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Animated Solution for Chemistry - Chemical Kinetics: Consider the reaction, The rate equation for this reaction is, Which of these mechanisms is/are consistent with this rate equation? (I) (slow) (fast) (II) (fast) (slow)

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Visualized Solution

\text{Rate Determining Step}

\text{Analyzing Mechanism (I)}

\text{Rate Law for Mechanism (I)}

\text{Analyzing Mechanism (II)}

\text{Initial Rate Law for Mechanism (II)}

\text{Eliminating the Intermediate}

\text{Final Rate Law for Mechanism (II)}

\text{Conclusion}

The Sigma Insight: Rate of Chemical Reaction

Solution Diagram
The study of chemical kinetics is not just about how fast a reaction goes, but how it gets there. A balanced chemical equation tells us the starting materials and the final products, but it rarely tells us the actual path the molecules take. This path is called the reaction mechanism.
In this problem, we are given an overall reaction and its experimentally determined rate law:
Our goal is to act as chemical detectives and determine which of the proposed mechanisms is consistent with this experimental fact.

The Golden Rule

The Rate-Determining Step
Imagine a relay race where one runner is significantly slower than the rest. The overall time of the team will be dictated almost entirely by that slow runner. In chemistry, this is known as the Rate-Determining Step (RDS). The slowest step in a multi-step mechanism acts as a bottleneck, and the rate law for the overall reaction is derived directly from this step.

Analyzing Mechanism (I)

Let's look at the first proposed mechanism: 1. (slow) 2. (fast)
The first step is explicitly labeled as the slow step. Therefore, it is our rate-determining step. According to the law of mass action, we can write the rate law directly from the stoichiometry of this elementary slow step:
This derived rate law perfectly matches the experimentally given rate equation! This is a strong indicator that Mechanism (I) is a plausible pathway for the reaction.

Analyzing Mechanism (II)

Now, let's evaluate the second mechanism. We must always check all options to avoid falling into traps. 1. (fast) 2. (slow)
Here, the second step is the slow one. Writing the rate law based on this step gives:
There is a major issue here: is an intermediate. It is produced in the first step and consumed in the second. Experimental rate laws must be expressed in terms of measurable quantities, typically the initial reactants, not fleeting intermediates.

Eliminating the Intermediate

To express the rate law in terms of the main reactants, we utilize the first step, which is a fast equilibrium. Because it's fast, we assume it reaches equilibrium quickly, allowing us to write an equilibrium constant expression:
We can rearrange this to solve for the concentration of our troublesome intermediate, :
Now, we substitute this expression back into our initial rate law for Mechanism (II):
Combining the constants and into a new constant , we get the final predicted rate law for Mechanism (II):

The Final Verdict

Comparing this derived rate law with the experimental one (), it is clear they do not match. The presence of in the derived law means that adding acid would slow down the reaction if Mechanism (II) were correct, which contradicts the given experimental rate law.
Therefore, we can confidently conclude that only Mechanism (I) is consistent with the observed kinetics of the reaction.

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