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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ck196

A mechanism for $2 A + B products$ is: fast equilibrium $A + B I$ (constant $K$ ), then slow $I + A products$ (constant $k_{2}$ ). Select the correct statement(s).

AHalving $[A]$ and doubling $[B]$ halves the rate.
BThe rate law is $k_{2} K [A]^{2} [B]$ , of overall order 3.
CThe rate is independent of $[B]$ because B is consumed in the fast step.
DThe slow step is termolecular.

Answer: A, B

Slow step: rate

= k_{2} [I] [A]

with

[I] = K [A] [B]

, so rate

= k_{2} K [A]^{2} [B]

, order

2 + 1 = 3

. Changing

[A]

by

0.5

and

[B]

by

2

multiplies the rate by

0. 5^{2} \times 2 = 0.5

. The slow step involves I and A, so it is bimolecular, and the rate depends on

[B]

through the pre-equilibrium.

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