The reaction Q+R2=R2Q is found to be first order in R2 and

The steps in a mechanism are elementary reactions, so the exponents in the rate laws are the coefficients in the steps.
No reaction can go faster than its slowest step.

(D) R₂ ⟶ 2R (slow); rate = k[R₂]

(A) is wrong. rate = k[Q][R₂]. Reaction is 1st order in Q.

(B) is wrong. rate = k[Q]². Reaction is 2nd order in Q.

(C) is wrong. rate = k[Q][Z]. Reaction is 1st order in Q and 1st order in Z.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-12-16T10:28:45+01:00

The correct mechanism of the reaction is shown in the sequence of steps in option (D).

A reaction is said to occur in an elementary step if it takes place in a single reactive encounter. However, some reactions do not take place in a single reactive encounter rather they take place in a sequence of reactions. The rate determining step is the slowest step in the reaction sequence.

Since the reaction is first order in R2 and zero order in Q, the only possible reaction mechanism is;

R2 -----> 2R(slow)

R + Q ----> RQ(fast)

RQ + R -----> R2Q (fast)

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