32. Kinetics: Reaction Mechanisms by MIT OpenCourseWare

Summary by www.lecturesummary.com: 32. Kinetics: Reaction Mechanisms by MIT OpenCourseWare

• • Preface: Importance of Kinetics and Reaction Mechanisms

    • The final unit of the semester is kinetics. Reaction mechanisms, a crucial aspect of kinetics, are the subject of today's lecture. Understanding the processes through which a reaction occurs is essential for comprehending reaction mechanisms.

      • Reactions rarely happen in a single step; they often involve multiple steps.
      • Elementary reactions, also known as complex reactions, are decomposed into a sequence of steps.
      • Finding the fast and slow steps in a mechanism is crucial for optimizing reactions (e.g., by re-engineering enzymes).
      • Applications such as the development of enzyme inhibitors (e.g., HIV protease) require an understanding of mechanisms.

      Examination of Earlier Ideas: Equations and Order

      This section explains a clicker question concerning important numbers and formulas.

      • A second-order equation was used in the clicker question for numbers three and four.
      • Finding the right equation for a problem from equation sheets will be a requirement for both the fourth and final exams.
      • The rate of radioactive decay is independent of the surrounding nuclei, which is why nuclear chemistry uses first-order equations.

      First Example: 2NO + O2 → 2NO2 - Overall Order and Experimental Rate Law

      • An example of a particular reaction is when 2NO and O2 combine to form 2NO2.
      • This reaction's experimental data produced the following rate law: rate = K_obs * [NO]^2 * [O2]^1.
      • In terms of NO and O2, the reaction is second order and first order, respectively.
      • Avoid losing exam points for basic addition (2 + 1 = 3).
      • It is unlikely that three distinct substances will come together simultaneously in a one-step reaction (termolecular).

      First Example: 2NO + O2 → 2NO2 - The Suggested Two-Step Method

      • The overall reaction is divided into two steps by the suggested mechanism.
      • Step 1: The reversible step 2NO ⇌ N2O2. N2O2 serves as an intermediary.
      • Step 2: N2O2 + O2 → 2NO2. Multi-step reactions frequently form an ephemeral intermediate.

      First Example: 2NO + O2 → 2NO2 - Creating Rate Laws for Basic Steps

      • Elementary reactions are another name for a sequence of actions.
      • The definition of an elementary reaction is that it happens exactly as written.
      • The stoichiometry can be used to directly write the rate law for a simple reaction.
      • Rate_forward1 = K1 * [NO]^2 is the rate law for Step 1's forward motion.
      • A biomolecular reaction is one that has an order of two.
      • Rate_reverse1 = K_minus1 * [N2O2]^1 is the rate law for Step 1 in the opposite direction.
      • Unimolecular reactions have an order of one.
      • Step 2's rate law is Rate_2 = K2 * [O2] * [N2O2]. Step 2 is in the order of two.

      First Example: 2NO + O2 → 2NO2 - Composing the General Rate Law

      • The final step (Step 2) can be used to write the total rate of NO2 formation.
      • Rate_overall = (1/2) * d[NO2]/dt = K2 * [O2] * [N2O2] (correcting for stoichiometry).
      • Due to the formation of two moles, the rate at which NO2 is formed is twice that of O2 and N2O2.
      • The stoichiometric coefficient of NO2 is accounted for by the '2' in the expression 2 * K2 * [O2] * [N2O2].
      • There cannot be an intermediate ([N2O2]) in the overall rate law.

      First Example: 2NO + O2 → 2NO2 - Finding the Intermediate: Steady State Approximation

      • The intermediate concentration ([N2O2]) must be solved for.
      • The rate at which the intermediate is formed less the rate at which it is consumed is its net rate of formation.
      • The steady state approximation is applied to solve for the intermediate.
      • The formula for [N2O2] is [N2O2] = (K1 * [NO]^2) / (K_minus1 + K2 * [O2]).

      First Example: 2NO + O2 → 2NO2 - Changing the Intermediate to the Overall Rate Law

      • Changing the Intermediate to the Overall Rate Law:

        • Enter [N2O2] as the replacement in the overall rate law: Rate_overall = 2 * K2 * [O2] * [N2O2].
        • Rearranged: Rate_overall = (2 * K1 * K2 * [O2] * [NO]^2) / (K_minus1 + K2 * [O2]).
        • Compare this derived rate law with the experimental rate law: Rate = K_obs * [NO]^2 * [O2].

        Due to the [O2] term in the denominator, the derived rate law does not match the experimental rate law. This disparity implies that there are fast and slow steps in the mechanism, necessitating a reevaluation of the expression in light of these steps.

        Example 1: Steps That Are Fast and Slow (Rate Determining Step)

        • Presenting the idea of the rate limiting step, also known as the rate determining step (RDS).
        • The overall rate of a reaction is determined by its slow step.
        • The RDS is demonstrated by the analogy of finding a table in the MIT library: finding the table (40 minutes) is the slow step that determines the total time, not the fast step of starting problems (2 seconds).
        • Being the slow person (RDS) at a dinner party is a personal example.
        • The mechanism's proposal states that Step 2 is slow and Step 1 is quick and reversible.
        • If K_minus1 is significantly larger than K2 * [O2], the K2 * [O2] term is insignificant and can be removed from the denominator of the intermediate expression.
        • [N2O2] ≈ (K1 * [NO]^2) / K_minus1 is the simplified intermediate expression.
        • [N2O2] / [NO]^2 = K1 / K_minus1, where K1 / K_minus1 is the first step's equilibrium constant K1.
        • A fast reversible step is practically in equilibrium if it is followed by a slow step.
        • The overall rate law is substituted with the simplified [N2O2] expression.
        • Rate_overall = 2 * K2 * [O2] * [(K1 * [NO]^2) / K_minus1].
        • Rearranged: Rate_overall = (2 * K1 * K2 / K_minus1) * [O2] * [NO]^2.
        • K_obs * [O2] * [NO]^2 = rate_overall. The experimental rate law is consistent with this derived rate law.
        • The suggested mechanism is consistent with experimental data.

        Example 2: 2NO + Br2 -> 2NOBr

        An additional illustration of the experimental rate law:

        • The formula for the experimental rate law is Rate = K_obs * [Br2]^1 * [NO]^1.
        • 1 + 1 = 2 is the overall order.

        Example 2: Rate Laws and the Suggested Two-Step Mechanism

        • The suggested mechanism is as follows:
        • Step 1: NO + Br2 <=> NOBr2 (reversible step). An intermediate is NOBr2.
        • Step 2: NOBr2 + NO -> 2NOBr.
        • The forward rate law: Rate_forward1 = K1 * [NO] * [Br2].
        • The rate law in reverse Step 1: Rate_reverse1 = K_minus1 * [NOBr2].
        • Step 2's rate law: Rate_2 = K2 * [NOBr2] * [NO].

        Example 2: The General Rate Law and Intermediate Solving

        • Overall rate law (derived from Step 2): Overall rate = (1/2) * d[NOBr]/dt = K2 * [NOBr2] * [NO].
        • It is necessary to eliminate intermediate ([NOBr2]) from the rate law.
        • For the intermediate [NOBr2], use the steady state approximation.
        • In Step 1: K1 * [NO] * [Br2], an intermediate is created.
        • K_minus1 * [NOBr2] is the intermediate used in reverse Step 1.
        • K2 * [NOBr2] * [NO] is the intermediate used in Step 2.

          Example 2: 2NO + Br2 → 2NOBr

          The General Rate Law and Intermediate Solving:

          • Overall rate law: Overall rate = (1/2) * d[NOBr]/dt = K2 * [NOBr2] * [NO] (using a product stoichiometry factor of 2).
          • Eliminate intermediate: It is necessary to eliminate intermediate ([NOBr2]) from the rate law.
          • Steady state approximation: For the intermediate [NOBr2], use the steady state approximation.
          • Step 1 creation: In Step 1: K1 * [NO] * [Br2], an intermediate is created.
          • Reverse Step 1: K_minus1 * [NOBr2] is the intermediate used in reverse Step 1.
          • Step 2 intermediate: K2 * [NOBr2] * [NO] is the intermediate used in Step 2.
          • Rate of formation: Rate of formation = Rate of consumption in a steady state. The formula for calculating [NOBr2] is K1 * [NO] * [Br2] = [NOBr2] * (K_minus1 + K2 * [NO]) + K_minus1 * [NOBr2] + K2 * [NOBr2] * [NO].
          • Final calculation: [NOBr2] = (K1 * [NO] * [Br2]) / (K_minus1 + K2 * [NO]).

          Example 2: 2NO + Br2 → 2NOBr - Changing Intermediate and Taking Fast/Slow Steps into Account:

          • Substitution in overall rate law: In the overall rate law, [NOBr2] should be substituted (Rate_overall = 2 * K2 * [NOBr2] * [NO]).
          • Rearranged rate law: Rate_overall = 2 * K2 * [NO] * [(K1 * [NO] * [Br2]) / (K_minus1 + K2 * [NO])].
          • Final rearrangement: Rearranged: Rate_overall = (2 * K1 * K2 * [NO]^2 * [Br2]) / (K_minus1 + K2 * [NO]).

          Example 2: 2NO + Br2 → 2NOBr Examining Fast/Slow Step Situations:

          • Scenario 1: Step 1 is slow, and Step 2 is quick.
          • Indication: K2 * [NO] >> K_minus1 is what this indicates. K_minus1 is insignificant in the denominator (K_minus1 + K2 * [NO]).
          • Denominator: K2 * [NO] becomes the denominator.
          • Overall rate approximation: Rate_overall ≈ (2 * K1 * K2 * [NO]^2 * [Br2]) / (K2 * [NO]).
          • Cancellation: K2 and one [NO] term are being cancelled: Overall rate ≈ 2 * K1 * [NO] * [Br2].
          • Expressing overall rate: K_obs * [NO] * [Br2] is one way to express this, where K_obs = 2 * K1. 1 + 1 = 2 is the overall order. This is consistent with the experimental rate law, which is first order in Br2 and first order in NO.
          • Consistency: Consequently, the experimental data is consistent with a slow first step and a fast second step.

          Scenario 2: Step 1 is quick and reversible, while Step 2 is sluggish.

          • Indication: K_minus1 >> K2 * [NO] is what this indicates.
          • Insignificance: K2 * [NO] is insignificant in the denominator (K_minus1 + K2 * [NO]).
          • Denominator approaches: The denominator approaches K_minus1.
          • Overall rate approximation: Rate_overall ≈ K_minus1 / (2 * K1 * K2 * [NO]^2 * [Br2]).
          • Expressing overall rate: K_obs * [NO]^2 * [Br2] is one way to express this, where K_obs = (2 * K1 * K2) / K_minus1. 2 + 1 = 3 is the overall order.
          • Inconsistency: The experimental rate law (order 2) is not consistent with this.

            Example 3: 2O3 -> 3O2 - The Laws of Mechanism and Rate

            The suggested mechanism for ozone decomposition is:

            1. Step 1: O3 <=> O2 + O (reversible)
            2. Step 2: O + O3 -> 2O2. The forward rate law is K1 * [O3].

            Rate Laws

            The rate law in reverse:

            • Step 1: K minus 1 * [O2] * [O].
            • Step 2 rate law: K2 * [O] * [O3].

            Example 3: 2O3 -> 3O2 - Steps Overview

            A quick, reversible first step is followed by a slow one:

            • Step 2 is slow while Step 1 is quick and reversible.
            • The slow step (Step 2) establishes the rate.
            • Overall rate = (1/3) * d[O2]/dt = K2 * [O] * [O3] (considering product stoichiometry).
            • The intermediate [O] must be solved for.
            • Step 1 can be regarded as being in equilibrium due to its quick and reversible nature.

            Applying the Equilibrium Expression

            Use the Step 1 equilibrium expression:

            • [O2] = K_equilibrium * [O] / [O3] = K1 / K_minus1.
            • Using the equilibrium expression, [O] = (K1 / K_minus1) * ([O3] / [O2]), find [O].
            • When there is a quick reversible first step, this is a more straightforward method of solving for the intermediate.

            Example 3: 2O3 -> 3O2 - Replacing Intermediate and Final Rate Law

            Replace [O] with the expression in the overall rate law:

            • Rate_overall = K2 * [(K1 / K_minus1) * ([O3] / [O2])] * [O3].
            • Rate_overall = (K1 * K2 / K_minus1) * [O3]^2 / [O2].
            • K_obs * [O3]^2 / [O2] can be expressed as follows: K_obs = K1 * K2 / K_minus1.
            • Rate_overall = K_obs * [O3]^2 * [O2]^-1.

            Example 3: Establishing Orders

            The orders for the reactants are:

            • The order for O3 is 2.
            • The order for O2 is -1.
            • The rate increases by a factor of four (2^2) if [O3] is doubled.
            • The rate will be cut in half if [O2] doubles (2^-1 = 1/2).
            • The total of the individual orders is the overall order: (-1) + 2 = 1.

            Conclusion - Significance and Future Plans

            Key takeaways include:

            • Different rate laws result from different assumptions about fast and slow steps, and one of these will match experimental data.
            • Understanding the steps that determine rates is crucial.
            • Exam 4's additional problems are lengthy.
            • Problem set 9 is due tomorrow.
            • Remind students of the last clicker competition and encourage them to begin problems early. Wednesday is the next class.