For the hypothetical reaction 2X + G to Q + 2 | vedclass

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(C) The rate-determining step $(RDS)$ is the slow step of the mechanism.For option $C$:Step $1$: $X + X \rightleftharpoons X_2$ (fast equilibrium)Step

Vedclass · Accepted Answer

$X + X ightleftharpoons X_2$ (fast)$X_2 	o Q + T$ (slow)$T + G 	o 2M$ (fast)

Vedclass · Answer

(C) The rate-determining step $(RDS)$ is the slow step of the mechanism.For option $C$:Step $1$: $X + X ightleftharpoons X_2$ (fast equilibrium)Step $2$: $X_2 	o Q + T$ (slow,$RDS$)Step $3$: $T + G 	o 2M$ (fast)The rate of the reaction is determined by the slow step: $r = k_2[X_2]$.From the fast equilibrium in step $1$,the equilibrium constant $K_C = \frac{[X_2]}{[X]^2}$,which gives $[X_2] = K_C[X]^2$.Substituting this into the rate expression: $r = k_2 K_C [X]^2 = k[X]^2$.This matches the given rate law.

For the hypothetical reaction $2X + G \to Q + 2M$,the rate expression is $\frac{d[Q]}{dt} = k[X]^2$. Which of the following is the most likely mechanism?

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