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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

If the rate constant $k = 4.5 \times 10^{-7} \text{ L}^2 \text{ mol}^{-2} \text{ s}^{-1}$, then what is the order of the reaction?

The given reaction $2NO + O_2 \to 2NO_2$ is an example of

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?

The rate of reaction,$A + B + C \longrightarrow P$ is given by
$r = K[A]^{1/2} [B]^{1/2} [C]^{1/4}$
The order of reaction is

The mechanism for the reaction is given below:
$2P + Q \to S + T$
$P + Q \to R + S$ (slow)
$P + R \to T$ (fast)
The rate law expression for the reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module