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).

Step by step solution

01

(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: \(A + B \longrightarrow C + X\)

02

(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.

03

(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: \(A + X \longrightarrow C + D\)

04

(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] = \(k[A] \cdot \frac{[A][B]}{[C]K_{eq}}\)

05

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

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

06

(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).

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