AP® Chemistry Kinetics Questions Free Response

This paragraph introduces an AP Chemistry exam-style kinetics problem, focusing on interpreting graphs to deduce the reaction order with respect to NO2. The speaker explains that typically, graphs show time on the x-axis and reactant concentration on the y-axis in one of three forms: linear concentration, natural log of concentration, or the inverse of concentration. The key to solving such problems is identifying which of these graphs is linear, as this indicates the reaction order. For instance, a linear concentration graph suggests a zero-order reaction, a linear ln[A] graph indicates a first-order reaction, and a linear 1/A graph points to a second-order reaction. The provided graphs show that the reaction in question is second order with respect to NO2, as the time versus 1/NO2 graph is linear.

The speaker proceeds to explain how to write a rate law for a reaction, assuming it is zero order with respect to CO. The standard rate law formula is presented as rate = k × [reactant]^order. Given that the reaction is second order with respect to NO2 from the previous analysis, and zero order with respect to CO, the rate law simplifies to rate = k × [NO2]^2, with CO's concentration not influencing the rate due to its zero-order status. This step is crucial for understanding how the rate of a reaction is mathematically expressed in relation to the concentrations of the reactants involved.

The paragraph discusses how to calculate the rate constant 'k' using the initial rate of the reaction and the initial concentration of NO2. The rate law derived in the previous section is rearranged to solve for k, resulting in k = rate / [NO2]^2. The initial rate is provided in the problem statement, and the initial concentration of NO2 is extracted from the kinetics graphs at time zero. The speaker emphasizes the importance of squaring the concentration value when performing the calculation, which is a common mistake among students. The calculated rate constant is found to be 0.35 L/mol·s, and the speaker notes the importance of the value over the unit when grading the AP exam.

This section delves into reaction mechanisms, specifically whether a proposed two-step mechanism aligns with the previously determined rate law. The speaker outlines key concepts about reaction mechanisms, such as the rate-determining step being the slowest and thus dictating the overall reaction rate. It is explained that this slow step must include all species present in the rate law, and the molecularity of the step is discussed, indicating how many molecules of a reactant are involved in the slow step. The provided mechanism is deemed consistent with the rate law, as the slow step only involves NO2, which is second order, aligning with the bimolecular nature of the step.

The final paragraph addresses how to sketch a potential energy diagram for a spontaneous reaction, with a focus on the concepts of Gibbs free energy (delta G) and activation energy. The speaker explains that the diagram should show the reaction progress on the x-axis and energy on the y-axis, with the reactants at a higher energy level than the products due to the negative delta G of a spontaneous reaction. Activation energy is depicted as energy 'bumps', with the rate-determining step having the higher bump. The diagram should reflect that the slow step comes first, followed by the fast step with a lower activation energy bump. The speaker assures that as long as the diagram is conceptually accurate and follows these guidelines, it should meet the AP exam's expectations.