Consider this overall reaction, which is experimentally observed to be second order in X and first order in Y: X + Y → XY. a. Does the reaction occur in a single step in which X and Y collide? b. Is this two-step mechanism valid? 2X →k1/k2 X2 (Fast) X2 + Y →k3 XY + X (Slow)

Verified step by step guidance

1. <Step 1: Understand the reaction order.> The reaction is given as second order in X and first order in Y. This means the rate law is rate = k[X]^2[Y].
2. <Step 2: Analyze the possibility of a single-step mechanism.> For a reaction to occur in a single step, the molecularity must match the reaction order. Here, the reaction order is third (2 in X and 1 in Y), which suggests a termolecular process. Termolecular reactions are rare due to the low probability of three molecules colliding simultaneously with the correct orientation and energy.
3. <Step 3: Evaluate the proposed two-step mechanism.> The mechanism consists of two steps: (1) 2X → X2 (fast) and (2) X2 + Y → XY + X (slow).
4. <Step 4: Determine the rate-determining step.> The slow step (step 2) is the rate-determining step. The rate law for this step is rate = k3[X2][Y].
5. <Step 5: Relate the mechanism to the observed rate law.> From step 1, X2 is formed quickly and is in equilibrium with 2X. Therefore, [X2] can be expressed in terms of [X] using the equilibrium constant for the fast step. Substitute [X2] in the rate law of the slow step to see if it matches the observed rate law rate = k[X]^2[Y]. If it does, the mechanism is valid.>