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Chemistry Question on Rate of a Chemical Reaction

Mechanism of a hypothetical reaction X2+Y2→2XYX _{2}+ Y _{2} \rightarrow 2 XY is given below: (i) X2→X+XX_{2} \rightarrow X+X (fast) (ii) X+Y2⇌XY+YX + Y _{2} \rightleftharpoons XY + Y ( slow ) (iii) X+Y→XYX + Y \rightarrow XY (fast) The overall order of the reaction will be

A

2

B

0

C

1.5

D

1

Answer

1.5

Explanation

Solution

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][Y2]= k [ X ]\left[ Y _{2}\right]...(1) k=k = rate constant of step (ii) Assuming step (i) to be reversible, its equilibrium constant, keq=[X]2[X2]k_{e q}=\frac{[X]^{2}}{\left[X_{2}\right]} ⇒[X]=keq12[X2]12\Rightarrow[X]=k_{e q}^{\frac{1}{2}}\left[X_{2}\right]^{\frac{1}{2}}...(2) Put (2) in (1) Rate =kkeq12[X2]12[Y2]= kk _{ eq }^{\frac{1}{2}}\left[ X _{2}\right]^{\frac{1}{2}}\left[ Y _{2}\right] Overall =12+1=32=\frac{1}{2}+1=\frac{3}{2}