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Q.

When the concentration of A is doubled, the rate for the reaction: 2A+B2C quadruples. When the concentration of B is doubled the rate remains the same. Which mechanism below is/are not consistent with the experimental observations? 

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a

Step I : 2A D ( fast equilibrium)

Step II :B+D E(slow)
Step III :E 2C (fast)

b

Step I :A+ B D (fast equilibrium)
Step II :A +D2C (slow)

c

Step I :A+B D (slow)
Step II :A+D 2C (fast equilibrium)

d

Step I : 2A D (slow)
Step II :B+D E (fast)
Step III :E 2C (fast)

answer is A, B, C.

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Detailed Solution

The information suggest that rate =k[A]2
This is satisfied by (D),
ln (A) and (C) B is involved in slow step,
ln (B), if we slove, B will appears in rate law.

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When the concentration of A is doubled, the rate for the reaction: 2A+B→2C quadruples. When the concentration of B is doubled the rate remains the same. Which mechanism below is/are not consistent with the experimental observations?