🤓 Based on our data, we think this question is relevant for Professor Moussa's class at VCU.

Solution: When the concentration of A is doubled, the rate for the reaction: 2 A + B → 2 C quadruples.When the concentration of B is doubled the rate remains the same. Which mechanism below is consistent with the experimental observations?      A) Step 1:  A + B → D (slow)Step 2:  A + D → 2 C (fast)      B) Step 1:   A → D (slow)Step 2:  A + B + D → E (fast)Step 3:  E → 2 C (fast) C) Step 1:  2 A → D (slow)Step 2:  B + D → E (fast)Step 3:  E → 2 C (fast)      D) Step 1:  A + B → D (slow)Step 2:  A + D ⇌ 2 C (fast equilibrium)

Problem

When the concentration of A is doubled, the rate for the reaction: 2 A + B → 2 C quadruples.

When the concentration of B is doubled the rate remains the same. Which mechanism below is consistent with the experimental observations?     

 

A) Step 1:  A + B → D (slow)

Step 2:  A + D → 2 C (fast)     

 

B) Step 1:   A → D (slow)

Step 2:  A + B + D → E (fast)

Step 3:  E → 2 C (fast)

 

C) Step 1:  2 A → D (slow)

Step 2:  B + D → E (fast)

Step 3:  E → 2 C (fast)     

 

D) Step 1:  A + B → D (slow)

Step 2:  A + D ⇌ 2 C (fast equilibrium)

Solution

We’re being asked to determine the reaction mechanism that is consistent with the given experimental observations for this reaction:

2 A + B  2 C


Recall that a reaction mechanism is a sequence of single steps that add up to the chemical reaction

The rate law of the reaction depends on the rate-determining step of the mechanism, which is the slow step in the mechanism


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