Chemistry 12 Assignment 11 Solutions

Chemistry 12 Assignment 11-Solutions Fall 2009 Chapter 15: Chemical Kinetics Problems #42, 43, 44, 47, 48, 75, 78 Reaction Mechanisms 42. a. An elementary step (reaction) is one in which the rate law can be written from the molecularity, i.e., from the coefficients in the balanced equation. b. The mechanism of a reaction is the series of elementary reactions that occur to give the overall reaction. The sum of all the steps in the mechanism gives the balanced chemical reaction. c. The rate determining step is the slowest elementary reaction in any given mechanism. 43. For elementary reactions, the rate law can be written using the coefficients in the balanced equation to determine the orders. a. Rate = k[CH₂NC] b. Ratek[O][NO] c. Rate = k[0] d. Rate = k[0][0] c. Rate = k[C] or Rate = kN where N the number of C atoms (convention) 44. Since the rate of the slowest elementary step equals the rate of a reaction, then: Rate rate of step 1 = k[NO₂] The sum of all steps in a plausible mechanism must give the overall balanced reaction. Summing all steps gives: NO₂+ NO, NO, + NO NO,+CONO,₂ + CO₂ NO₂+ CONO + CO₂ 47. Let's determine the rate law for each mechanism. If the rate law derived from the mechanism is the same as the experimental rate law, then the mechanism is possible. When deriving rate laws from a mechanism, we must substitute for all intermediate concentrations. a. Rate = k[NO][0₂] not possible b. Ratek[NO,][NO] and [NO,] [NO][0] Rate kK [NO][0,] possible -Kk/k, or [NO,] = K[NO][0] c. Rate k[NO] not possible d. Rate = k[N₂O₂] and [N₂O₂] = K[NO]² where Kqk/k.₁, RatekK [NO] not possible a. This rate law occurs when the first step is rate determining. Rate = k[Br₂] = k'[Br₂] b. This rate law occurs when the second step is rate determining and the first step is a fast equilibrium step. Rate = k₂[Br][H₂]; From the fast equilibrium first step (rate forward = rate reverse): k₁[Br₂] = k₋₁[Br]², [Br] = (k₁/k₋₁)^(1/2)[Br₂]^(1/2) Substituting into the rate equation: Rate = k₂(k₁/k₋₁)^(1/2)[Br₂]^(1/2)[H₂] = k''[H₂][Br₂]^(1/2) c. From a, k' = k₁; From b, k'' = k₂(k₁/k₋₁)^(1/2) Additional Exercises Rate = k[I⁻][OCl⁻][OH⁻]ᶻ; Comparing the first and second experiments: 18.7 x 10⁻³ / 9.4 x 10⁻³ = k(0.0026)^(x)(0.012)^(y)(0.10)^(z) / k(0.0013)^(x)(0.012)^(y)(0.10)^(z), Comparing the first and third experiments: 2.0 = 2.0^(x), x = 1 9.4 x 10⁻³ / 4.7 x 10⁻³ = k(0.0013)(0.012)^(y)(0.10)^(z) / k(0.0013)(0.0060)^(y)(0.10)^(z), Comparing the first and sixth experiments: 2.0 = 2.0^(y), y = 1 4.7 x 10⁻³ / 9.4 x 10⁻³ = k(0.0013)(0.012)(0.20)^(z) / k(0.0013)(0.012)(0.10)^(z), 1/2 = 2.0^(z), z = -1 Rate = k[I⁻][OCl⁻] / [OH⁻]; The presence of OH⁻ decreases the rate of the reaction. For the first experiment: 9.4 x 10⁻⁵ mol / L s = k(0.0013 mol/L)(0.012 mol/L) / (0.10 mol/L) For all experiments, k_mean = 60 s⁻¹. k = 60.3 s⁻¹ ≈ 60 s⁻¹