2 NO (g) + O₂ (g) --> 2 NO₂ (g) The proposed mechanism for the reaction above is 1) NO(g) + NO (g) --> N₂O₂ (g) 2) N₂O₂(g) + O₂ (g) --> 2 NO₂ (8) Write the rate law for the mechanism if the first step is fast and the second step is slow.

The isomerization of cyclopropane, C3H6, is believed to occur by the mechanism shown in the equations above. Here  C3H6* is an excited cyclopropane molecule. At low pressure, Step 1 is much slower than Step 2. Derive the rate law for this mechanism at low pressure?

A proposed mechanism for the decomposition of hydrogen peroxide consists of three elementary steps: (1) H₂O2(g) →>> 2OH(g) (2) H₂O2(g) + OH(g) → H₂O(g) + HO₂(g) (3) HO₂(g) + OH(g) → H₂O(g) + O₂(g) If the rate law for the reaction is first order in H₂O2, which step in the mechanism is the rate-determining step? O Step 1 O None of the steps O Step 3 O Impossible to determine Steps 1 and 2 O All of the steps O Step 2 Hint: The slowest step in a mechanism determines the rate law. From each elementary reaction we can write a rate-law expression. Submit Question

4. The rate law for the reaction H₂ (g) + I₂ (g) → 2 HI (g) was determined to be rate = k [H₂] [1₂], which led to a simple single-step bimolecular mechanism where an H₂ molecule simply collided with an I₂ molecule. In the 1960s, though, different teams found spectroscopic evidence for this mechanism: (1) (2) (3) I₂ (g) →21 (g) H₂(g) + I (g) → H₂I (g) H₂I (g) + I (g) → 2 HI (g) (fast) (fast) (slow) Show that this mechanism is consistent with the rate law: rate = k [H₂] [1₂]. To do this, assign individual rate constants for ALL forward and reverse reactions in the mechanism (I'm guessing k₁, k-1, k2, k-2 and k3). Then show that the overall rate law from combining individual rate laws for the mechanism steps yields rate = (some combination of k's) [H₂] [1₂], and thus k= (some combination of k's). Remember that there should be no reaction intermediates in the rate law, only reactants. \