$A k_{1} - \to B k_{2} - \to C k_{3} - \to D, k_{3} > k_{2} > k_{1}$ , then the rate determining step of the reaction is:

A \to B

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

C \to D

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

B \to C

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

A \to D

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

Open in App

Solution

The correct option is A $A \to B$
Slowest Step is the Rate Determining Step. The one with lower rate constant is the Rate determining step.
So, $A \to B$ is the rate determining step.

Suggest Corrections

Similar questions

A k_{3} - \to B k_{2} - \to C k_{1} - \to D,

k_{3} > k_{2} > k_{1}

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

Which of the following expression is/are correct?

2 A \to B + 3 C; i f - \frac{d [A]}{d t} = k_{1} [A]^{2}; \frac{d [B]}{d t} = k_{2} [A]^{2}; \frac{d [C]}{d t} = k_{3} [A]^{2}

k_{1}, k_{2} a n d k_{3}

2 A ⟶ B + 2 C

- \frac{d [A]}{d t} = k_{1} [A]^{2}

- \frac{d [B]}{d t} = k_{2} [A]^{2}

- \frac{d [C]}{d t} = k_{3} [A]^{2}

k_{1}

k_{2}

k_{3}

Join BYJU'S Learning Program