Home
Class 12
CHEMISTRY
The rate of reaction is determined by sl...

The rate of reaction is determined by slow step reaction. The step is called

A

Reaction rate

B

Activation step

C

Rate determining step

D

None of the above

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the term used for the step in a reaction that determines the rate of the reaction, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Reaction Mechanisms**: - In a chemical reaction, there can be multiple steps involved. Each step can have a different rate at which it occurs. 2. **Identifying the Slow Step**: - Among these steps, one step is typically slower than the others. This slow step is crucial because it limits the overall rate of the reaction. 3. **Defining the Term**: - The step that determines the rate of the reaction is known as the **rate-determining step** (RDS). This term is used to describe the slowest step in a reaction mechanism. 4. **Implications of the Rate-Determining Step**: - Since the rate of the entire reaction cannot exceed the rate of the slowest step, the rate-determining step effectively controls how fast the overall reaction proceeds. 5. **Conclusion**: - Therefore, the answer to the question is that the step which determines the rate of the reaction is called the **rate-determining step**.
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

In a multi-step reaction, the rate is determined by conisdering the………….step.

What is the rate determining step of a reaction ?

Knowledge Check

  • The steps of dark reactions are

    A
    regeneration `to` carboxylation `to` reduction
    B
    reduction `to` oxidation `to` hydrogenation
    C
    carboxylation `to` reduction `to` regeneration
    D
    reduction `to` carboxylation `to` regneration

  • The rate determining step of a reaction is the step

    A
    which involves the minimum number of molecules
    B
    which involves the maximum number of molecules
    C
    which is the slowest
    D
    which is the fastest

  • Assertion : Complex reaction takes place in different steps and the slowest step determines the rate of reaction. Reason : Order and molecularity of a reaction are always equal.

    A
    If both assertion and reason are true and reason is the correct explanation of assertion.
    B
    If both assertion and reason are true but reason is not the correct explanation of assertion.
    C
    If assertion is true but reason is false.
    D
    If both assertion and reason are false.

    • Similar Questions

    Explore conceptually related problems

    Determining the rate law from a mechanism with an initial slow step: Ozone reacts with nitrogen dioxide to produce oxygen and dinitrogen pentoxide O_(3)(g)+2NO_(2)(g)rarrO_(2)(g)+N_(2)O_(5)(g) The proposed mechanism is O_(3)(g)+NO_(2)(g)overset(slow)rarr NO_(3)(g)+O_(2)(g) NO_(3)(g)+NO_(2)(g)overset("fast")rarrN_(2)O_(5)(g) What is the rate law predicted by this mechanism ? Strategy : The designations "slow" and "fast" indicate the relative rates of the steps. The rate is determined completely by the slow step, or rate-determining sterp.

    X overset("Step"1)to Y overset("StepII")to Z is a complex reaction. Total order of reaction is 2 and step II is a slow step. What is the molecularity of step II?

    1st step (reaction) of glycolysis is -

    The rate law expresison is given for a typical reaction, n_(1)A + n_(2) B rarrP as r = k[A]^(n)[B]^(n2) . The reaction completes only in one step and A and B are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for S_(N)l reaction r = k[RX] . If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction: [[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]] If we increase the concentration of I_(2) two times, then the rate of formation of HI will

    The rate law expresison is given for a typical reaction, n_(1)A + n_(2) B rarrP as r = k[A]^(n)[B]^(n2) . The reaction completes only in one step and A and B are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for S_(N)l reaction r = k[RX] . If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction: [[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]] Which of the following expresison is correct?

    Home
    Profile