Question

The mechanism for the reaction 2X + Y → C + D is as follows :-
2X  X2 [fast]
X2 + Y → C + D [Slow]
The rate law expression is

(1)K[X]2[Y](2)K[X]2(3)K[Y](4)K[X]2[X2]

Answer 1

Slow step is rate determining step

in this rxn slow step is
2X + Y -----------> C + D

Rate = k[X]²[Y]

SO option (1) is correct

Answer 2

Answer : The correct option is, (1) $K[X]^2[Y]$

Explanation :

As we are given the mechanism for the reaction :

Step 1 : $2X\rightarrow X_2$    (fast)

Step 2 : $X_2+Y\rightarrow C+D$     (slow)

Overall reaction : $2X+Y\rightarrow C+D$

The rate law expression for overall reaction will be:

$Rate=K[X]^2[Y]$

Now we have to determine the rate law from the slow step 2.

The expression for law will be,

$Rate=K'[X_2][Y]$       .............(1)

Now applying steady state approximation for $X_2$, we get:

$\frac{d[X_2]}{dt}=K"[X]^2$      .........(2)

Now put equation 2 in 1, we get:

$Rate=K'K"[X]^2[Y]$

$K'K"=K$

Hence, the rate law expression will be:

$Rate=K[X]^2[Y]$