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: $$ \begin{array}{l} \text { Step } 1: \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X} \\\ \text { Step } 2: \mathrm{A}+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D} \end{array} $$ \(\mathrm{X}\) 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[A][B]. (b) If Step 2 is rate-determining, the predicted rate law expression is: Rate = \(\frac{k}{K_{eq}} [A]^2[B]/[C]\). (c) Your result for part (b) might be considered surprising for reason (iii): Both the concentration of a product is in the rate law and there is a negative reaction order in the rate law.

Step-by-step solution

(Step 1: Identify the rate-determining step in part (a))

In part (a), it asks us to find the rate law expression if step 1 is rate-determining. When a step is rate-determining, it means that step is the slowest step in the mechanism and is controlling the overall rate of the reaction. In this case, step 1 is given by: \(A + B \longrightarrow C + X\)

(Step 2: Write the rate law expression for part (a))

Since step 1 is the rate-determining step, the rate law expression can be written by considering the reactants of this step. We have: Rate = k[A][B] Where k is the rate constant and [A] and [B] are the concentrations of species A and B, respectively.

(Step 3: Identify the rate-determining step in part (b))

In part (b), it asks us to find the rate law expression if step 2 is rate-determining. In this case, step 2 is given by: \(A + X \longrightarrow C + D\)

(Step 4: Write the rate law expression for part (b))

When step 2 is rate-determining, the rate law expression can be written by considering the reactants of this step. However, since X is an intermediate, we need to eliminate it from the expression. From step 1, we can write the following equilibrium expression: \(K_{eq} = \frac{[C][X]}{[A][B]}\) Where \(K_{eq}\) is the equilibrium constant. Now, solve the expression for X concentration. [X] = \(\frac{[A][B]}{[C] K_{eq}}\) Now substitute the concentration of X in the rate law expression for step 2: Rate = k[A][X] = \(k[A] \cdot \frac{[A][B]}{[C]K_{eq}}\)

(Step 5: Simplify the rate law expression for part (b))

Now we can simplify the rate law expression for part (b) as: Rate = \(\frac{k}{K_{eq}} [A]^2[B]/[C]\)

(Step 6: Evaluate the result for part (c))

For part (c), we need to evaluate if the result found in part (b) might be considered surprising for some reasons. Let's analyze each possibility: (i) The concentration of a product is in the rate law. Considering the result of part (b), we can see that the concentration of species C, which is a product, is indeed present in the rate law expression. So, this statement is true. (ii) There is a negative reaction order in the rate law. As the reaction order for species C is -1 (due to its presence in the denominator), this statement is also true. (iii) Both reasons (i) and (ii). Since statements (i) and (ii) are true, this statement is also true. (iv) Neither reasons (i) nor (ii). As explained earlier, both reasons (i) and (ii) are true, so this statement is not true. Thus, the answer for part (c) is (iii).

Key concepts

Rate Law Expression

In reaction kinetics, the rate law expression is vital for understanding how the concentration of reactants affects the rate at which a reaction proceeds. The rate law is typically determined by the slowest or rate-determining step in a reaction mechanism. It expresses the reaction rate as a function of the concentration of the reactants, each raised to a power corresponding to their order in the rate law.For example, if Step 1 in our reaction is rate-determining, the rate law for the reaction can be expressed as:- Rate = \(k[A][B]\)This indicates that the rate depends linearly on the concentration of both reactant A and B. The rate constant \(k\) is specific to the reaction and changes with temperature and presence of a catalyst.

Intermediate Species

In multi-step reaction mechanisms, intermediate species are formed in one step and consumed in another. They don't appear in the overall balanced equation. However, they play a crucial role in the mechanism of the reaction.For this particular reaction, species \(\text{X}\) is formed as an intermediate in Step 1 and used up in Step 2:- Step 1: \(\text{A} + \text{B} \rightarrow \text{C} + \text{X}\)- Step 2: \(\text{A} + \text{X} \rightarrow \text{C} + \text{D}\)Though intermediate species like \(\text{X}\) are transient and short-lived, their concentration is crucial when deriving rate laws for reactions where they are involved.

Rate-Determining Step

The rate-determining step in a reaction mechanism is the slowest step, acting like a bottleneck for the entire process. In chemical reactions, this step governs the rate at which the reaction proceeds.When Step 1 is the rate-determining step:- The rate law is based directly on the reactants in this step. Hence, Rate = \(k[A][B]\).When Step 2 is the rate-determining step and involves the intermediate \(\text{X}\), additional care is needed:- The concentration of \(\text{X}\) must be expressed in terms of other reactants/products using equilibrium expressions, ensuring that intermediates don't appear in the final rate law.This concept highlights the importance of identifying the slowest step in predicting reaction kinetics accurately.

Equilibrium Constant

The equilibrium constant \(K_{eq}\) helps in relating the concentrations of reactants and products in a reversible reaction. It finds relevance in mechanisms with fast equilibrium steps, where intermediate species' concentrations need elimination from the rate law.In the given reaction:- Step 1 is assumed to reach an equilibrium where \(K_{eq} = \frac{[C][X]}{[A][B]}\).- Solving for \([X]\), we find \([X] = \frac{[A][B]}{[C]K_{eq}}\).This rearrangement allows us to replace \(\text{X}\) when writing the rate law for Step 2, ensuring accurate expression without intermediates. Such manipulations showcase how equilibrium constants bridge the understanding of reaction dynamics between elementary steps and overall reaction kinetics.