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Chapter 12: Problem 82

Considering both the enthalpy effect and the entropy effect for the Haber- Bosch process, explain why choosing the temperature at which to run this reaction is very important.

Short Answer

Expert verified
Temperature affects equilibrium (enthalpy) and reaction rate (entropy), requiring a compromise for efficiency.

Step by step solution

01

Understanding the Haber-Bosch Process

The Haber-Bosch process synthesizes ammonia (NH鈧�) from nitrogen (N鈧�) and hydrogen (H鈧�) gases. The balanced chemical equation is: \[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \] This is an exothermic reaction, meaning it releases heat.
02

Analyzing the Enthalpy Effect

Enthalpy change (\(\Delta H\)) is negative for this reaction since it's exothermic. Lower temperatures favor the production of ammonia because they shift the equilibrium towards the products according to Le Chatelier's principle.
03

Considering the Entropy Effect

The reaction results in fewer moles of gas (4 moles of reactants to 2 moles of products), leading to a decrease in entropy (\(\Delta S < 0\)). This means higher temperatures can undesirably shift the equilibrium towards the reactants, due to the T\(\Delta S\) term in Gibbs free energy.
04

Balancing Enthalpy and Entropy Effects

Since \(\Delta G = \Delta H - T\Delta S\), and both \(\Delta H\) is negative and \(\Delta S\) is negative, a balance needs to be struck. Lower temperatures favor \(\Delta H\), but excessively low temperatures can slow the reaction kinetics, making the process inefficient.
05

Choosing an Optimal Temperature

In practice, a compromise temperature, often around 400-500掳C, is chosen to optimize the production rate of ammonia while maintaining favorable equilibrium conditions. This allows for enough kinetic energy for reactions without shifting equilibrium unfavorably.
06

Conclusion

The temperature for the Haber-Bosch process must balance the exothermic nature of the reaction (enthalpy) with the entropy considerations, ensuring efficient and economical ammonia production.

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Key Concepts

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

Enthalpy Effect
In the Haber-Bosch process, enthalpy change (\( \Delta H \)) plays a critical role. This process is exothermic, meaning it releases heat when converting nitrogen and hydrogen gases into ammonia. The reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \) results in the release of energy, indicating a negative enthalpy change. Lower temperatures tend to favor the production of ammonia because, according to Le Chatelier's principle, the equilibrium shifts towards the products to release more heat. This makes low temperatures desirable from an enthalpy perspective. However, extreme low temperatures can result in inefficient reaction rates.
  • Exothermic reactions have negative \( \Delta H \).
  • Low temperatures shift equilibrium towards the products.
  • Balancing efficiency and temperature is crucial in industrial applications.
Entropy Effect
Entropy change (\( \Delta S \)) reflects the disorder in a system. In the Haber-Bosch process, the reaction converts four moles of gas (three moles of hydrogen and one mole of nitrogen) into two moles of ammonia. This reduction in moles leads to a decrease in entropy. Since \( \Delta S \) is negative, increasing the temperature can shift the equilibrium undesirably towards the reactants, as \( T\Delta S \) becomes increasingly significant in the Gibbs free energy equation. This effect must be managed by choosing a temperature that does not excessively take the system towards reduced product formation.
  • Decrease in moles means negative \( \Delta S \).
  • High temperatures increase the effect of \( T\Delta S \) term.
  • Careful temperature management is needed to maintain product formation.
Gibbs Free Energy
Gibbs free energy (\( \Delta G \)) is a critical concept when studying chemical reactions. For the Haber-Bosch process, it provides insight into the spontaneity of the reaction. The equation \( \Delta G = \Delta H - T\Delta S \) combines enthalpic and entropic effects. A negative \( \Delta G \) indicates a spontaneous reaction; hence, we aim for conditions where \( \Delta G \) remains negative.
To achieve this:
  • Choose temperatures that balance \( \Delta H \) and \( T\Delta S \).
  • Optimize for negative \( \Delta G \), ensuring efficient ammonia synthesis.
  • Consider industrial conditions that practically achieve this balance.
Le Chatelier's Principle
Le Chatelier's principle helps predict how a change in conditions affects chemical equilibrium. In the context of the Haber-Bosch process, it suggests that when a change occurs (like temperature adjustment), the equilibrium will shift to counteract that change. For this exothermic reaction:
  • Increasing temperature shifts the equilibrium towards the reactants to absorb heat.
  • Decreasing temperature shifts the equilibrium towards the products, releasing heat.
  • Using this principle helps determine optimal operating conditions.
Applying Le Chatelier's principle assists in deciding the appropriate temperature for efficient ammonia production. This consideration is vital for large-scale industrial applications where maintaining a balance between production and energy efficiency is key.

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Most popular questions from this chapter

Draw a nanoscale (particulate) level diagram for an equilibrium mixture of \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \quad K_{\mathrm{c}}=4.00\)

A solid sample of benzoic acid, a carboxylic acid, is in equilibrium with an aqueous solution of benzoic acid. A tiny quantity of \(\mathrm{D}_{2} \mathrm{O},\) water containing the isotope \({ }^{2} \mathrm{H}\), deuterium, is added to the solution. The solution is allowed to stand at constant temperature for several hours, after which some of the solid benzoic acid is removed and analyzed. The benzoic acid is found to contain a tiny quantity of deuterium, D, and the formula of the deuterium-containing molecules is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOD}\). Explain how this can happen.

The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) is endothermic. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ (a) Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. (b) If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a sealed flask in which this equilibrium exists, how is the equilibrium affected? (c) What if some additional \(\mathrm{NH}_{3}\) is placed in a sealed flask containing an equilibrium mixture? (d) What will happen to the partial pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{~S}\) is removed from the flask?

\(\mathrm{H}_{2}\) gas and \(\mathrm{I}_{2}\) vapor are mixed in a flask. The flask is sealed and heated to \(700^{\circ} \mathrm{C}\). The initial concentration of each gas is \(0.0088 \mathrm{~mol} / \mathrm{L},\) and \(78.6 \%\) of the \(\mathrm{I}_{2}\) has reacted when equilibrium is achieved according to the equation $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ Calculate \(K_{c}\) for this reaction.

A small sample of \(c i s\) -dichloroethene in which one carbon atom is the radioactive isotope \({ }^{14} \mathrm{C}\) is added to an equilibrium mixture of the cis and trans isomers at a certain temperature. Eventually, \(40 \%\) of the radioactive molecules are found to be in the trans configuration at any given time. (a) Determine the value of \(K_{\mathrm{c}}\) for the cis \(\rightleftharpoons\) trans equilibrium. (b) What would have happened if a small sample of radioactive trans isomer had been added instead of the cis isomer?
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