The reaction,X + 2Y + Z to N occurs by the fo | vedclass

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(D) The rate of the reaction is determined by the slow step,which is the rate-determining step.From step $(ii)$,the rate expression is: Rate $= k_1[M]

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Rate $= k[X][Y][Z]$

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(D) The rate of the reaction is determined by the slow step,which is the rate-determining step.From step $(ii)$,the rate expression is: Rate $= k_1[M][Z]$ ...... $(1)$Since step $(i)$ is a very rapid equilibrium,we can write the equilibrium constant expression:$K_{eq} = \frac{[M]}{[X][Y]}$Therefore,$[M] = K_{eq}[X][Y]$ ...... $(2)$Substituting the value of $[M]$ from equation $(2)$ into equation $(1)$:Rate $= k_1 K_{eq} [X][Y][Z]$Let $k = k_1 K_{eq}$,then the rate law becomes:Rate $= k[X][Y][Z]$

The reaction,$X + 2Y + Z \to N$ occurs by the following mechanism:
$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)
$(ii)$ $M + Z \to P$ (slow)
$(iii)$ $P + Y \to N$ (very fast)
What is the rate law for this reaction?

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The reaction,$X + 2Y + Z \to N$ occurs by the following mechanism:$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)$(ii)$ $M + Z \to P$ (slow)$(iii)$ $P + Y \to N$ (very fast)What is the rate law for this reaction?