## Chemistry and Chemical Reactivity (9th Edition)

$Rate = (K_Fk_S)[A]^2[B]$
Slow, determining rate: $r=r_S=k_S[I][B]$ Rate of formation of I $r_{-F}=k_{-F}[I]=r_F$ (equilibrium) $r_{F}=k_F[A]^2$ $K_F=k_F/k_{-F}\rightarrow [I]=K_F[A]^2$ Therefore: $Rate = (K_Fk_S)[A]^2[B]$