Experimentally Determined Rate Law: rate = k[NO][Cl₂]   Step 1: NO + Cl2 ---> NOCl2 (slow) Step 2: NOCl2 + NO ---> 2NOCl (fast) Step 1: NO+ Cl2 ---> NOCl2 (slow) Step 2: NOCl2 ---> NOCl + Cl (fast) Step 3: NO+ Cl ---> NOCl (fast) Step 1: NO + NO ---> N2O2 (slow) Step 2: N2O2 + Cl2 ---> 2NOCl (fast) Step 1: NO + Cl2 ---> NOCl + Cl (slow) Step 2: NOCl + NO ---> NOCl2 (fast) Step 1: Cl2 ---> 2Cl (slow) Step 2: 2NO + 2Cl ---> 2NOCl (fast) Step 1: NO + Cl2 ---> NOCl+ Cl (slow) Step 2: Cl + NO ---> NOCl (fast)

Suppose the decomposition of ozone proceeds by the following mechanism: step elementary reaction rate constant 1 03(g) O₂(g) + 0(g) k₁ 2 03(g) + O(g) → 20₂(g) k₂ Suppose also k₁ »k₂. That is, the first step is much faster than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally- observable rate law for the overall chemical reaction. rate = k Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of K₁, K2, and (if necessary) the rate constants k.1 and k-2 for the reverse of the two elementary reactions in the mechanism. = 0 k = 00 X 5 S ?

For the following reaction Mechanism write the overall reaction and write the rate law.  H3O++ I- → HI + H2O (fast)   H2O2 + HI → H2O + HOI (fast)  HOI + H3O++ I- →2H2O + I2 (slow)   I2 + I- → I3-(fast)

For the following reaction: 2N2O5(g) ---> 4 NO2(g) + O2(g) The rate law is first order with respect to N2O5 and has a rate constant of 4.82 x 10-3 sec-1 .  Calculate the rate of reaction when the [N2O5] = 0.25M.

The rate for the overall reaction NO2 (g) + CO (g) → NO (g) + CO2 (g) is Rate = k[NO2]2 a. Consider this mechanism: i. 2NO2 (g) → N2 (g) + 2O2 (g) [slow] ii. 2CO (g) + O2 (g) → 2CO2 (g) [fast] iii. N2 (g) + O2 (g) →2NO (g) [fast] d. Draw a reaction energy diagram for this mechanism, given that the overall reaction’s ΔH°rxn = -226 kJ/mol.

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Writing the rate law implied by a simple mechanism with an initial slow... Suppose the formation of nitryl fluoride proceeds by the following mechanism: elementary reaction step 1 NO₂ (g) +F₂ (g) 2 k₁ k₂ 2 F (g) + NO₂(g) → NO₂F (g) Suppose also k₁ « k₂. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction. NO₂F (g) +F (g) Note: your answer should not contain the concentrations of any intermediates. rate constant rate = k 0+ 00 X Ś

A proposed mechanism for the reaction is: - Step 1: H₂(g) + 2NO(g) → N₂O(g) + H₂O(g) Step 2: N₂O(g) + H₂(g) → N₂(g) + H₂O(g) What is the rate law? More information is needed to answer this question. Or = K[H₂][NO] Or = K[H₂]²[NO]² Or = K[H₂][NO]² Or=K[H₂]²[NO] (slow) (fast)

The rate law for the reaction: NH4+ + NO₂ N₂ + 2H₂O is rate =k[NH4+][NO₂] It is found that the rate constant for this reaction is 3.05x10-4 M-1s-1. Calculate the rate of the reaction, including units, if [NH4*] is 0.265M and [NO₂] is 0.0803M. Attach File Browse Local Files

Consider the gas phase reaction 2 NO + H₂ → N₂O + H₂O The experimental rate law is found to be R = [NO]²[H₂] Which of the mechanisms shown below is consistent with the information about this reaction given in this question? Step 1: NO + H₂ → N + H₂O (slow) Step2: N + NO → N₂O (fast) Step 1: NO + NO 2 NO₂ + N (slow) (fast) Step2: NO₂ + H₂ Step 3: N + NO → NO + H₂O N₂O (fast) Step 1: NO + NO>>> N₂O₂ Step2: N₂O₂ + H2 →→ N₂O + H₂O 2H (fast) Step 1: H₂ Step2: H+2 NO → Step 3: H+ OH → N₂O + OH (slow) H₂O (fast) (fast) (slow)