is 0.0205 M-2 min-1. What is the rate of this reaction when [A] is 0.005M and [B] is 3.02M. 35. The reaction H2O2(aq) → H2O(l) +1/2O2(g) yields the following data when decomposed at 600K for 500s t=0s [H2O2]=2.00M Time(s) [A] (M) 43. Since zero-order and second-order reactions both depend on the initial concentration and the rate constant, why does the half-life for one get longer as the initial concentration increases while the other decreases? (Make sure to label which reaction is which) 45. Explain
47. For the reversible reaction A + B ⇄ A + B the enthalpy change of the forward reaction is +34kJ/mol. The activation energy of the forward reaction is 66 kJ/mol.
49. Determine the following from the reaction profile for the reaction A to E given
51. Given for following data for the reaction A+B→C at different temperatures. Find the activation energy for the reaction. 450k, k=4.5x10-4 M-1s-1; 625k, k=1.9x10-2 M-1s-1. 61. Compare and contrast the catalytic activity of rhodium and of an enzyme. 63. What reaction conditions are necessary to account for a linear relationship between enzyme concentration and reaction rate? 67. Hypothetically, the reaction H2 + 2ICl → I2 + 2HCl is first order in [H2] and second order in [ICl]. Based on the first fast step given below and the fact that the second step is the slow step, propose a second-step mechanism, and show that it conforms to the experimentally determined reaction order. Assume that [I2] does not affect the reaction rate. H2 + 2ICl → I2 + 2HCl Fast: 2ICl → I2 + 2Cl 87. The following is a two-step reaction for the reaction of Nitric oxide and Oxygen. k-1 k1 (1) NO+ NO <--- ---> N2O2 k2 (2) N2O2 +O2 --->2NO2
The rate constants have been determined to be k1=3.4*102; k-1 = 2.8*102;k2 = 3.48*102 Determine k and the rate law.
Solutions1. In the reaction 3A + 2B → C + D, reactant A is found to disappear at the rate of -8.2 × 10-4 M s-1.
a) reaction rate = -(1/3)d[A]/dt = -(1/2)d[B]/dt = d[C]/dt = d[D]/dt To review reaction rate laws, please visit the page "The Rate Law". 3. average rate= delta[B]/delta(t)= (0.567M-0.356M)/(50.3s-31.6s)=0.011M/s To review reaction rate laws, please visit the pages "The Rate Law" and "Rate of Reaction" 11. The initial rate of the reaction 2A+B →3C+D can be determined by the following table. Using the given information to (1)find the order of reaction with respect to A and to B. (2) solve the overall reaction order. (3). Solve for the rate constant,k Expt 1 [A],M 0.150 [B],M 0.123 Initial rate,Ms-1 3.52x10-3
1)using the equation rate of reaction=k[A]x[B]y we can solve for x and y R1=k[A]x[B]y=k(0.150)x(0.123)y=3.52x10-3 To review reaction rate laws, please visit the page "The Rate Law". 13. R1=k[A]x[B]y=k(.155)x(.265)y=2.27*10-5 R2=k[A]x[B]y=k(.155)x(.053)y=4.55*10-5 R2/R1=k(.155)x(.265)y/k(.155)x(.053)y=.053y*2y/.053y=2y=4.55*10^-5/2.27*10-5=1 y=1 R3=k[A]x[B]y=k(.310M)x(.0265)y=9.08*10^-5 R2/R3=k(.155)x(.053)y/k(.0310M)x(.0265)y=2y/2x=2/2x 2/4.55*10^-5/9.08*10^-5=4 x=2 R1=k[CO]2[F2]=R1/[CO][F2]=2.27*10-5/(.0155M)2*.0265M)=3.57M-2/min-1 k=3.57M-2/min-1 R=3.57M-2/min-1[CO]2[F2] To review reaction rate laws, please visit the page "The Rate Law". 17. For first order reactions t1/2 = ln2/k = .693/k if the half life is t1/2 = 180s, then the value of k for the reaction is k= t1/2/ln2 = .693/720s = 9.625x10-4s-1 a) since ln[At] = -kt + ln[A]o, then ln([A]t/[A]t) = -kt, you can say [A]t/[A]o = e-kt this is convenient since the percent remaining is the “amount at time t” / “initial amount” which is equal to [A]t/[A]o therefore % remaining = e-kt = e^(-(9.625x10-4s-1)(720s)) = .50007 (100%) = 50% b) given [A] = .25M, and we know that rate of reaction = -Δ[A]/Δt = k[A] k[A] = 9.625x10-4s-1(.25M) = 2.406x10-4M/s To review the rate laws, visit the page "The Rate Law". 19. The reaction A → products is first order in A. Step 1: Find k. For a first order reaction: half life (t1/2) = ln2/k. In this reaction, 8 minutes = ln2/k. k = ln2/8minutes = 0.0866 minutes-1 Step 2: Use the following equation to find the amount of A at time = 1hour: ln(At/Ao) = -kt ln(At/12.24g) = -(0.0866 minutes-1*60 minutes) To review half life and first-order reactions , please visit the pages "Half Life" and "First-Order Reactions".
To review half life and first-order reactions , please visit the page "Half Life" To reivew first-order reactions, visit the page "First-Order Reactions". 27. a) 1 The amount that it decreases by per second (slope) is a constant value so its zero order. b) 2 The amount that it decreases by per second is a set percentage. c) 3 The slope (1/A1-1/A2)/(t1-t2) is the same value at all points. To review rate laws, visit the page "The Rate Law". 29. We need to solve for t1/2 = .693/k. So solving for k in the integrated rate law for first order reactions... ln([A]t/[A]o) = -kt k = ln([A]t/[A]o)/-t now we use any point of time and its corresponding concentration, like t = 100s and [A] = .4M k = ln(.4/1)/(-100) = 9.16x10-3s-1 then use this in the half-life equation t1/2 = .693/k = .693/9.16x10-3s-1 = 75.66s To review first-order reactions visit the page "First-Order Reactions". 33. The reaction A + B → C + D is in second order in A and first order in B. The value of k is 0.0205 M-2 min-1. What is the rate of this reaction when [A] is 0.005M and [B] is 3.02M. To review second-order reactions, please visit the pages "Second-Order Reactions". 35. Average rate of reaction: delta[H2O2]/delta(t)=(2M-1.26M)/(500s)=0.00148M/s To review zero-order reactions, please visit the page "Zero-Order Reactions" 37. For the following reaction A→B the following information was obtained. Time(s) [A] (M) To review zero-order reaction, visit the page "Zero-Order Reactions". 43. The half life of zero-order reactions increase with the initial concetration, because the concentration is decreasing with the constant slope -k, so a larger value will be reduced by a smaller percent. For instance 10-1=9/10*100=90% of the original, while 5-1=4/5*100%=80% of the original. The half life of second-order reactions decrease with the initial concentration, because when the concentration is higher, there is a higher probability that the molecules will interact. This causes the concentration to decrease faster. This relationship is shown by the equation t1/2=1/(k*A0). To review zero-order reactions, visit the page "Zero-Order Reactions". 45. a) The collision frequency only determines the frequency of molecules to collide. If each collision of molecules resulted in a reaction, then the reaction’s rate would be extremely fast and would be determined solely by the collision frequency. But each collision does not result in a reaction, and therefore the reaction rate is only a fraction of the calculated collision frequency. b) A raise in temperature directly affects the gas molecules speed. This rise in speed makes a higher chance for molecules to collide and therefore raises the collision frequency of a reaction. The rise in temperature dramatically rises the reaction’s rate, however. This is because the reaction’s rate is already a very small fraction of the reactions collision frequency, and is directly affected by both the collision frequency and the fraction of molecules that become “activated”. Adding heat helps overcome the activation energy, so it not only raises the collision frequency but also the amount of activated molecules. This causes a much more dramatic rise to reaction rate with respect to collision frequency. c) A catalysts main purpose is to lower the activation energy in a reactions pathway. The lower the activation energy, the larger the fraction of energetic collisions and the faster the reaction. This speeds up reaction rate, regardless of how fast the molecules are sped up by an increase in temperature. They are independent of one another. To review this topic, visit the page "The Rate Law". 47. For the reversible reaction A + B ⇄ A + B the enthalpy change of the forward reaction is +34kJ/mol. The activation energy of the forward reaction is 66 kJ/mol.
To review the potential energy profile of reactions, please visit the page "Potential Energy Profile". 49. To review the potential energy profile of reactions, please visit the page "Potential Energy Profile". Using the modified Arrhenius Equation ln(k2/K1)=-Ea/R(1/T2-1/T1) To review activation energy, visit the page "Arrhenius Equation". 61. Enzymes are much more specific with what reaction they can be catalyts for, while metals like rhodium can be catalysts for a wide variety of reactions. Enzymes also normally only operate at a very specific tempearture range. With enzymes, the reactant attaches tot he enzyme, the reaction occurs, and the products are released. With catalysts like rhodium (heterogeneous catalysis), tje catalysis absorbs the reactants, diffues them across its occurs, and then is desorbed. To review enzymes, visit the page "Catalytic Efficiency of Enzymes". 63. The reaction must have a substrate that reacts with the enzyme at an active site to form the enzyme-substrate complex. This must happen for the complex to decompose and form products and regenerate the enzyme, as well as the enzyme must be operating at the correct temperature. To review enzymes, visit the page "Catalytic Efficiency of Enzymes". 67. Hypothetically, the reaction H2 + 2ICl → I2 + 2HCl is first order in [H2] and second order in [ICl]. Based on the first fast step given below and the fact that the second step is the slow step, propose a second-step mechanism, and show that it conforms to the experimentally determined reaction order. Assume that [I2] does not affect the reaction rate.
Fast: 2ICl → I2 + 2Cl ------------------------- The reaction rate is only as fast as its slowest step (aka, the rate limiting step). Thus, reaction rate = slow rate = k2[H2][Cl]2 Since Cl is no in the overall equation, we need to find out how to put it in terms of [ICl] since [ICl] is in the overall equation. The fast rate is nearly in equilibrium since it produces I2 and 2Cl faster than the slow reaction can use up the Cl. Thus, the forward rate of the fast reaction equals its own reverse rate. Fast rate = k1[ICl2]2 = k-1[I2][Cl]2 Since all the k’s are constants, they can be combined into the overall reaction constant, k. Also according to the problem, the [I2] is assumed not to affect the reaction rate.Thus, the reaction rate can be described as: To review rate determining steps, visit the page "Rate Determining Step". 78. We have observed that the units of k depend on the overall reaction order. Derive a general expression for the units of k using the units of the order of the reaction (o) units of concentration (M) and time (s) The units of k change on the overall reaction To review reaction rates, visit the page "Measuring Reaction Rates". 87. reaction rate=k2[N2O2][O2] disappearance of N2O2 = k-1[N2O2]+k2[N2O2][O2] formation of N2O2 = k1[NO]2 We can assume that the formation of N2O2 is equal to its disappearance, so k1[NO]2 = k-1[N2O2]+k2[N2O2][O2] [N2O2]=k1[NO]2 /(k-1+k2[O2]) substituting for the reaction rate we get, assuming molarities are one, rate=k1[NO]2 /(k-1+k2[O2])*k2*[O2] = 172 Since k1 < k2, NO+ NO ← ---> N2O2 is the slower reaction so, rate=k[NO]2 k= 172 rate = 172[NO]2 To review reaction rates, visit the page "Measuring Reaction Rates". This page viewed 11180 times
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