Mechanism of a hypothetical reaction X 2 + Y 2 → 2 XY is given below: (i) X 2 → X + X fast (ii) X + Y 2 ⇌ XY + Y slow (iii) X + Y → XY fast The overall order of the reaction will be

Question

Aicharya · Accepted Answer

The solution of this question is given by assuming step

(i)

to be reversible which is not given in question

Overall rate = Rate of slowest step (ii)

= k [X] [Y_{2}]

......

(i)

K

=

rate constant of step

(ii)

Assuming step

(i)

to be reversible, its equilibrium constant,

k_{e q} = \frac{{[X]}^{2}}{[X_{2}]} \Rightarrow [X] = k_{e q}^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}}

.....

(ii)

Put

(ii)

in

(i)

Rate

= k k_{e q}^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}} [Y_{2}]

Overall order

= \frac{1}{2} + 1 = \frac{3}{2}

Mechanism of a hypothetical reaction X 2 + Y 2 → 2 XY is given below: (i) X 2 → X + X fast (ii) X + Y 2 ⇌ XY…

Solution