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For a two step reaction. `AhArrR+B" "R+Coverset(k_(2))toP` (where, R is a reactive intermediate whose concentration is maintained at some low steady state throughout the reaction). If the concentration of C is very high then the order of reaction for formation of "P" isA. 2B. 0C. 1D. `1//2` |
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Answer» Correct Answer - C Rate law `=(dp)/(dt)=K_(2)[R][C]` at steady state `(d[R])/(dt)=0` `(d[R])/(dt)=K_(1)[A]-K_(2)[R][B]-K_(2)[R][C]=0` `K_(1)[A]={K_(2)[B]+K_(2)[C]}[R]` `[R]=(K_(1)[A])/(K_(2){[B]+[C]})` Rate `=K_(2)xx(K_(1))/(K_(2))([A])/([B]+[C])xx[C]` `ul(K_(1)[A])` `=([B])/([C])+1` where C is very high Rate `=K_(1)[A]` `:.` order of reaction` = 1 " "]` |
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