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The reaction between ozone and nitrogen dioxide at \(231 \mathrm{K}\) is first- order in both \(\left[\mathrm{NO}_{2}\right]\) and \(\left[\mathrm{O}_{3}\right]\) $$ 2 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ (a) Write the rate equation for the reaction. (b) If the concentration of \(\mathrm{NO}_{2}\) is tripled (and \(\left[\mathrm{O}_{3}\right]\) is not changed , what is the change in the reaction rate? (c) What is the effect on reaction rate if the concentration of \(\mathbf{O}_{3}\) is halved (with no change in \(\left.\left[\mathrm{NO}_{2}\right]\right) ?\)

Short Answer

Expert verified
(a) Rate = k[NO2][O3]; (b) Reaction rate triples; (c) Reaction rate is halved.

Step by step solution

01

Understanding the Rate Equation

The rate equation of a reaction gives the relationship between the rate of reaction and the concentrations of the reactants. For this reaction, the rate is first-order in both \([\mathrm{NO}_2]\) and \([\mathrm{O}_3]\). This means the rate equation can be written as \[\text{Rate} = k [\mathrm{NO}_2]^1 [\mathrm{O}_3]^1 = k [\mathrm{NO}_2][\mathrm{O}_3]\]where \(k\) is the rate constant.
02

Evaluating the Effect of Tripling \([NO_2]\)

If the concentration of \([\mathrm{NO}_2]\) is tripled, we substitute \([\mathrm{NO}_2] = 3 \times [\mathrm{NO}_2]_0\) into the rate equation. The new rate will be:\[\text{Rate}_{\text{new}} = k [3 \times \mathrm{NO}_2]_0 [\mathrm{O}_3]_0 = 3k [\mathrm{NO}_2]_0[\mathrm{O}_3]_0\]Compared to the original rate:\[\text{Rate}_{\text{original}} = k [\mathrm{NO}_2]_0 [\mathrm{O}_3]_0\]The reaction rate will therefore triple.
03

Evaluating the Effect of Halving \([O_3]\)

If the concentration of \([\mathrm{O}_3]\) is halved, we use \([\mathrm{O}_3] = \frac{1}{2} \times [\mathrm{O}_3]_0\) in the rate equation. The new rate will be:\[\text{Rate}_{\text{new}} = k [\mathrm{NO}_2]_0 \left(\frac{1}{2} [\mathrm{O}_3]_0\right) = \frac{1}{2} k [\mathrm{NO}_2]_0 [\mathrm{O}_3]_0\]This shows that the reaction rate is halved compared to its original value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-order Reaction
A first-order reaction is characterized by its dependency on the concentration of a single reactant. In our case, the reaction between ozone ( O_3) and nitrogen dioxide ( NO_2) is first-order with respect to each reactant, meaning:
  • The rate at which the reaction proceeds is directly proportional to the concentration of both NO_2 and O_3 .
  • When you double the concentration of one reactant, the reaction rate doubles as well.
This directly influences how we construct the rate equation, which shows the interplay between the rate constant ( k ) and reactant concentrations. The simplicity of first-order reactions allows for straightforward calculations of changes in reaction conditions.
Concentration Change Effect
Concentration changes impact the reaction rate significantly in chemical kinetics. For this specific reaction:
  • If the concentration of NO_2 is tripled, the reaction rate triples, as demonstrated in the step-by-step solution.
  • Alternatively, halving the O_3 concentration decreases the reaction rate by half.
These effects stem from the first-order nature of the reaction concerning each reactant. Since both reactants individually affect the rate linearly, simple arithmetic changes in concentrations directly translate to proportional changes in the reaction rate.
Rate Constant k
The rate constant ( k ) is a crucial part of the reaction rate equation. It is specific to a particular reaction at a given temperature and provides the link between the concentrations of reactants and the rate of the reaction. In our current study:
  • The value of k remains constant if temperature conditions remain unchanged.
  • It helps predict how fast the reaction will occur for given concentrations of the reactants.
By understanding the role of k , chemists can compare the speeds of different reactions or the same reaction under varied conditions.
Chemical Kinetics
Chemical kinetics involves the study of reaction rates and how they change under different conditions. In our reaction:
  • We analyze how first-order rates change with varying reactant concentrations.
  • The principles of kinetics aid in determining the speed of transformation from reactants to products.
Understanding kinetics is vital for optimizing reactions in industrial applications, environmental modeling, and even biochemical pathways. It informs us not only about how quickly a reaction is proceeding but also about the mechanisms driving the transformation.

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Most popular questions from this chapter

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step \(1: \quad\) Fast, reversible: \(\quad \mathrm{HA} \rightleftarrows \mathrm{H}^{+}+\mathrm{A}^{-}\) Step \(2: \quad\) Fast, reversible: \(\quad \mathrm{X}+\mathrm{H}^{+} \rightleftharpoons \mathrm{XH}^{+}\) Step 3: Slow \(\mathrm{XH}^{+} \rightarrow\) products What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?

You want to study the hydrolysis of the beautiful green, cobalt-based complex called transdichlorobis-(ethylenediamine)cobalt(III) ion, In this hydrolysis reaction, the green complex ion trans- \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: \(\left.\quad \text { trans-ICo(en) }_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ Fast: \(\quad\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq}) $$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first-order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k\). Use the data below to find the activation energy. green $$ \left.\underset{\text { red }}{\operatorname{Co}(\mathrm{en})_{2}}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker.

Radioactive gold- 198 is used in the diagnosis of liver problems. The half- life of this isotope is 2.7 days. If you begin with a 5.6 -mg sample of the isotope, how much of this sample remains after 1.0 day?

The following statements relate to the reaction for the formation of HI: $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left|\mathrm{I}_{2}\right| $$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

Compare the lock-and-key and induced-fit models for substrate binding to an enzyme.

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