The mechanism of the reaction : $A + 2 B + C \to D$ is
(step-1)(fast) equilibrium $A + B ⇌ X$
(step-2)(slow) $X + C ⟶ Y$
(step-3)(fast) $Y + B ⟶ D$
Which rate law is correct :

(A)

v = k [C]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(B)

v = k [A] {[B]}^{2} [C]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(C)

v = k [A] [B] [C]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(D)

v = k [D]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D (C) $v = k [A] [B] [C]$
$r = k [X] [C]$
$k^{'} = \frac{[X]}{[A] [B]}$
$r = k^{'} [A] [B] [C]$

Suggest Corrections

Similar questions

Q. A reaction

A ⟶ D

, involves following mechanism:
Step 1:

A k_{1} - -------- \to B

(fast)
Step 2:

B k_{2} - -------- \to C

(slow)
Step 3:

C k_{3} - -------- \to D

(fast)
The rate law of the reaction may be given as:

A reaction : $A_{2} + B \to$ Products, involves the following mechanism :
$A_{2} ⇌ 2 A (f a s t)$ (A being the intermediate)
$A + B \to k_{2} (s l o w)$ .The rate law consistent to this mechanism is :

Q. In the sequence of reaction

A_{1} B_{2} C_{3} D k_{3} > k_{2} > k_{1}

then the rate determining step of the reaction

Q. The following mechanism has been proposed for a reaction:

A + B \to C + D

(Slow)

A + C \to E

(Fast)

2 A + B \to D + E

The rate law expression for the reaction is:

For a single step reaction to the type A + 2B → E + 2F

Join BYJU'S Learning Program

The mechanism of the reaction :A+2B+C→D is(step-1)(fast) equilibrium A+B⇌X(step-2)(slow) X+C⟶Y(step-3)(fast) Y+B⟶DWhich rate law is correct :

The correct option is D (C)v=k[A][B][C]r=k[X][C]k′=[X][A][B]r=k′[A][B][C]

The mechanism of the reaction : $A + 2 B + C \to D$ is
(step-1)(fast) equilibrium $A + B ⇌ X$
(step-2)(slow) $X + C ⟶ Y$
(step-3)(fast) $Y + B ⟶ D$
Which rate law is correct :

The correct option is D (C) $v = k [A] [B] [C]$
$r = k [X] [C]$
$k^{'} = \frac{[X]}{[A] [B]}$
$r = k^{'} [A] [B] [C]$