Question

Consider the following elementary steps that make up the mechanism of a certain reaction: 3X→E+F E+M→F+N
Part A
What is the overall reaction? Express your answer as a chemical equation.
Part B
Which species is a reaction intermediate?
Part C
What is the rate law for step 1 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C]3 type k*[A]*[C]^3.
Part D
What is the rate law for step 2 of this reaction?

Answer 1

Answer:

Part A- The overall reaction is: 3X + M → 2 F + N.

Part B- The intermediate of this reaction is E.

Part C- Rate law for step 1 = k[X]³ = k*[X]^3.

Part D- Rate law for step 2 = k[E][M] = k*[E]*[M].

Explanation:

Part A: What is the overall reaction? Express your answer as a chemical equation.

• We can get the overall reaction by summing the two steps of the reaction.

Step 1: 3X → E + F.

Step 2: E + M → F + N.

The E component in the products side in step 1 will cancel that in the reactants side of step 2.

∴ The overall reaction is: 3X + M → 2 F + N.

Part B: Which species is a reaction intermediate?

• We should identify the intermediate firstly to determine the reaction intermediate.

The intermediate is the species that produced in a step of the reaction and consumed in next steps and do not appear in the overall reaction

So, the intermediate of this reaction is E.

Part C: What is the rate law for step 1 of this reaction?

The rate of the reaction is directly proportional to the concentration of the reactants.

Rate = k[reactants]ˣ.

where, k is the rate constant of the reaction,

x is the no. of moles of the reactants that involved in the reaction mechanism before the rate determining step.

So, The rate law for step 1 of this reaction (neglecting the mechanism) is:

Rate = k[X]³ = k*[X]^3.

Part D: What is the rate law for step 2 of this reaction?

Also, as in part C:

The rate law for step 2 of this reaction (neglecting the mechanism) is:

Rate = k[E][M] = k*[E]*[M].

Answer 2

Considering the provided elementary steps, the overall equation is:

3X + M → 2 F + N

E is the reaction intermediate.

The rate law for the Step 1 is:

$rate = k_1 [X]^{3}$

The rate law for the Step 2 is:

$rate = K_2 [E][M]$

Let's consider the following elementary steps that make up the mechanism of a certain reaction:

Step 1: 3X → E + F

Step 2: E + M → F + N

We can find the overall equation by adding both steps, side to side, and canceling what is repeated on both sides.

3X + E + M → E + 2 F + N

3X + M → 2 F + N

The reaction intermediate is the species that appears in the mechanism but not in the overall equation, i.e. E.

The rate law for each step can be expressed through the following expression.

$rate = k [A]^{x} [B]^{y}$

where,

• rate is the reaction rate
• k is the rate constant
• A and B are the reactants
• x and y are the reaction orders. For elementary steps, they coincide with the stoichiometric coefficients.

The rate law for Step 1 is:

$rate = k_1 [X]^{3}$

The rate law for Step 2 is:

$rate = K_2 [E][M]$

Considering the provided elementary steps, the overall equation is:

3X + M → 2 F + N

E is the reaction intermediate.

The rate law for the Step 1 is:

$rate = k_1 [X]^{3}$

The rate law for the Step 2 is:

$rate = K_2 [E][M]$

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