Question

A reaction mechanism is defined as the sequence of reaction steps that define the pathway from reactants to products. Each step in a mechanism is an elementary reaction, which describes a single molecular event of usually one or two molecules interacting.

The rate law for an overall reaction is the rate law for the slowest step in the mechanism, which is directly related to the stoichiometric coefficients of the reactants.

The exception to this rule occurs when the slowest step contains intermediates. In these cases, the slowest step is usually preceded by an equilibrium step, which can be used to substitute for the intermediates in the overall rate law.

Part A

What is the rate law for the following mechanism in terms of the overall rate constant k?

Step1:Step2:A+BB+C⇌→CD(fast)(slow)

Express your answer in terms of k and the necessary concentrations (e.g., k*[A]^3*[D]).

View Available Hint(s)

rate =	nothing

Submit

Part B

Consider the reaction

2XY2+Z2⇌2XY2Z

which has a rate law of

rate= k[XY2][Z2]

Select a possible mechanism for the reaction.

View Available Hint(s)

Consider the reaction

which has a rate law of

Select a possible mechanism for the reaction.

A Step1:Step2:Step3:Z2→Z+Z           (slow)XY2+Z→XY2Z   (fast) XY2+Z→XY2Z   (fast)

B Step1:Step2:XY2+Z2→XY2Z+Z   (slow)XY2+Z→XY2Z   (fast)

C Step1:Step2:XY2+Z2→XY2Z2   (slow)XY2Z2 →XY2Z+Z   (fast)

D Step1:Step2:2XY2⇌X2Y4   (fast)X2Y4+Z2→2XY2Z   (slow)

E Step1:2XY2+Z2→2XY2Z   (slow)

Answer 1

part-A step-1 A+B A (fast) Step-2 Btc kz L D (slow) rate is determined by slowest step called rate determining step (eds) : .: Та е Гв) (c) 2 = Const ok k [B] [c7 aka= rate second step here [c] is intermediate concentration, so we have to replace it with the help of step-1. We have to assume that step-1 is reversible and its late constant is ki then we can write

as ki= [A] [B] [ ] TA] [B] part-B .: [c] = ki [A] (18) put value of [c] in rate law expression de= K [c] [B] = kg KiTA) (3]CR) = kika [A] [B]2 rate Law r = k. [A] [B] ²1 fik=kik} expression au given chemical reaction, 2XY 2 + Z2 2 X Y Z and rate = k [XY₂] [22] correct option is B. solution step-1 XY₂ + Z₂ X Y Z + Z (slow) step-2 X Y ₂ + 2 - XY2Z (fast) Netran a 2x Y2 + 2₂ → 2X12 Z rate lan is written with the help of slowest step. : rate = k [XY2] [22] at is same as given rate law expression ③ is correct option