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Is the Haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at \(25^{\circ} \mathrm{C} ?\) At what temperature \(\left({ }^{\circ} \mathrm{C}\right)\) does the changeover occur? \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H^{\circ}=-92.2 \mathrm{~kJ} ; \Delta S^{\circ}=-199 \mathrm{~J} / \mathrm{K}\)

Short Answer

Expert verified
The Haber process is spontaneous at 25°C, and the changeover occurs at about 190°C.

Step by step solution

01

Understand the Gibbs Free Energy Formula

To determine if the reaction is spontaneous, use the Gibbs free energy formula: \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \), where \( \Delta G^{\circ} \) is the change in free energy, \( \Delta H^{\circ} \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S^{\circ} \) is the change in entropy.
02

Convert Units

Convert \( \Delta S^{\circ} \) from \( \mathrm{J/K} \) to \( \mathrm{kJ/K} \) to match the units of \( \Delta H^{\circ} \).\[ \Delta S^{\circ} = -199 \mathrm{~J/K} = -0.199 \mathrm{~kJ/K} \]
03

Calculate \( \Delta G^{\circ} \) at 25°C

Substitute the values into the Gibbs free energy equation at \( 25^{\circ} \mathrm{C} \) which is \( 298 \mathrm{~K} \).\[ \Delta G^{\circ} = -92.2 \mathrm{~kJ} - (298 \mathrm{~K}) \times (-0.199 \mathrm{~kJ/K}) \]\[ \Delta G^{\circ} = -92.2 \mathrm{~kJ} + 59.302 \mathrm{~kJ} \]\[ \Delta G^{\circ} = -32.898 \mathrm{~kJ} \]
04

Interpret \( \Delta G^{\circ} \) at 25°C

Since \( \Delta G^{\circ} < 0 \), the process is spontaneous at \( 25^{\circ} \mathrm{C} \).
05

Determine the Temperature for Changeover

Find the temperature where \( \Delta G^{\circ} = 0 \). Set up the equation:\[ 0 = \Delta H^{\circ} - T \Delta S^{\circ} \]\[ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} \]\[ T = \frac{-92.2 \mathrm{~kJ}}{-0.199 \mathrm{~kJ/K}} \approx 463.32 \mathrm{~K} \]
06

Convert Changeover Temperature to Celsius

Convert the temperature from Kelvin to Celsius.\[ T_{\mathrm{changeover}} (^{\circ}C) = 463.32 \mathrm{~K} - 273.15 \approx 190.17^{\circ}C \]

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Key Concepts

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

Gibbs Free Energy
The concept of Gibbs Free Energy (\( \Delta G^{\circ} \)) is crucial for understanding the spontaneity of chemical reactions, like the Haber process. Gibbs Free Energy links several thermodynamic quantities and measures the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure. To predict reaction spontaneity, we use:
\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]
Here,
  • \( \Delta G^{\circ} \): Change in Gibbs free energy
  • \( \Delta H^{\circ} \): Change in enthalpy
  • \( T \): Absolute temperature in Kelvin
  • \( \Delta S^{\circ} \): Change in entropy

When \( \Delta G^{\circ} < 0 \), the process is spontaneous under the given conditions. If \( \Delta G^{\circ} > 0 \), it is nonspontaneous. For the Haber process, \( \Delta G^{\circ} \) is negative at 25°C, indicating spontaneous ammonia synthesis under these conditions.
Enthalpy Change
Enthalpy change (\( \Delta H^{\circ} \)) is the total heat absorbed or released in a reaction at constant pressure. It plays a major role in determining the spontaneity of reactions. For the Haber process:
\[ \Delta H^{\circ} = -92.2 \text{ kJ/mol} \]
This indicates that the reaction releases heat, making it exothermic. Exothermic reactions often favor spontaneity since they lower the energy of the system. The negative \( \Delta H^{\circ} \) is one reason why the Haber process can proceed spontaneously at certain temperatures. This release of heat helps drive the formation of ammonia.
Entropy Change
Entropy change (\( \Delta S^{\circ} \)) refers to the disorder or randomness within a system. It is crucial for understanding how the spontaneity of a reaction is influenced by temperature. In the Haber process:
\[ \Delta S^{\circ} = -199 \text{ J/K}\]
This negative value indicates a decrease in disorder as nitrogen and hydrogen gases form solid ammonia. Generally, a decrease in entropy acts against spontaneity, especially at higher temperatures where the \( T \Delta S^{\circ} \) term becomes significant.
Thus, while a negative \( \Delta S^{\circ} \) doesn't favor spontaneity, it is balanced by the exothermic nature (negative \( \Delta H^{\circ} \)), allowing spontaneity under standard conditions.
Spontaneity
Spontaneity is a vital concept for predicting whether a reaction will proceed on its own under set conditions. For the Haber process, determining spontaneity involves evaluating the \( \Delta G^{\circ} \).
At standard conditions (25°C), the negative Gibbs free energy (\( \Delta G^{\circ} = -32.898 \text{ kJ/mol} \)) implies that the Haber process is spontaneous.
However, as temperature shifts, so does spontaneity potential due to entropy's temperature dependence. The changeover temperature, calculated to be approximately 190.17°C, marks the point where the balance of \( \Delta H^{\circ} \) and \( T \Delta S^{\circ} \) shifts and \( \Delta G^{\circ} \) reaches zero, i.e., the process is no longer spontaneous. Understanding such balance aids in optimizing industrial conditions for ammonia synthesis.

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Most popular questions from this chapter

Hess's law can be used to calculate reaction enthalpies for hypothetical processes that can't be carried out in the laboratory. Set up a Hess's law cycle that will let you calculate \(\Delta H^{\circ}\) for the conversion of methane to ethylene: \(2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g)\) You can use the following information: \(2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{\circ}=-3120.8 \mathrm{~kJ}\) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{\circ}=-890.3 \mathrm{~kJ}\) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) \quad \Delta H^{\circ}=-136.3 \mathrm{~kJ}\) \(\mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\mathrm{f}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol}\)

The familiar "ether" used as an anesthetic agent is diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\). Its heat of vaporization is \(+26.5 \mathrm{~kJ} / \mathrm{mol}\) at its boiling point. How much energy in kilojoules is required to convert 100 mL of diethyl ether at its boiling point from liquid to vapor if its density is \(0.7138 \mathrm{~g} / \mathrm{mL}\) ?

Tell whether reactions with the following values of \(\Delta H\) and \(\Delta S\) are spontaneous or nonspontaneous and whether they are exothermic or endothermic: (a) \(\Delta H=-128 \mathrm{~kJ} ; \Delta S=35 \mathrm{~J} / \mathrm{K}\) at \(500 \mathrm{~K}\) (b) \(\Delta H=+67 \mathrm{~kJ} ; \Delta \mathrm{S}=-140 \mathrm{~J} / \mathrm{K}\) at \(250 \mathrm{~K}\) (c) \(\Delta H=+75 \mathrm{~kJ} ; \Delta S=95 \mathrm{~J} / \mathrm{K}\) at \(800 \mathrm{~K}\)

How much heat in kilojoules is evolved or absorbed in the reaction of \(2.50 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) with enough carbon monoxide to produce iron metal? Is the process exothermic or endothermic? $$\begin{aligned}\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) \\ \Delta H^{\circ}=-24.8 \mathrm{~kJ}\end{aligned}$$

Given \(400.0 \mathrm{~g}\) of hot tea at \(80.0{ }^{\circ} \mathrm{C}\), what mass of ice at \(0{ }^{\circ} \mathrm{C}\) must be added to obtain iced tea at \(10.0^{\circ} \mathrm{C}\) ? The specific heat of the tea is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\), and \(\Delta H_{\text {fusion }}\) for ice is \(+6.01 \mathrm{~kJ} / \mathrm{mol}\).

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