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Chapter 14: Problem 77

The decomposition of nitrogen dioxide at a high temperature $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is second-order in this reactant. (a) Determine the rate constant for this reaction if it takes 1.76 min for the concentration of \(\mathrm{NO}_{2}\) to fall from 0.250 mol/L to 0.100 mol/L. (b) If the chemical equation is written as $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ what is the value of the rate constant?

Short Answer

Expert verified
(a) Rate constant \( k = 0.000947 \) L/mol/s. (b) Revised rate constant \( k = 0.001894 \) L/mol/s.

Step by step solution

01

Identify the Rate Law

For a second-order reaction with respect to a single reactant \( NO_2 \), the rate law is given by \[ ext{Rate} = k[ ext{NO}_2]^2 \] where \( k \) is the rate constant.
02

Apply the Integrated Rate Law for Second-Order Reaction

The integrated rate law for a second-order reaction is \[ \frac{1}{[ ext{NO}_2]_t} = kt + \frac{1}{[ ext{NO}_2]_0} \] where \( [ ext{NO}_2]_t \) is the concentration at time \( t \), and \( [ ext{NO}_2]_0 \) is the initial concentration.
03

Substitute Known Values into the Equation

Given \( t = 1.76 \) min (convert to seconds by multiplying by 60, since rate constant units are often per second), \( [ ext{NO}_2]_0 = 0.250 \) mol/L, and \( [ ext{NO}_2]_t = 0.100 \) mol/L, substitute these values into the integrated rate law:\[ \frac{1}{0.100} = k(1.76 imes 60) + \frac{1}{0.250} \]
04

Solve for the Rate Constant \( k \)

Simplify and solve the equation for \( k \): \[ 10 = k(105.6) + 4 \] \[ 6 = k(105.6) \] \[ k = \frac{6}{105.6} \] \[ k \approx 0.0568 ext{ L/mol/min} \] Convert to L/mol/s: \[ k \approx 0.000947 ext{ L/mol/s} \]
05

Adjust for the Stoichiometry Change

In the reformulated chemical reaction \( 2 ext{NO}_2(g) \rightarrow 2 ext{NO}(g) + ext{O}_2(g) \), the stoichiometry factor changes the rate constant by a factor of 2 (as reaction coefficients are doubled). Thus, \[ k \text{ (new)} = 2 imes 0.000947 = 0.001894 ext{ L/mol/s} \]

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Key Concepts

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

Second-order reaction
In chemistry, a second-order reaction is a type where the rate depends on the concentration of one reactant raised to the second power, or possibly two different reactants, each raised to the first power. For our specific example involving \(\mathrm{NO}_2\) decomposition, the rate of reaction is dictated by the square of the concentration of \(\mathrm{NO}_2\). This is expressed in the rate law as:
  • Rate = \(k[\mathrm{NO}_2]^2\)
This means that any change in the concentration of \(\mathrm{NO}_2\) will have a squared effect on the rate of reaction. So, if the concentration doubles, the rate of reaction increases by four times. This squared relationship is what characterizes second-order reactions.
Rate constant
The rate constant, denoted as \(k\), is a crucial component in the rate law equation. It is a constant that correlates the concentration of reactants to the reaction rate and can be determined experimentally. For our problem, the rate constant for the decomposition of \(\mathrm{NO}_2\) was calculated using the integrated rate law:
  • \(k = \frac{6}{105.6} \)
  • \(k \approx 0.0568 \text{ L/mol/min} \)
  • Converted to \(\text{L/mol/s}: \k \approx 0.000947 \text{ L/mol/s}\)
The rate constant is specific to a particular reaction and gives insight into the speed of the reaction. In our reaction, it signifies how fast \(\mathrm{NO}_2\) decomposes per unit time.
Integrated rate law
The integrated rate law is a mathematical expression that relates the concentrations of reactants to time, and it varies depending on the order of reaction. For a second-order reaction, like the decomposition of \(\mathrm{NO}_2\), the integrated rate law is:
  • \(\frac{1}{[\mathrm{NO}_2]_t} = kt + \frac{1}{[\mathrm{NO}_2]_0}\)
Here, \(\frac{1}{[\mathrm{NO}_2]_t}\) is the reciprocal of the concentration of \(\mathrm{NO}_2\) at time \(t\), and \(\frac{1}{[\mathrm{NO}_2]_0}\) is the reciprocal of the initial concentration. By plugging in known values, you can solve for the rate constant \(k\), or predict how long it will take for concentrations to change under certain conditions.
Reaction stoichiometry
Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It helps predict amounts of substances consumed and produced. In our exercise, we initially had the equation:
  • \(\mathrm{NO}_2(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2}\mathrm{O}_2(g)\)
Changing this to conform with more typical stoichiometric coefficients:
  • \(2 \mathrm{NO}_2(g) \rightarrow 2 \mathrm{NO}(g) + \mathrm{O}_2(g)\)
This adjustment changes the overall stoichiometry and thus the rate constant by a factor of \(2\), since the coefficients of the reacting species have doubled. So, the new rate constant becomes:
  • \(k \text{ (new)} = 2 \times 0.000947 = 0.001894 \text{ L/mol/s}\)
Alterations in stoichiometry can significantly impact how reaction rates and mechanisms are interpreted.

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Most popular questions from this chapter

In the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g}),\) the rate of for- mation of \(\mathrm{O}_{2}\) is \(1.5 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). What is the rate of decomposition of \(\mathrm{O}_{3} ?\)

The substitution of \(\mathrm{CO}\) in \(\mathrm{Ni}(\mathrm{CO})_{4}\) by another molecule L Iwhere L is an electron-pair donor such as \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\) ) was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal-CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, Vol. \(90,\) p. \(6927,1968 .\) A detailed study of the kinetics of the reaction led to the following mechanism: Slow: \(\quad \mathrm{Ni}(\mathrm{CO})_{4} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{CO}\) Fast: \(\quad \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{L} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3} \mathrm{L}\) (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) increased the reaction rate by a factor of 2 Doubling the concentration of L had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when \(\mathrm{L}=\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) is \(9.3 \times 10^{-3} \mathrm{s}^{-1}\) at \(20^{\circ} \mathrm{C}\) If the initial concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(0.025 \mathrm{M},\) what is the concentration of the product after 5.0 minutes?

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?

Nitramide, \(\mathrm{NO}_{2} \mathrm{NH}_{2}\), decomposes slowly in aqueous solution according to the following reaction: $$ \mathrm{NO}_{2} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ The reaction follows the experimental rate law $$ \text { Rate }=\frac{k\left[\mathrm{NO}_{2} \mathrm{NH}_{2}\right]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} $$ (a) What is the apparent order of the reaction in a pH buffered solution? (In a pH buffered solution, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is a constant.) (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. (Note that when writing the expression for \(K,\) the equilibrium constant, \(\left.\left[\mathrm{H}_{2} \mathrm{O}\right] \text { is not involved. See Chapter } 15 .\right)\) Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 $$ \mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{\overleftarrow{k_{2}^{\prime}}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} $$ \(\mathrm{NO}_{2} \mathrm{NH}_{3}^{+} \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+} \quad\) (rate-limiting step) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\underset{(\text { rapid equilibrium })}{\mathrm{H}_{3} \mathrm{O}^{+}}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \stackrel{k_{5}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rate-limiting step) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (d) Based on the experimental rate law, will the reaction rate increase or decrease if the pH of the solution is increased?

Egg protein albumin is precipitated when an egg is cooked in boiling \(\left(100^{\circ} \mathrm{C}\right)\) water. \(E_{\mathrm{a}}\) for this first- order reaction is \(52.0 \mathrm{kJ} / \mathrm{mol} .\) Estimate the time to prepare a 3 -minute egg at an altitude at which water boils at \(90^{\circ} \mathrm{C}\).
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