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Chapter 13: Problem 80

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ equilibrium constants \(\left(K_{\mathrm{p}} \text { values ) as a function of temperature }\right.\) are\(\begin{array}{ll}{300^{\circ} \mathrm{C},} & {4.34 \times 10^{-3}} \\ {500^{\circ} \mathrm{C},} & {1.45 \times 10^{-5}} \\\ {600^{\circ} \mathrm{C},} & {2.25 \times 10^{-6}}\end{array}\) Is the reaction exothermic or endothermic?

Short Answer

Expert verified
The reaction is exothermic, as the equilibrium constant \(K_p\) decreases with increasing temperature, indicating a negative \(\Delta H^{\circ}\).

Step by step solution

01

Calculate the temperature in Kelvin

First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. For the given equilibrium constants, we have: \[\begin{array}{ll}{T_1 = 300^{\circ} \mathrm{C} + 273.15 = 573.15 \mathrm{K},} & {K_{p1} = 4.34 \times 10^{-3}} \\{T_2 = 500^{\circ} \mathrm{C} + 273.15 = 773.15 \mathrm{K},} & {K_{p2} = 1.45 \times 10^{-5}} \\{T_3 = 600^{\circ} \mathrm{C} + 273.15 = 873.15 \mathrm{K},} & {K_{p3} = 2.25 \times 10^{-6}}\end{array}\] Now, we will use the Van't Hoff equation to analyze the relationship between \(K_p\) and temperature.
02

Analyze the Van't Hoff equation

We want to see how the equilibrium constant \(K_p\) changes with temperature. If the reaction were exothermic, increasing temperature would result in a decrease in \(K_p\). Conversely, if the reaction were endothermic, increasing temperature would result in an increase in \(K_p\). By looking at the given equilibrium constants and the associated temperatures, we see that as temperature increases, the equilibrium constant \(K_p\) decreases: - At \(T_1 = 573.15\,\mathrm{K}\), \(K_{p1} = 4.34 \times 10^{-3}\) - At \(T_2 = 773.15\,\mathrm{K}\), \(K_{p2} = 1.45 \times 10^{-5}\) - At \(T_3 = 873.15\,\mathrm{K}\), \(K_{p3} = 2.25 \times 10^{-6}\) Since \(K_p\) decreases as temperature increases, this means \(\frac{d(\ln K_p)}{dT} < 0\). We can now look at the Van't Hoff equation and determine the sign of \(\Delta H^{\circ}\).
03

Determine the sign of \(\Delta H^{\circ}\)

Recall the Van't Hoff equation: \[\frac{d(\ln K_p)}{dT} = \frac{\Delta H^{\circ}}{RT^2}\] We found that \(K_p\) decreases as temperature increases (\(\frac{d(\ln K_p)}{dT} < 0\)). Therefore, the right side of the equation must also be negative: \[\frac{\Delta H^{\circ}}{RT^2} < 0\] Since \(R\) and \(T^2\) are always positive, we can conclude that \(\Delta H^{\circ} < 0\). Now that we know the sign of \(\Delta H^{\circ}\) is negative, we can classify the reaction.
04

Classify the reaction

Since \(\Delta H^{\circ} < 0\), the reaction is exothermic. In other words, the reaction releases heat as it proceeds.

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Key Concepts

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

Equilibrium Constants in the Haber Process
The Haber process involves the reaction of nitrogen and hydrogen to create ammonia and is represented by the equation \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\). An important aspect of this reaction is the equilibrium constant, denoted as \(K_p\) when dealing with gases: * It indicates the ratio of the concentrations of products to reactants at equilibrium.* The specific values of \(K_p\) depend on the temperature at which the reaction occurs.In the original exercise, several equilibrium constants were provided for different temperatures. As the temperature changes, these constants give us insight into the nature of the reaction. At 300掳C, the equilibrium constant \(K_p\) is 4.34 脳 10鈦宦�, whereas at 600掳C, it drops to 2.25 脳 10鈦烩伓. This variation shows how shifts in temperature influence the proportion of ammonia produced. Understanding equilibrium constants is crucial for determining the yield of ammonia under varying conditions.
Understanding the Van't Hoff Equation
The Van't Hoff equation is a fundamental tool for studying how equilibrium constants change with temperature. It relates the change in the logarithm of the equilibrium constant (\(K_p\)) to the change in temperature:\[\frac{d(\ln K_p)}{dT} = \frac{\Delta H^{\circ}}{RT^2}\]Where:* \(\Delta H^{\circ}\) is the standard enthalpy change,* \(R\) is the universal gas constant,* \(T\) is the temperature measured in Kelvin.In the exercise, the constants decrease as temperature increases, indicating that the reaction is exothermic (since \(\Delta H^{\circ}\) is negative). By observing that \(\frac{d(\ln K_p)}{dT} < 0\), we confirm an exothermic process. This understanding allows chemists to predict how changes in temperature will affect the production of ammonia, which is integral for optimizing industrial production.
Exothermic Reactions Explained
An exothermic reaction is one that releases heat to its surroundings. In the context of the Haber process, it signifies that as ammonia forms, the system loses heat:* The negative sign of \(\Delta H^{\circ}\) indicates this heat release.* As the temperature increases, the equilibrium shifts to produce less ammonia, evidenced by a decreasing \(K_p\).Exothermic reactions are vital in various industrial processes because they can drive reactions naturally due to the heat released:
  • These reactions can often occur spontaneously under suitable conditions.
  • The release of heat can be used to drive other reactions, improving energy efficiency.
Understanding whether a reaction is exothermic or endothermic helps engineers and chemists tailor conditions to maximize efficiency, such as adjusting pressures and temperatures in the Haber process to optimize ammonia yield while considering energy consumption.

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Most popular questions from this chapter

Calculate a value for the equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ given $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) \quad K=6.8 \times 10^{-49}$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad K=5.8 \times 10^{-34}$$ (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Explain why the development of a vapor pressure above a liquid in a closed container represents an equilibrium. What are the opposing processes? How do we recognize when the system has reached a state of equilibrium?

Consider the reaction $$\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q)$$ How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\operatorname{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? [Fe(OH) \(_{3}\) is insoluble. \(]\) d. Fe(NO \(_{3} )_{3}(a q)\) is added?

Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$ a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

In which direction will the position of the equilibrium $$2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)$$ be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\operatorname{HI}(g)\) is removed. d. In a rigid reaction container, some Ar(g) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).
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