The mechanism of the reaction: `A + 2B + C rarr D` is
(step 1) (fast) equilibrium `A+B hArr X`
(step 2)(slow) `X + C rarr Y`
(step 3) (fast) `Y + B rarr D`
Which rate law is correct ?

For the single step reaction of type A + 2B rarr E + 2F the rate law is.

Mechanism of a hypothetical reaction X_(2) + Y_(2) rarr 2XY is given below: (i) X_(2) hArr X + X (fast) (ii) X+Y_(2)rarr XY+Y (slow) (iii) X + Y rarr XY (fast) The overall order of the reaction will be :

Mechanism of a hypothetical reaction X_(2) + Y_(2) rarr 2XY is given below: (i) X_(2) rarr X + X (fast) (ii) X+Y_(2) hArr XY+Y (slow) (iii) X + Y rarr XY (fast) The overall order of the reaction will be :

Mechanism of a hypothetical reaction X_(2) + Y_(2) rarr 2XY is given below: (i) X_(2) rarr X + X (fast) (ii) X+Y_(2) hArr XY+Y (slow) (iii) X + Y rarr XY (fast) The overall order of the reaction will be :

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

The mechanism of the reaction A+2B rarr C+D is: Step I: A+B underset(k_(2))overset(k_(1))hArr I Step II: B+I overset(k_(3))rarr C+D In the first step is a fast equilibruim , then the incorrect statement 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