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 (fast equilibrium)
Step 2: A + D → 2 C (slow)
B) Step 1: A + B → D (slow)
Step 2: A + D ⇌ 2 C (fast equilibrium)
C) Step 1: 2 A → D (slow)
Step 2: B + D → E (fast)
Step 3: E → 2 C (fast)
D) Step 1: 2 A ⇌ D (fast equilibrium)
Step 2: B + D → E (slow)
Step 3: E → 2 C (fast)
Correct Answer:
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