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) 1₂ (g) → 21 (g) (fast) (2) H₂ (g) + I (g) → H₂I (g) (fast) (3) H₂I (g) + I (g) → 2 HI (g) (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. \
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) 1₂ (g) → 21 (g) (fast) (2) H₂ (g) + I (g) → H₂I (g) (fast) (3) H₂I (g) + I (g) → 2 HI (g) (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. \
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section11.7: Reaction Mechanisms
Problem 11.12E
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