# Consider the following three-step representation of a reaction mechanism. Step 1: A + B mc016-1.jpg AB (fast) and the rate = k[A][B] Step 2: AB + B mc016-2.jpg AB2 (slow) and the rate = k[AB][B] Step

Consider the following three-step representation of a reaction mechanism. Step 1: A + B mc016-1.jpg AB (fast) and the rate = k[A][B] Step 2: AB + B mc016-2.jpg AB2 (slow) and the rate = k[AB][B] Step 3: AB2 + B mc016-3.jpg AB3 (fast) and the rate = k[AB2][B] Overall: A + 3B mc016-4.jpg AB3 and the rate = k[A][B]2 Which explains why the rate law for the overall equation is not the same as the rate equation for the rate-determining step?

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Consider the following three-step representation of a reaction mechanism. Step 1: A + B mc016-1.jpg AB (fast) and the rate = k[A][B] Step 2: AB + B mc016-2.jpg AB2 (slow) and the rate = k[AB][B] Step
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