Test: Collision Theory & Reaction Mechanism - NEET MCQ

By slow step II,

By reversible fast step

∴ 
∴ 

 In I, slow step is

Thus,
 

∴ 
In fast reversible step,

∴ 
 
∴
 

I is reaction intermediate while II is transition complex.

Slow step I is the rate-determining step.

By slow step II,

By fast step l,
 
∴ 
 
Net reaction is 

Thus, rate law is based on the slow step.

(a) Slow step is the rate-determining step, thus by
rate law = k[A][B]
Net reaction,

Thus, this mechanism is not consistent with the stoichiometry, but is consistent with rate law.
(b) By slow step,
Rate law = k₃[A][B₂]
By fast step, 

∴ 
∴ 
Thus, this mechanism is consistent with the stoichiometry of the net reaction but not with the rate law.
Thus, (b) is correct.

If step II is fast and reversible, then

∴ 
∴ 

H₂O is a weak nucleophile and cannot attack tertiary carbon of maximum electron density directly. Thus, rate-determining slow step is
step is
   
and rate law is  
and [H₂O] is not involved.
It is S_N1 reaction.

Slow step II is the rate-determining step.
...(1)
By step I in equilibrium,
 
∴ 
Put the value of [N₂O₂] in Eq. (i), we get
∴ 

Slow step is the rate-determining step. 

and  

∴ 
which is not the rate law.

From slow step,

From equilibrium step,

∴ 

It is second order in A and first order in B thus, total order = 3. 
II. 
∴ 

Thus, it is second order in A and first order in B.
Total order = 3

Thus, it is second order in A and first order in B.
Thus total order = 3

Slow step (II) is the rate-determining step.

By fast step I in equilibrium.

∴ 
∴  

Step / is fast and reversible

Step // is slow and rate-determining step.
∴ 
∴ 
∴ 
where, k₁ k_eq 

Step / is fast and reversible

Step // is slow and rate-determining step.
∴ 
∴ 
∴
 
where, 

Half-life period is independent of [N₂O₅]. Hence, it is first order reaction.

Half-life period is independent of [N₂O₅]. Hence, it is first order reaction.

Molecularity of each step is two thus, true.
Thus, (a) is correct.
The limiting step of a complex reaction is always the slowest, thus (b) is correct.
It involves more than one elementary step, thus, it is a complex reaction. Thus, (d) is also correct.
HOCl appears in step I and disappears in step II, Thus it is an intermediate.
HOI appears in step II and disappears in step III. Thus, HOI is also an intermediate.
Thus, (c) is incorrect.
Slow step il is the rate-determining step.

Note H₂O as reactant is not taken in equilibrium and rate expression.

Molecularity of each step is two thus, true.
Thus, (a) is correct.
The limiting step of a complex reaction is always the slowest, thus (b) is correct.
It involves more than one elementary step, thus, it is a complex reaction. Thus, (d) is also correct.
HOCi appears in step I and disappears in step II, Thus it is an intermediate.
HOI appears in step II and disappears in step III. Thus, HOI is also an intermediate.
Thus, (c) is incorrect.
Slow step il is the rate-determining step.


Note H₂O as reactant is not taken in equilibrium and rate expression.

Rate determining step is 




The reaction is third order in A and first order in B.
Total order = 3 + 1 = 4

Thus, is fast and rate-determining step is