Home
Class 12
CHEMISTRY
A following mechanism has been proposed ...

A following mechanism has been proposed for a reaction
`2A +B rarr D rarr E`
`A+ B rarrC+D `(slow)
`A+C rarr E` (fast)
The rate law expression for the reaction by RDS methd is:

A

`r=k[A]^(2)[B]`

B

`r=k[A][B]`

C

`r=k[A]^(2)`

D

`r=k[A][C]`

Text Solution

AI Generated Solution

The correct Answer is:
To derive the rate law expression for the given reaction mechanism using the Rate Determining Step (RDS) method, we can follow these steps: ### Step 1: Identify the Rate Determining Step (RDS) The first step in the mechanism is given as: \[ A + B \rightarrow C + D \] (slow) Since this step is the slowest, it is the Rate Determining Step (RDS). The rate of the overall reaction will be determined by this step. ### Step 2: Write the Rate Law for the RDS The rate law for a reaction can be expressed in terms of the concentration of the reactants involved in the RDS. For the RDS: \[ A + B \rightarrow C + D \] The rate law can be written as: \[ \text{Rate} = k [A]^m [B]^n \] where \( m \) and \( n \) are the stoichiometric coefficients of A and B in the RDS. ### Step 3: Determine the Stoichiometric Coefficients In the RDS, the stoichiometric coefficients for A and B are both 1. Therefore, we can substitute these values into the rate law: \[ \text{Rate} = k [A]^1 [B]^1 \] This simplifies to: \[ \text{Rate} = k [A] [B] \] ### Step 4: Final Rate Law Expression Thus, the final rate law expression for the overall reaction based on the RDS is: \[ \text{Rate} = k [A][B] \] ### Summary The rate law expression for the reaction based on the Rate Determining Step method is: \[ \text{Rate} = k [A][B] \] ---
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETIC & NUCLEAR CHEMISTRY

    NARENDRA AWASTHI|Exercise Level 1 (Q.3 To Q.32)|2 Videos
  • CHEMICAL KINETIC & NUCLEAR CHEMISTRY

    NARENDRA AWASTHI|Exercise Level 1 (Q.33 To Q.62)|3 Videos
  • SURFACE CHEMISTRY

    NARENDRA AWASTHI|Exercise Level 3 - Assertion - Reason Type Questions|1 Videos

Similar Questions

Explore conceptually related problems

A following mechanism has been proposed for a reaction: 2A+Brarr D+E A+B rarr C+D (slow) A+C rarr E (fast) The rate law expresison for the reaction is

2A + B to D +E , for the reaction proposed mechanism A + B to C +D (slow), A + C to E (fast). The rate law expression for the reaction is

2A+2B to D +E for the reaction following mechanism has been proposed . A+2B to 2C +d ( slow) A+2C to E ( fast ) The rate law expression for the reaction is

A chemical reaction A + 2B rArr AB_(2) follows in two steps A + B rArr AB(slow) AB + B rArr AB_(2) (fast) Then the order of the reaction is

For the reaction : A + 2B rarr C + D , the expression of rate of reaction will be :

A reaction is found to proceed in two steps as A + B rarr E (slow),, A+E rarr C+D (fast). Write the law expresison and overall balanced equation.

The mechanism for the overall reaction is A_(2) + B rarr C A_(2) rarr 2A (slow) 2A + B rarr X (fast) If a catalyst D changes the mechanism to A_(2) + D rarr A_(2) D (slow) A_(2) D rArr 2A+D (fast) 2A +B rarr C (fast) Which is the rate expresison for the reaction with and without a catalyst ?

The reaction , X + 2Y + Z rarr N occurs by the following mechanism (i) X+Y hArr M (very rapid equilibrium) (ii) M + Z rarr O (slow) (iii) O + Y rarr N (very fast) What is the rate law for this reaction

For the reaction 2NO_(2)+F_(2)rarr 2NO_(2)F , following mechanism has been provided: NO_(2)+F_(2) overset("slow")(rarr) NO_(2)F+F NO_(2)+Foverset("fast")(rarr) NO_(2)F Thus rate expression of the above reaction can be writtens as: