Test: Collision Theory of Chemical Reactions & Reaction Mechanism (November 25) - NEET MCQ

By slow step II,

By reversible fast step

∴ 
∴ 

I is reaction intermediate while II is transition complex.

By slow step II,

By fast step l,
 
∴ 
 
Net reaction is 

Thus, rate law is based on the slow step.

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 is the rate-determining step. 

and  

∴ 
which is not the rate law.

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, 

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.

Thus, is fast and rate-determining step is