Question

Consider this overall reaction, which is experimentally observed to be second order in \(\mathrm{AB}\) and zero order in \(\mathrm{C}\) . $$\mathrm{AB}+\mathrm{C} \longrightarrow \mathrm{A}+\mathrm{BC}$$ Is the following mechanism valid for this reaction? $$\mathrm{AB}+\mathrm{AB} \longrightarrow \mathrm{AB}_{2}+\mathrm{A} \quad Slow$$ $$\mathrm{AB}_{2}+\mathrm{C} \longrightarrow \mathrm{AB}+\mathrm{BC}\quad Fast$$

Short answer

Yes, the proposed mechanism is valid as it correctly predicts the experimentally observed rate law: second order in \(\mathrm{AB}\) and zero order in \(\mathrm{C}\).

Step-by-step solution

Determine Rate Law from Mechanism

Examine the given mechanism to determine the rate law implied by the individual steps. The slow step usually determines the rate law because it is the rate-determining step (RDS). For the mechanism provided, the slow step is as follows:\[\mathrm{AB} + \mathrm{AB} \longrightarrow \mathrm{AB}_2 + \mathrm{A}\] The rate law implied by this step would be second order with respect to \(\mathrm{AB}\), since two \(\mathrm{AB}\) molecules are involved.

Analyze Orders of Reactants

Compare the overall reaction rate law given in the problem statement with the rate law determined from the mechanism. The stated overall reaction is second order in \(\mathrm{AB}\) and zero order in \(\mathrm{C}\). The rate law for the slow step is consistent with this, as it depends on \(\mathrm{AB}\) but not on \(\mathrm{C}\). Hence, the slow step agrees with the overall reaction order.

Consider Intermediate Reactant

Look at the role of \(\mathrm{AB}_2\) in the mechanism. It is an intermediate formed in the slow step and consumed in the fast step. Since \(\mathrm{C}\) is involved in the fast step, which is not the rate-determining step, it's reasonable that \(\mathrm{C}\) does not appear in the rate law for the overall reaction. The rate law is consistent with being zero order with respect to \(\mathrm{C}\), even though \(\mathrm{C}\) is involved in the fast step.

Validate the Mechanism with the Experimental Rate Law

It is essential to ensure that the mechanism proposed can lead to the experimentally determined rate law. In this case, the mechanism implies that the rate law should involve \(\mathrm{AB}\) in second order as the slow step involves two \(\mathrm{AB}\) molecules, and zero order for \(\mathrm{C}\) because it is involved only in the fast step. Since this aligns with the given experimental observation, we can suggest that the mechanism is probable.

Key concepts

Rate Law Determination

Understanding the rate of a chemical reaction is crucial because it tells us how fast or slow a reaction occurs. To determine the rate law, we observe how the concentration of reactants affects the reaction rate.

The rate law is an equation that quantifies the effect of reactant concentrations on the rate of a reaction. It's usually determined experimentally but can be predicted by examining the reaction mechanism, which outlines the step-by-step sequence of elementary reactions leading to the product formation.

To find the rate law from a mechanism, focus on the slowest step, known as the rate-determining step (RDS), since it has the greatest impact on the overall reaction rate. Take the example where \(\mathrm{AB} + \mathrm{AB} \rightarrow \mathrm{AB}_2 + \mathrm{A}\) is the slow step. Here, the rate law is of the form \(rate = k[\mathrm{AB}]^2\), indicating that the reaction rate depends on the concentration of \(\mathrm{AB}\) to the second power. This step explains why the reaction is second order in \(\mathrm{AB}\).

Through rate law determination, chemists gain insight into the kinetics of a reaction, which is essential for controlling and optimizing industrial chemical processes, pharmaceutical development, and understanding natural phenomena.

Reaction Order

The reaction order is a fundamental concept that indicates how the reaction rate is affected by the concentration of reactants. Specifically, it tells us the power to which the concentration of a reactant is raised in the rate law equation.

For instance, if the rate law is \(rate = k[\mathrm{AB}]^2[\mathrm{C}]^0\), the reaction is second order with respect to \(\mathrm{AB}\) and zero order in \(\mathrm{C}\). The 'second order' means that if you double the concentration of \(\mathrm{AB}\), the rate of reaction increases by a factor of four (since \(2^2 = 4\)). Meanwhile, 'zero order' indicates that changes in the concentration of \(\mathrm{C}\) have no effect on the rate. Thus, even if the concentration of \(\mathrm{C}\) is doubled or halved, the rate remains constant.

Understanding the reaction order helps predict how changes in conditions will alter the reaction rate, which is crucial in scenarios ranging from cooking to the manufacturing of chemicals.

Rate-Determining Step

The rate-determining step (RDS) is the slowest step in a reaction mechanism and hence, it controls the overall rate of the reaction. You can think of it as the 'bottleneck' in a production line; no matter how fast the other steps are, the overall pace is dictated by this slowest step.

In the given example, \(\mathrm{AB} + \mathrm{AB} \rightarrow \mathrm{AB}_2 + \mathrm{A}\) represents the RDS because it progresses more slowly than the other step in the mechanism. Identifying the RDS allows chemists to manipulate reaction conditions to increase efficiency or to design catalysts that target the slowest step to speed up the overall process.

The presence of intermediates, like \(\mathrm{AB}_2\) in our example, can also provide clues about the RDS. Since intermediates are involved in subsequent steps of a mechanism, they do not appear in the final rate law for the reaction if they are formed and consumed after the RDS. This is why, despite \(\mathrm{C}\) reacting with the intermediate \(\mathrm{AB}_2\), it doesn't influence the rate law, confirming the zero order with respect to \(\mathrm{C}\).