Question

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}\) Step 2: A \(+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}\) \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

Short answer

(a) If step 1 is rate determining, the predicted rate law expression is: \(Rate = k_1[\mathrm{A}][\mathrm{B}]\) (b) If step 2 is rate determining, the predicted rate law expression is: \(Rate = \frac{k_1 k_2}{k_{-1}}[\mathrm{A}]^2[\mathrm{B}]\) (c) The correct answer is (iv) Neither reasons (i) nor (ii).

Step-by-step solution

Determine the rate law expression for Step 1 being rate determining

When Step 1 is rate determining, the overall reaction rate is mainly dependent on the rate of Step 1. The rate law expression can be written directly from this step. We have: Step 1: \(\mathrm{A + B \longrightarrow C + X}\) The rate law expression for Step 1 is: \(Rate = k_1[\mathrm{A}][\mathrm{B}]\) The predicted rate law expression for the overall reaction when Step 1 is rate determining is: \(Rate = k_1[\mathrm{A}][\mathrm{B}]\)

Determine the rate law expression for Step 2 being rate determining

When Step 2 is rate determining, we need to take the intermediate X into account. We have: Step 2: \(\mathrm{A + X \longrightarrow C + D}\) Now, we need to eliminate the intermediate X from the rate law expression. We can use the equilibrium expression for Step 1: \(\mathrm{X} = \frac{k_1}{k_{-1}}[\mathrm{A}][\mathrm{B}]\) Now we can write the rate law expression for Step 2: \(Rate = k_2[\mathrm{A}][\mathrm{X}]\) Substitute the expression for X: \(Rate = k_2[\mathrm{A}]\left(\frac{k_1}{k_{-1}}[\mathrm{A}][\mathrm{B}]\right)\) This simplifies to: \(Rate = \frac{k_1 k_2}{k_{-1}}[\mathrm{A}]^2[\mathrm{B}]\)

Analyzing the rate law expressions

Analyzing the rate law expressions for the cases when step 1 and step 2 are rate determining, we can notice that the concentration of the intermediate species X is not present in the rate laws indicating that we have successfully found the rate law expressions for both cases. Now, we'll examine which of the given reasons make part (b) surprising: (i) The concentration of a product is in the rate law. This is not true for part (b), as only reactants A and B are present in the rate law expression for part (b). (ii) There is a negative reaction order in the rate law. This is also not true for part (b), as the reaction orders for both reactants A and B are positive. Hence, the correct answer is: (c) (iv) Neither reasons (i) nor (ii).

Key concepts

Rate Law

The rate law is a mathematical expression that describes the relationship between the concentrations of reactants and the rate of a chemical reaction. It is determined from the slowest, or rate-determining, step of the reaction mechanism. The rate law for a reaction can often be expressed in the form:

• Rate = k \( [A]^x [B]^y \)

Where:

• \( k \) is the rate constant.
• \( [A] \) and \( [B] \) are the concentrations of the reactants.
• \( x \) and \( y \) are the reaction orders with respect to the reactants, usually determined experimentally.

To determine the rate law for a given step, one must analyze which step in the mechanism controls the overall reaction speed. If Step 1 is rate determining, then, for example, the rate law depends on the concentrations of reactants involved in Step 1.

Reaction Mechanism

A reaction mechanism details the step-by-step process by which reactants transform into products. It is a sequence of elementary steps, each representing a single stage of the reaction at the molecular level. Understanding a reaction mechanism is crucial for predicting the rate law and is often used to infer the overall path of a chemical transformation.

• Each step in the mechanism can have its own reactants, products, and intermediates.
• Elementary steps are characterized by molecularity, indicating how many molecules participate in that step.

For the given hypothetical reaction:

• Step 1: \( A + B \rightarrow C + X \)
• Step 2: \( A + X \rightarrow C + D \)

The reaction takes place through two elementary steps, each contributing to our understanding of how the products \( C \) and \( D \) are formed. Step 1 and Step 2 involve different reactants and play their roles in governing the reaction speed and mechanism.

Rate Determining Step

The rate determining step is the slowest step in a reaction mechanism and effectively sets the pace for the entire reaction. Because it acts as a bottleneck, the rate determining step is crucial for calculating the overall rate law of the reaction.The interaction of reactants in this step dictates how quick or slow the entire reaction will appear:

• If Step 1 is the rate determining step, the reaction rate is dependent on this first step.\[ Rate = k_1 [A][B] \]
• If Step 2 is rate determining, we must consider intermediates, and the expression becomes:\[ Rate = \frac{k_1 k_2}{k_{-1}} [A]^2[B] \]

Recognizing which step is the rate determining step allows chemists to accurately predict the rate law and understand how different conditions can affect the rate.

Intermediate Species

Intermediates are species formed in one step of a reaction mechanism and consumed in a subsequent step. They are crucial to reaction mechanisms as they provide insight into the reaction's pathway but do not appear directly in the rate law of the overall reaction.Consider the reaction mechanism discussed:

• The intermediate \( X \) is produced in Step 1 and consumed in Step 2.
• Its transient nature makes it important for bridging the two steps together.

In deriving the rate law for Step 2 as rate determining, it is essential to express the intermediate in terms of known reactants to eliminate it from the final rate law. This often involves equilibrium expressions that relate intermediate concentrations to reactant concentrations. Here, \( X \)'s concentration was expressed as: \[ X = \frac{k_1}{k_{-1}} [A][B] \]Thus, intermediates function as key players in facilitating reactions without appearing in their final expressions.