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Chapter 12: Problem 64

The mechanism for the gas-phase reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be $$ \begin{array}{c}{\mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}} \\ {\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}}\end{array} $$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction?

Short Answer

Expert verified
The rate law for the given reaction mechanism is: Overall Rate = (k₂/k₁)(Rate₁)[CO] Where Rate₁ = k₁[NO₂]² and k₁ and k₂ are the rate constants for the elementary steps. The overall balanced equation for the reaction is: 2NO₂ + CO → 2NO + CO₂

Step by step solution

01

Write the rate law for each elementary step

To find the rate law for each elementary step, we need to look at the coefficients of the reactants in the step. For an elementary reaction, the rate law can be directly obtained from the reaction equation. 1. For the first elementary step, NO₂ + NO₂ → NO₃ + NO, we have two molecules of NO₂ involved. Rate₁ = k₁[NO₂]² 2. For the second elementary step, NO₃ + CO → NO₂ + CO₂, we have one molecule of NO₃ and one molecule of CO involved. Rate₂ = k₂[NO₃][CO]
02

Identify the reaction intermediates

Reaction intermediates are species that are produced during one elementary step and consumed during another, so they don't appear in the overall balanced equation. In this case, NO₃ is the reaction intermediate, as it is produced in the first step and consumed in the second step.
03

Write the overall rate law

Since NO₃ is a reaction intermediate, we need to express [NO₃] in terms of other concentrations. From the first step, we have: [NO₃] = (Rate₁)/(k₁[NO₂]) Now we can substitute this into the Rate₂ expression: Overall Rate = Rate₂ = k₂[(Rate₁)/(k₁[NO₂])][CO] This simplifies to: Overall Rate = (k₂/k₁)(Rate₁)[CO]
04

Write the overall balanced equation

Now we can find the overall balanced equation for the reaction by adding the two elementary steps: (NO₂ + NO₂ → NO₃ + NO) + (NO₃ + CO → NO₂ + CO₂) The NO₃ intermediate is produced in the first step and consumed in the second step. Therefore, it does not appear in the overall balanced equation. The equation simplifies to: NO₂ + NO₂ + CO → NO + NO₂ + CO₂ Further simplifying it, we have: 2NO₂ + CO → 2NO + CO₂ So, the overall balanced equation is: 2NO₂ + CO → 2NO + CO₂

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Most popular questions from this chapter

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text {Rate} =-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\) \([\mathrm{B}]_{0}=3.0 M,\) and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after 8.0 seconds, the concentration of \(\mathrm{A}\) is $3.8 \times 10^{-3} \mathrm{M}$ a. Calculate the value of k for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

Provide a conceptual rationale for the differences in the halflives of zero-, first-, and second-order reactions.

Upon dissolving \(\operatorname{In} \mathrm{Cl}(s)\) in $\mathrm{HCl}, \operatorname{In}^{+}(a q)$ undergoes a disproportionation reaction according to the following unbalanced equation: $$ \operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of \(\operatorname{In}^{+}(a q)\) after 1.25 \(\mathrm{h}\) if the initial solution of \(\operatorname{In}^{+}(a q)\) was prepared by dis- solving 2.38 \(\mathrm{g} \operatorname{InCl}(s)\) in dilute \(\mathrm{HCl}\) to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution? What mass of In \((s)\) is formed after 1.25 \(\mathrm{h}\) ?

For a first order gas phase reaction \(\mathrm{A} \longrightarrow\) products, \(k=\) \(7.2 \times 10^{-4} \mathrm{s}^{-1}\) at \(660 . \mathrm{K}\) and $k=1.7 \times 10^{-2} \mathrm{s}^{-1}\( at \)720 . \mathrm{K} .$ If the initial pressure of \(\mathrm{A}\) is 536 torr at \(295^{\circ} \mathrm{C},\) how long will it take for the pressure of \(\mathrm{A}\) to decrease to 268 torr?

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\) , with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l}{k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{s}^{-1}} \\\ {k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1}}\end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) 4.00\([\mathrm{B}]\) ?
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