In my Chemistry textbook (Chemistry: The Central Science), it says:Step 1: NO(g) + Br(2)(g) <---> NOBr2(g) (fast step) Step 2: NOBr2(g) + NO ---> 2NOBr(g) (slow step) It says that because step 2 is the slow step, we can assume that most of the NOBr2 falls apart, and thus we have both a forward and reverse reaction.   The rate law is rate = k[NO]^2[Br2].  Even though the rate law includes reactants from both elementary steps, the two steps are not the same rate, according to the book.  The book states that the rate law is taken from the slwo step.  It's actually k[NOBr2][NO].  It says about this, "Note that NOBr2 is an intermediate generated in the forward reaction of step 1. Intermediates are usually unstable and have a low, unknown concentration. Thus, the rate law depends on the unknown concentration of an intermediate, which isn’t desirable. We want instead to express the rate law for a reaction in terms of the reactants, or the products if necessary, of the reaction." So it defines the rate law as rate = k[NO]^2[Br2].  It says, "Solving for [NOBr2], we have [NOBr2] = (k1/k-1)[NO][Br]."  Substituting that back into the equation rate = k[NOBr2][NO], we get rate = k[NO][Br][NO], which is k[NO]^2[Br].Okay, so there's that whole explanation in our book about rate laws when the second step is the slow step.  Then in a class problem, the question is Nitric oxide (NO) reacts with H2 to form N2O and H2O. It is known that there are two elementarysteps to this reaction:Step I: NO(g) + NO(g) ---> N2O2(g)Step II: N2O2(g) + H2(g) ---> N2O(g) + H2O(g)The measured rate law for this overall reaction is:Rate = k[NO]2 [H2]What can we conclude about the above elementary reaction steps based on this rate law?(A) Steps I & II are about the same rate.(B) Step I is much faster than step II.(C) Step II is much faster than step I.(D) Not enough information to tell. Everyone tells me that the answer is A.  How is this problem different from the problem in the textbook?  Why does the textbook say that the "rate = k[NO]^2[Br2]" means that the second step is the slow step, but in this problem "rate = k[NO]2 [H2]" means that they're the same rate?  What is the difference between these two problems that indicates that one means the second step is the slow step while the other indicates that they're the same rate?

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter13: Chemical Equilibrium
Section: Chapter Questions
Problem 129MP: A gaseous material XY(g) dissociates to some extent to produce X(g) and Y(g): XY(g)X(g)+Y(g) A...
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In my Chemistry textbook (Chemistry: The Central Science), it says:

Step 1: NO(g) + Br(2)(g) <---> NOBr2(g) (fast step)

Step 2: NOBr2(g) + NO ---> 2NOBr(g) (slow step)

It says that because step 2 is the slow step, we can assume that most of the NOBr2 falls apart, and thus we have both a forward and reverse reaction.  

The rate law is rate = k[NO]^2[Br2].  Even though the rate law includes reactants from both elementary steps, the two steps are not the same rate, according to the book.  The book states that the rate law is taken from the slwo step.  It's actually k[NOBr2][NO].  It says about this, "Note that NOBr2 is an intermediate generated in the forward reaction of step 1. Intermediates are usually unstable and have a low, unknown concentration. Thus, the rate law depends on the unknown concentration of an intermediate, which isn’t desirable. We want instead to express the rate law for a reaction in terms of the reactants, or the products if necessary, of the reaction." 

So it defines the rate law as rate = k[NO]^2[Br2].  It says, "Solving for [NOBr2], we have [NOBr2] = (k1/k-1)[NO][Br]."  Substituting that back into the equation rate = k[NOBr2][NO], we get rate = k[NO][Br][NO], which is k[NO]^2[Br].

Okay, so there's that whole explanation in our book about rate laws when the second step is the slow step.  Then in a class problem, the question is 
Nitric oxide (NO) reacts with H2 to form N2O and H2O. It is known that there are two elementary
steps to this reaction:
Step I: NO(g) + NO(g) ---> N2O2(g)
Step II: N2O2(g) + H2(g) ---> N2O(g) + H2O(g)
The measured rate law for this overall reaction is:
Rate = k[NO]2 [H2]
What can we conclude about the above elementary reaction steps based on this rate law?
(A) Steps I & II are about the same rate.
(B) Step I is much faster than step II.
(C) Step II is much faster than step I.
(D) Not enough information to tell.

Everyone tells me that the answer is A.  How is this problem different from the problem in the textbook?  Why does the textbook say that the "rate = k[NO]^2[Br2]" means that the second step is the slow step, but in this problem "rate = k[NO]2 [H2]" means that they're the same rate?  What is the difference between these two problems that indicates that one means the second step is the slow step while the other indicates that they're the same rate?  

 

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