3: Chemical Kinetics Name: HW 6: Review for Unit Test KEY Class: Date: A Products

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1 3: Chemical Kinetics Name: HW 6: Review for Unit Test KEY Class: Date: Page 1 of 9 AP Multiple Choice Review Questions The reaction rate is defined as the change in concentration of a reactant or product per unit time. If we graph [A] vs. t, it would usually look like the solid line in the following plot. A Products Rate = [A] t An instantaneous rate is the slope of a tangent line to the graph of [A] vs. t. We can determine the instantaneous rate at any time during the reaction. On the plot, tangent lines at t 0 and t = t1 are drawn. The slope of these tangent lines would be the instantaneous rates at t 0 and t = t1. We call the instantaneous rate at t 0 the initial rate. The average rate is measured over time. For example, the slope of the dashed line connecting points a and c is the average rate of the reaction over the entire length of time 0 to t2 (average rate = Δ[A]/Δt). An average rate is determined over some time period, whereas an instantaneous rate is determined at one specific time. The rate that is largest is generally the initial rate. At t 0, the slope of the tangent line is greatest, which means the rate is largest at t 0. The initial rate is used by convention so that the rate of reaction only depends on the forward reaction; at t 0, the reverse reaction is insignificant because no products are present yet. 2. The differential rate law (often called just the rate law) describes the dependence of the rate on the concentration of reactants. The integrated rate law expresses reactant concentrations as a function of time. The rate constant k is a constant that allows one to equate the rate of a reaction to the concentration of reactants. The order is the exponent that the reactant concentrations are raised to in the rate equation.

2 Page 2 of 9 3. The method of initial rates uses the results from several experiments where each experiment is carried out at a different set of initial reactant concentrations and the initial rate is determined. Two experiments are compared where only one reactant concentration changes by taking the ratio of their rate laws to solve for the order. This is repeated until all the orders are determined. After the orders are determined, choose the values from any experiment to plug into the rate law and solve for the rate law constant (k). The units on k depend on the orders in the rate law. Because there are many different rate laws, there are many different units for k. For example, if Rate = k[a] n ; If [A] is And the rate The order is Tripled (x3) Tripled (x3) First Quadrupled (x4) Increases by a factor of 16 (x16) Second Doubled (x2) Increases by a factor of 2 3 (x8) Third Changed Remains the same Zero 4. Some calculus (don t worry if you can t do this yet).

3 Page 3 of 9 5. The integrated rate laws can be put into the equation for a straight line, y = mx + b where x and y are the x and y axes, m is the slope of the line, and b is the y-intercept. Zero A plot of [A] vs. time will be linear with a negative slope [A] = kt + [A] order 0 equal to k and a y-intercept equal to [A]. First A plot of ln[a] vs. time will be linear with a negative slope Ln[A] = kt + Ln[A] 0 order Second order Order 1 [A] = kt + 1 [A] 0 equal to k and a y-intercept equal to ln[a]0. A plot of 1/[A] vs. time will be linear with a positive slope equal to k and a y-intercept equal to 1/[A]0. For reactions with multiple reactants, only one of the reactants is allowed to change by having a large excess of the other reactant(s) as compared to the reactant studied; so large that the concentration of the other reactant(s) stays effectively constant during the experiment. The slope of the straight-line plot equals k (or k) multiplied by the other reactant concentrations raised to the correct orders. Once all the orders are known for a reaction, then any of the slopes can be used to determine k. 6. At t = t½, [A] = (1/2)[A]0; Plugging these terms into the integrated rate laws yields the following half-life expressions: Derive (See CW 5 How does t½ depend on What is the second half-life? for step by step) [A]0? Zero t 1/2 = [A] 0 2k First t 1/2 = k Second t 1/2 = 1 k[a] 0 directly related to the concentration independent of concentration inversely related to concentration The second half-life will be 10. s, because the half-life has a direct relationship with concentration. The second half-life will be 20. s because the half-life is concentration independent. The second half-life will be 40. s because half-life is inversely related to [A]0. 7. Definitions: a. An elementary step (reaction) is one for which the rate law can be written from the molecularity, i.e., from coefficients in the balanced equation. b. The molecularity is the number of species that must collide to produce the reaction represented by an elementary step in a reaction mechanism. c. The mechanism of a reaction is the series of proposed elementary reactions that may occur to give the overall reaction. The sum of all the steps in the mechanism gives the balanced chemical reaction. d. An intermediate is a species that is neither a reactant nor a product but that is formed and consumed in the reaction sequence. e. The rate-determining step is the slowest elementary reaction.

4 Page 4 of 9 8. For a mechanism to be acceptable, the sum of the elementary steps must give the overall balanced equation for the reaction, and the mechanism must give a rate law that agrees with the experimentally determined rate law. A mechanism can never be proven absolutely. We can only say it is possibly correct if it follows the two requirements. Most reactions occur by a series of steps. If most reactions were unimolecular, then most reactions would have a first order overall rate law, which is not the case. 9. The premise of the collision model is that molecules must collide to react, but not all collisions between reactant molecules result in product formation. a. The larger the activation energy, the slower the rate. b. The higher the temperature, the more molecular collisions with enough energy to convert to products and the faster the rate. c. The greater the frequency of collisions, the greater the opportunities for molecules to react, and, hence, the greater the rate. d. For a reaction to occur, it is the reactive portion of each molecule that must be involved in a collision. Only some of all the possible collisions have the correct orientation to convert reactants to products. The activation energy for the reverse reaction will be the energy difference between the products and the transition state at the top of the potential energy hill. For an exothermic reaction, the activation energy for the reverse reaction (Ea,r) is larger than the activation energy for the forward reaction (Ea), so the rate of the forward reaction will be greater than the rate of the reverse reaction. For an endothermic reaction, Ea,r < Ea so the rate of the forward reaction will be less than the rate of the reverse reaction (with other factors being equal).

5 10. The Arrhenius Equation: k = Ae E a RT ; 1 T + Ln(A) Ln(k) = E a R y = m x + b The data needed is the value of the rate constant as a function of temperature: plot ln(k) versus 1/T to get a straight line (temperature in kelvin). The slope of the line is equal to Ea/R and the y-intercept is equal to ln(a). The R value is J/K mol. Page 5 of 9 If the rate constant is known at two different temperatures, then the following equation allows determination of Ea: ln ( k 2 k 1 ) = E a R ( 1 T 1 1 T 2 ) 11. A catalyst increases the rate of a reaction by providing reactants with an alternate pathway (mechanism) to convert to products. This alternate pathway has a lower activation energy, thus increasing the rate of the reaction. A homogeneous catalyst is one that is in the same phase as the reacting molecules, and a heterogeneous catalyst is in a different phase than the reactants. The heterogeneous catalyst is usually a solid, although a catalyst in a liquid phase can act as a heterogeneous catalyst for some gas phase reactions. Since the catalyzed reaction has a different mechanism than the uncatalyzed reaction, the catalyzed reaction most likely will have a different rate law.

6 Page 6 of 9 Review Questions 1 11 on Pages 507 and D. Higher concentrations increase the rate of a chemical reaction because the increased number of particles causes increased collisions, so 3M HCl will react faster than 1M HCl. Greater surface area will also increase the rate of a chemical reaction (more collisions) and since the problem states that the total mass of magnesium is held constant, there should be two 0.5 cm 2 squares used for each 1.0 cm 2 square, so the 0.5 cm 2 squares have more surface area. 2. D. In trials 1 and 2, [A] is constant, [B] goes from to increasing by a factor of 2; the rate of formation goes from to , increasing by a factor of 4 and since 2 2 = 4, [B] is 2nd order. In trials 1 and 3, [B] is constant, [A] goes from to increasing by a factor of 2; the rate of formation goes from to , increasing by a factor of 2 and since 2 1 = 2, [A] is 1st order. Rate = k[a] 1 [B] 2 3. C. In trial one and two, the [Y] is constant and the [X] goes from 6 to 12, increasing by a factor of 2. Since X is first order, the rate will need to go up by a factor of 2 1, so 0.10 Ms = 0.20 M s A. In trial one and three, the [X] is constant and the [Y] goes from 6 to 3, down by a factor of 2; the rate remains unchanged, so the order with respect to Y is zero. 5. C. k = = = t 1/ s s Math note: Since you do not have a calculator: moving the decimal point one place to the right on the 0.693, you have k = = , however, since the decimal 23.1 s 3s was moved one place to the right on the to become 6.93, the decimal point on the answer will need to be moved back one decimal place to left, so the answer is approximately The only two answer choices close to this is C and D. The only one with the correct unit of s 1 is letter C. 6. A. For first order kinetics, t 1/2 = 0.693, and this relationship between half-life and the k rate constant does not depend on the amount (there is no A0 term). The only thing that changes the rate constant for a first order reaction is a change in temperature and the problem states that temperature is constant

7 Page 7 of 9 7. D. Mechanism A: Slow step is bimolecular and will result in the rate law of: Rate = k[q][r2] Mechanism B: Slow step is bimolecular and will result in the rate law of: Rate = k[q] 2 Mechanism C: Z must be a catalyst, as it is reactant in an early step in the mechanism and then reappears as a product in a later step. The slow step is bimolecular and will result in the rate law of: Rate = k[q][z] Mechanism D: Slow step is unimolecular and will result in the rate law of: Rate = k[r2] The mechanism that is first order in R2 and zero order with respect to Q is D. 8. B. Since the graph has only one hump, indicating that this reaction occurs as a single step, with no intermediate. The top of the humps indicates the transition states (temporary high energy state) in the mechanism. A graph with two humps would indicate one intermediate and two transition states. 9. D. As the temperature increases, the particles are moving faster and they (i) collide more often and (ii) the collisions are more energetic. If the molecules are more energetic, more of them will have the required activation energy to get the reaction to take place. 10. D. The catalyst does not affect the reactants and the products; it simply speeds up the process. The catalyst would cancel out in the mechanism as it is on the reactant side of an initial step and on the product side of a later step. It is usually written above the arrow. 11. C. Since this is an elementary reaction, it occurs in a single step, exactly as written in the equation. It is bimolecular and will result in the rate law of: Rate = k[a][b], which is first order with respect to A and first order with respect to B, so second order overall. The units on the rate constant can always be found using the following: overall order 1 L mol overall order 1 time 12. A. Since the rate law given is for a second order reaction, the integrated rate law linear graph would be [A] 1 on the y-axis and time on the x-axis. This graph is for [A] versus time, so it is not linear for a 2nd order reaction, but concentration will decrease over time. 13. B. H3O + must be a catalyst, as it is reactant in an early step in the mechanism and then reappears as a product in a later step.

8 Page 8 of D. The slow step is bimolecular and will result in the rate law of: Rate = k[w][x2] However, W is a reaction intermediate and cannot be included in the rate law. W will need to be removed from the rate law and something equivalent must replace it. In step 1, the forward reaction is unimolecular and the reverse reaction is bimolecular and will result in the following rate laws: Rateforward = kforward[w2] Ratereverse = kreverse[w] 2 Because step 1 is a fast equilibrium, the rateforward = ratereverse. kreverse[w] 2 = kforward[w2]. Solving for [W] results in [W] = k forward[w 2 ] = ( k 1/2 forward[w 2 ] ) k reverse k reverse Substituting this result into the original rate law from above: Rate = k ( k 1/2 forward[w 2 ] ) [X k 2 ] reverse And simplifying: Rate = k new [W 2 ] 1/2 [X 2 ] Note: knew is the combined rate constants k, kforward and kreverse. 15. C. The reaction with the highest activation energy will have the slowest rate. 16. B. The coefficient for CO2 is 6 and the coefficient for O2 is 3, so the CO2 is appearing twice as fast at the O2 is disappearing.

9 Ln[A] Short AP Free Response Question Graphical methods are used to determine certain kinetics information. a. On a separate sheet of graph paper, draw a line graph that illustrates how concentration and time data can be used to determine the half-life and specific rate constant, k, for a first order chemical or nuclear decay reaction. Label each axis, including appropriate units. First Order Graph: Ln[A] vs. t Slope = k Page 9 of 9 time b. Explain how your graph can be used to determine: a. The half-life of the reaction. b. The value of the specific rate constant, k. For a first order reaction: Ln[A] = kt + Ln[A] 0 t 1/2 = k The slope of the linear fit to Ln[A] vs.t gives k. This value can then be plugged into the half-life equation to determine the half-life. c. The decay of a radioactive isotope follows first-order kinetics. Iodine-131, used to treat cancer of the thyroid gland, has a half-life of 8.1 days. If 5.00 g of Iodine-131 is present in a sample initially, what mass of Iodine-131 is present exactly 24.0 hours later? Obtain the value of k from the first order half-life equation: t 1/2 = k = = k t 1/2 8.1 = day Plug values into the linear first order equation: Ln[A] = kt + Ln[A] 0 Where x = mass of Iodine-131 Ln(x) = (0.0856)(1 day) + Ln(5.00g) Ln[x] = e Ln[x] = e [x] = e = 4.59 g remains We replaced concentrations with the masses