/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Nitric oxide emitted from the en... [FREE SOLUTION] | 91Ó°ÊÓ

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Nitric oxide emitted from the engines of supersonic aircraft can contribute to the destruction of stratospheric ozone: $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \stackrel{k_{\mathrm{L}}}{\rightleftharpoons} \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ This reaction is highly exothermic \((\Delta H=-201 \mathrm{~kJ})\), and its equilibrium constant \(K_{c}\) is \(3.4 \times 10^{34}\) at \(300 \mathrm{~K}\) (a) Which rate constant is larger, \(k_{\mathrm{E}}\) or \(k_{r} ?\) (b) The value of \(k_{\mathrm{f}}\) at \(300 \mathrm{~K}\) is \(8.5 \times 10^{6} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). What is the value of \(k_{r}\) at the same temperature? (c) A typical temperature in the stratosphere is \(230 \mathrm{~K}\). Do the values of \(k_{\mathrm{p}} k_{\mathrm{r}}\) and \(K_{\mathrm{c}}\) increase or decrease when the temperature is lowered from \(300 \mathrm{~K}\) to \(230 \mathrm{~K}\) ?

Short Answer

Expert verified
(a) \( k_f \) is larger. (b) \( k_r \approx 2.5 \times 10^{-28} \ \mathrm{s}^{-1} \). (c) \( K_c \) increases; \( k_f/k_r \) may remain constant.

Step by step solution

01

Understand the Reaction and Equilibrium

The reaction given, \( \mathrm{NO}(g) + \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{NO}_{2}(g) + \mathrm{O}_{2}(g) \), is at equilibrium, and the equilibrium constant \( K_c = 3.4 \times 10^{34} \) indicates that the forward reaction greatly favors the formation of products. This high value implies that the forward reaction has a larger rate constant compared to the reverse reaction.
02

Compare Rate Constants \( k_f \) and \( k_r \)

For the equilibrium constant \( K_c \), the relationship is \( K_c = \frac{k_f}{k_r} \). Given \( k_f = 8.5 \times 10^{6} \ \mathrm{M}^{-1} \mathrm{s}^{-1} \) and \( K_c = 3.4 \times 10^{34} \), solve for \( k_r \): \( k_r = \frac{k_f}{K_c} = \frac{8.5 \times 10^{6}}{3.4 \times 10^{34}} \). Calculate \( k_r \) by performing this division.
03

Perform the Calculation for \( k_r \)

Calculate \( k_r = \frac{8.5 \times 10^{6}}{3.4 \times 10^{34}} \approx 2.5 \times 10^{-28} \ \mathrm{s}^{-1} \). This shows that the rate constant for the reverse reaction is extremely small compared to the forward reaction.
04

Analyze Temperature Effects on Equilibrium and Rate Constants

Since the reaction is highly exothermic \((\Delta H = -201 \text{ kJ})\), according to Le Chatelier's principle, lowering the temperature favors the exothermic (forward) reaction. As a result, \( K_c \) will increase. The rate constant for the forward reaction \( k_f \) increases with decreased temperature, and since \( k_r = \frac{k_f}{K_c} \), it may remain roughly constant or slightly decrease since \( K_c \) increases significantly.

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Key Concepts

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

Rate Constants
Rate constants are key to understanding how fast a chemical reaction proceeds. In the given reaction of nitric oxide (\( \mathrm{NO} \)) with ozone (\( \mathrm{O}_{3} \)), distinct constants dictate the rate of the forward and reverse reactions. The forward rate constant (\( k_f \)) is given as \( 8.5 \times 10^{6} \, \mathrm{M}^{-1}\, \mathrm{s}^{-1} \). This suggests a rapid formation of products — \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \). Oppositely, the reverse rate constant (\( k_r \)) is \( 2.5 \times 10^{-28} \, \mathrm{s}^{-1} \), implying the backward formation of reactants is exceptionally slow at 300 K. These constants relate to equilibrium through \( K_c = \frac{k_f}{k_r} \). When you solve the equilibrium constant equation, you get the small \( k_r \).

Understanding rate constants helps predict reactions under different conditions, knowing whether products or reactants are favored. If \( k_f \) greatly exceeds \( k_r \), the equilibrium position skews towards products. This knowledge is not only instrumental in theoretical predictions but also crucial in industrial applications where reaction rates directly impact product yields.
Le Chatelier's Principle
Le Chatelier's Principle is pivotal when assessing how a chemical equilibrium reacts to changes in conditions, like temperature. The principle dictates how a system at equilibrium will attempt to counteract any stress applied to it. For the reaction of \( \mathrm{NO} \) with \( \mathrm{O}_{3} \), this understanding is crucial.

In the exercise, as the temperature decreases from 300 K to 230 K, Le Chatelier's Principle suggests that the equilibrium will shift to favor the exothermic direction, which in this case, is the formation of \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \). This is because the forward reaction releases heat, and lowering temperature promotes reactions that produce heat to balance the temperature decrease.

Thus, this principle helps to predict that at lower temperatures, the equilibrium constant (\( K_c \)) will increase. This means more products compared to reactants. Awareness of how Le Chatelier's Principle applies enables students to understand how reaction conditions impact equilibrium, assisting in both academic and practical chemical scenario planning.
Exothermic Reactions
Exothermic reactions, like the reaction of nitric oxide with ozone, play a significant role in chemistry. They are characterized by the release of heat to the surroundings, indicated by a negative change in enthalpy (\( \Delta H < 0 \)). For the reaction in question, \( \Delta H = -201 \, \mathrm{kJ} \), signifies how much energy is released as \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \) form.

Understanding the nature of exothermic reactions is essential, especially in the context of temperature changes and equilibrium. When the temperature is decreased, as covered in Le Chatelier's principle, exothermic reactions usually become more favorable. The reason is that releasing heat helps to offset the temperature drop, pushing the equilibrium position toward the products, thus increasing \( K_c \).

Recognizing the exothermic nature also plays a part in real-world applications, as many industrial processes seek these reactions for their efficiency in heat management. Whether for energy production or synthesis, knowing how to maximize the benefits of exothermic reactions is crucial for chemists and engineers alike.

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Most popular questions from this chapter

The following reaction shows the equilibrium reaction that occurs when carbon monoxide enters the blood. (Note: The \(K\) value uses partial pressures of \(\mathrm{O}_{2}\) and \(\mathrm{CO}\) in the equilibrium expression.) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Hb}(\mathrm{CO})(a q)+\mathrm{O}_{2}(g) \quad K=207\) at \(37^{\circ} \mathrm{C}\) (a) What is the ratio of \([\mathrm{Hb}(\mathrm{CO})]\) to \(\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)\right]\) in air that contains \(20 \% \mathrm{O}_{2}\) and \(0.15 \%\) CO by volume? (b) The treatment for mild carbon monoxide poisoning is breathing pure oxygen. Use Le Châtelier's principle to explain why this treatment is effective.

Equilibrium constants at \(25^{\circ} \mathrm{C}\) are given for the sequential equilibrium reactions in the binding of oxygen to hemoglobin. \(\mathrm{Hb}+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right) \quad \mathrm{K}_{\mathrm{cl}}=1.5 \times 10^{4}\) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{2} \quad \mathrm{~K}_{\mathrm{c} 2}=3.5 \times 10^{4}\) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{2}+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{3} \quad \mathrm{~K}_{\mathrm{c} 3}=5.9 \times 10^{4}\) \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{3}+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4} \quad \mathrm{~K}_{\mathrm{c} 4}=1.7 \times 10^{6}\) (a) How does the value of the equilibrium constant change for each sequential binding step \(\left(K_{c 1}\right.\) through \(K_{c 4}\) )? (b) Do the equilibrium reactions become more product-or reactant-favored with the binding of each oxygen? (c) How do the sequential equilibrium steps affect the oxygencarrying capacity of hemoglobin?

Consider the sublimation of mothballs at \(27^{\circ} \mathrm{C}\) in a room having dimensions \(8.0 \mathrm{ft} \times 10.0 \mathrm{ft} \times 8.0 \mathrm{ft}\). Assume that the mothballs are pure solid naphthalene (density \(\left.1.16 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and that they are spheres with a diameter of \(12.0 \mathrm{~mm}\). The equilibrium constant \(K_{c}\) for the sublimation of naphthalene is \(5.40 \times 10^{-6}\) at \(27{ }^{\circ} \mathrm{C}\). $$ \mathrm{C}_{10} \mathrm{H}_{8}(s) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{\mathrm{8}}(g) $$ (a) When excess mothballs are present, how many gaseous naphthalene molecules are in the room at equilibrium? (b) How many mothballs are required to saturate the room with gaseous naphthalene?

Consider the following gas-phase reaction: \(2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) An equilibrium mixture of reactants and products is subjected to the following changes: (a) A decrease in volume (b) An increase in temperature (c) Addition of reactants (d) Addition of a catalyst (e) Addition of an inert gas Which of these changes affect the composition of the equilibrium mixture but leave the value of the equilibrium constant \(K_{c}\) unchanged? Which of the changes affect the value of \(K_{c}\) ? Which affect neither the composition of the equilibrium mixture nor \(K_{c} ?\)

An equilibrium mixture of \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) at \(700 \mathrm{~K}\) contains \(0.036 \mathrm{M} \mathrm{N}_{2}\) and \(0.15 \mathrm{M} \mathrm{H}_{2}\). At this temperature, \(K_{c}\) for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is \(0.29 .\) What is the con- centration of \(\mathrm{NH}_{3}\) ?

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