Answer : The rate law for the overall reaction is, [tex]R=\frac{K[A]^2[D]}{[C]}[/tex]
Explanation :
As we are given the mechanism for the reaction :
Step 1 : [tex]2A\rightleftharpoons B+C[/tex] (equilibrium)
Step 2 : [tex]B+D\rightarrow E[/tex] (slow)
Overall reaction : [tex]2A+D\rightarrow C+E[/tex]
First we have to determine the equilibrium constant from step 1.
The expression for equilibrium constant will be,
[tex]K'=\frac{[B][C]}{[A]^2}[/tex]
Form this, the value of [B] is,
[tex][B]=\frac{K'[A]^2}{[C]}[/tex] ............(1)
Now we have to determine the rate law from the slow step 2.
The expression for law will be,
[tex]Rate=K''[B][D][/tex] .............(2)
Now put equation 1 in 2, we get:
[tex]Rate=K''\frac{K'[A]^2}{[C]}[D][/tex]
[tex]Rate=\frac{K[A]^2[D]}{[C]}[/tex]
Therefore, the rate law for the overall reaction is, [tex]R=\frac{K[A]^2[D]}{[C]}[/tex]
