## Question 51 in the homework [ENDORSED]

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

104422816
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### Question 51 in the homework

15.51 The following mechanism has been proposed for the reaction between nitric oxide and bromine:
Step 1: NO -> Br2 + NOBr2 (slow)
Step 2: NOBr2 + NO + NOBr -> NOBr (fast)
Write the rate law for the formation of NOBr implied by this mechanism.

The explanation for the answer: k(NO)(Br2) says that this is a second order reaction but i thought second order equation is (a)^2

Reine Nakamura 1C
Posts: 38
Joined: Fri Sep 25, 2015 3:00 am

### Re: Question 51 in the homework

It's a second order reaction because the overall order of the reaction is the sum of the orders for [NO] and [Br2]. In this case both are first order, making the overall order 2.

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### Re: Question 51 in the homework  [ENDORSED]

Why is the speed of the rxn relevant in this problem? (slow vs fast)

Chem_Mod
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### Re: Question 51 in the homework

This is to show that the first step is the most important overall for determining reaction rate, as it is slower than the second, making it rate-limiting.

Dang Lam
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Joined: Thu Jul 27, 2017 3:01 am

### Re: Question 51 in the homework

Chem_Mod wrote:This is to show that the first step is the most important overall for determining reaction rate, as it is slower than the second, making it rate-limiting.

so which reaction (slow or fast) should we use to write our rate law?