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In. overall CO(g)+Cl_(2)(g)rarrCOCl_(2)(g)(1) Cl_(2)(K_(1))/(k_(k-1))2aFast equilibrium(2) u+CO(k_(2))/(sqrt(k_(-2)))CouFast equilibrium(3) COCl+u_(2)longrightarrow^(k_(3))COCl_(2)+u quad slow(4) 2Cllongrightarrow^(K_(4))Cl_(2)quad Fast.Now, we know that rate law is detemined by the slowest step.{:[rarr(d[CoCl_(2)])/(dt)=k_(3)[C_(2)][col]rarr" (A) but cocl doson't "],[" tape part in overall "],[" reaction "]:}from (2) (k_(2))/(k_(-2))=([Col])/([Q][oo])" so "[cos u]=(k_(2))/(k_(-2))[a][co]rarr" (B) "From (A) and (B) we have(d(coCl_(2)])/(dt)=(K_(2)K_(3))/(k_(-2))[Cl_(2)][co][Cl]rarr" ... See the full answer