Mechanism of a hypothetical reaction, is given below : The overall order of the reaction will be [2017]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Mechanism of a hypothetical reaction, $X_{2} + Y_{2} ⟶ 2 X Y,$ is given below :

(i)   $X_{2} ⟶ X + X (fast)$
(ii)   $X + Y_{2} ⇄ X Y + Y (slow)$
(iii)   $X + Y ⟶ X Y (fast)$

The overall order of the reaction will be [2017]

1 2

2 0

3 1.5

4 1

Ans.
(3)

Note: Correct the reactions given in question as

$X_{2} ⇌ X + X$ (fast)

$X + Y_{2} \to X Y + Y$ (slow)

Slow step is the rate determining step.

$Rate = k [X] [Y_{2}]$ ...(i)

Equilibrium constant for fast step, $K = \frac{{[X]}^{2}}{[X_{2}]}$

$[X] = \sqrt{K [X_{2}]}$

By substituting [X] in equation (i), we get $Rate = k \sqrt{K [X_{2}]}$ $[Y_{2}] = k' {[X_{2}]}^{1 / 2} [Y_{2}]$

$∴ Order of reaction = \frac{1}{2} + 1 = \frac{3}{2} = 1.5$

Want Us to Email you About Special Offers & Updates?