Question
upstudy study bank question image url

13.39 The reaction of NO with \( \mathrm{O}_{2} \) to give \( \mathrm{NO}_{2} \) is an important process in the formation of smog in any large city: \[ 2 \mathrm{NO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2} \] Experiments show that this reaction is third order overall. The following mechanism has been proposed: \[ \mathrm{NO}+\mathrm{NO} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}} \mathrm{~N}_{2} \mathrm{O}_{2} \] \[ \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \xrightarrow{k_{2}} \mathrm{NO}_{2}+\mathrm{NO}_{2} \] (a) If the second step is rate-determining, what is the rate law? (b) Is this rate law consistent with the overall third-order behaviour? Explain. (c) Draw molecular pictures that show different ways the intermediate species might bind together, and identify the one that is most reasonable with respect to the second step of the mechanism.

Ask by Haynes Little. in Canada
Mar 09,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

**(a)** The rate law is \( \text{Rate} = k [\mathrm{NO}]^2 [\mathrm{O_2}] \). **(b)** Yes, the rate law is consistent with the overall third-order behavior. **(c)** The most reasonable binding is when the two \(\mathrm{NO}\) molecules form a head-to-head bond via their nitrogen atoms, allowing the intermediate \(\mathrm{N_2O_2}\) to react easily with \(\mathrm{O_2}\).

Solution

**(a)** The proposed mechanism is \[ \begin{aligned} \text{Step 1:} \quad & \mathrm{NO} + \mathrm{NO} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}} \mathrm{N_2O_2} \quad (\text{fast equilibrium})\\[1mm] \text{Step 2:} \quad & \mathrm{N_2O_2} + \mathrm{O_2} \xrightarrow{k_{2}} 2\,\mathrm{NO_2} \quad (\text{slow, rate-determining}) \end{aligned} \] Since the second step is the rate-determining step, the rate law is given by \[ \text{Rate} = k_{2} [\mathrm{N_2O_2}][\mathrm{O_2}]. \] To express \([\mathrm{N_2O_2}]\) in terms of \([\mathrm{NO}]\), we use the equilibrium established in the first step. The equilibrium constant \(K\) for step 1 is \[ K = \frac{[\mathrm{N_2O_2}]}{[\mathrm{NO}]^2} = \frac{k_1}{k_{-1}}, \] so \[ [\mathrm{N_2O_2}] = K [\mathrm{NO}]^2. \] Substituting this back into the rate law gives \[ \text{Rate} = k_{2} K [\mathrm{NO}]^2 [\mathrm{O_2}]. \] Thus, the rate law is \[ \boxed{\text{Rate} = k_{\text{obs}} [\mathrm{NO}]^2 [\mathrm{O_2}], \quad \text{with } k_{\text{obs}} = k_{2}K.} \] **(b)** The derived rate law is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O_2}\). This means the overall reaction order is \[ 2 + 1 = 3, \] which is consistent with the experimentally observed third-order behavior. **(c)** The first step involves the dimerization of two \(\mathrm{NO}\) molecules to form the intermediate \(\mathrm{N_2O_2}\). There are several possible ways in which the two \(\mathrm{NO}\) molecules might associate. Two common conceptual pictures are: 1. **Head-to-Head Binding via Nitrogen Centers:** Each \(\mathrm{NO}\) molecule has an unpaired electron primarily on the nitrogen atom. The two nitrogen atoms can form a bond, leading to a structure like \[ \overset{\bullet}{\mathrm{N}}=\mathrm{O} \,\,\,\cdots\,\,\, \overset{\bullet}{\mathrm{N}}=\mathrm{O} \quad \longrightarrow \quad \mathrm{N\!-\!N} \text{ bond}, \] where the oxygen atoms remain free. This arrangement is advantageous because when \(\mathrm{O_2}\) approaches in the second step, it can easily access the reactive site on the \(\mathrm{N_2O_2}\) intermediate. 2. **Side-by-Side or End-On Association:** Alternatively, the \(\mathrm{NO}\) molecules might interact via their oxygen atoms or in a more parallel fashion. However, such an arrangement could cause steric hindrance or reduce the accessibility of the active sites needed for the reaction with \(\mathrm{O_2}\). The most reasonable picture, especially with respect to facilitating the second step, is the head-to-head binding via the nitrogen atoms. In this configuration, the intermediate \(\mathrm{N_2O_2}\) has a structure where the two nitrogen atoms are bonded together while the oxygen atoms are still available for further reaction. This arrangement minimizes steric hindrance and allows \(\mathrm{O_2}\) to attack readily, which is consistent with the observed kinetics of the second step. A simple diagram of the favorable binding is: \[ \begin{array}{cccc} \mathrm{O} & & \mathrm{O} \\ \quad | & & | \\ \overset{\bullet}{\mathrm{N}} & - & \overset{\bullet}{\mathrm{N}} \\ \quad & & \\ \end{array} \] Here, the \(\mathrm{N\!-\!N}\) bond is formed between the two nitrogen atoms (each indicated with an unpaired electron), leaving the oxygen atoms free. This orientation facilitates the approach of \(\mathrm{O_2}\) to react with the intermediate in the rate-determining second step. \[ \boxed{\text{The mechanism with head-to-head (nitrogen-to-nitrogen) association is the most reasonable.}} \]

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

Bonus Knowledge

In the proposed mechanism, the rate-determining step (the slow step) is the second one: \( \mathrm{N}_{2} \mathrm{O}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{NO}_{2} + \mathrm{NO}_{2} \). The rate law for this step, focusing on the reactants, would be: \[ \text{Rate} = k_2 [\mathrm{N}_{2}\mathrm{O}_{2}][\mathrm{O}_{2}] \] (b) To check the consistency with the overall third-order behavior, we need to consider the concentration of the intermediate \( \mathrm{N}_{2}\mathrm{O}_{2} \). Given that the first step is reversible and reaches a quasi-steady state, \( [\mathrm{N}_{2}\mathrm{O}_{2}] \) can be expressed in terms of \( [\mathrm{NO}] \). This leads to the overall rate being proportional to \( [\mathrm{NO}]^2 [\mathrm{O}_{2}] \), which indeed corresponds to third order (2+1=3). (c) To illustrate the possible binding configurations for the intermediate \( \mathrm{N}_{2}\mathrm{O}_{2} \), we can depict two nitric oxide (NO) molecules pairing in various angles or orientations. The optimal binding configuration would likely show the oxygen atoms of the two NO molecules oriented toward each other, facilitating the effective collision with oxygen molecules \( O_2 \) in the subsequent step. This alignment maximizes the likelihood of reaction when encountering \( O_{2} \), enhancing the formation of \( \mathrm{NO}_{2} \).

Related Questions

Latest Chemistry Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy