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Explain why the \(\Delta G^{\circ}\) for the following reaction becomes more positive as the temperature increases. $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)$$

Short Answer

Expert verified
\(\Delta G^{\circ}\) becomes more positive with increasing temperature due to the negative \(\Delta S^{\circ}\), making \(-T\Delta S^{\circ}\) more positive.

Step by step solution

01

Understanding the Reaction

The given reaction is the synthesis of ammonia from nitrogen and hydrogen gases: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\). We need to understand why the standard Gibbs free energy change, \(\Delta G^{\circ}\), becomes more positive with increasing temperature.
02

Analyzing \(\Delta G^{\circ}\)

The standard Gibbs free energy change \(\Delta G^{\circ}\) is related to the enthalpy change \(\Delta H^{\circ}\) and the entropy change \(\Delta S^{\circ}\) of the reaction by the equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\). To find out why \(\Delta G^{\circ}\) becomes more positive, we need to look at \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\).
03

Evaluating \(\Delta H^{\circ}\)

The formation of ammonia is an exothermic reaction, meaning \(\Delta H^{\circ}\) is negative. This indicates the release of heat during the reaction, which typically favors product formation at lower temperatures.
04

Evaluating \(\Delta S^{\circ}\)

In this reaction, four moles of gas are converted into two moles of ammonia gas. This results in a decrease in disorder or randomness, making \(\Delta S^{\circ}\) negative because there are fewer gas molecules in the products than in the reactants.
05

Effect of Increasing Temperature

Since \(\Delta S^{\circ}\) is negative, the term \(-T\Delta S^{\circ}\) becomes positive and larger as temperature \(T\) increases. This causes \(\Delta G^{\circ}\) to increase, as the \(-T\Delta S^{\circ}\) term becomes more positive.

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Key Concepts

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

Enthalpy Change
In a chemical reaction, enthalpy change ( abla H^{ }}) tells us how much heat is absorbed or released. For the ammonia synthesis reaction, where nitrogen and hydrogen gases combine to form ammonia ( abla NH_{3}) ( abla H^{ }} is negative.
This means the reaction releases heat into the surroundings, making it an example of an exothermic reaction.
Exothermic reactions tend to favor product formation at lower temperatures because the release of heat supports the progress of the reaction.

This negative abla H^{ }} is critical, as it influences the reaction's spontaneity at different temperatures. In simple terms, it helps explain why ammonia synthesis might proceed favorably at cooler temperatures. At higher temperatures, the enthalpy contribution alone wouldn't drive the reaction forward, making it less spontaneous.
Entropy Change
Entropy change ( abla S^{ }}) measures the change in disorder or randomness in a system during a reaction.
During the synthesis of ammonia, we start with four moles of gas—one mole of nitrogen and three moles of hydrogen—and end with two moles of ammonia gas.
This results in a decrease in the number of gas molecules, reducing randomness or disorder.
Therefore, abla S^{ }} is negative for this reaction.

A negative abla S^{ }} means the reaction results in a more ordered state, which is less favored at higher temperatures.
The equation for Gibbs free energy, abla G^{ }}, incorporates the entropy term in abla G^{ }} = abla H^{ }} - T abla S^{ }}.
With a negative abla S^{ }}, the abla -T abla S^{ }} term becomes positive as temperature increases, leading to a higher abla G^{ }} value, meaning the reaction becomes less spontaneous.
Exothermic Reaction
An exothermic reaction is one where heat is released to the surroundings.
The synthesis of ammonia from nitrogen and hydrogen is exothermic, which is indicated by a negative enthalpy change ( abla H^{ }}).
Such reactions are driven by the fact that the system releases energy, making low temperatures more favorable for reaction advancement.

However, as temperature rises, the driving force of this heat release diminishes because other thermodynamic factors start to matter more. The increase in temperature can overcome the effect of the negative abla H^{ }}.
Ultimately, high temperatures can make the reaction less favorable due to competing entropic effects that organization brings in a system with reduced gas particles.
Ammonia Synthesis
Ammonia synthesis is the well-known chemical process of combining nitrogen ( abla N_2}) and hydrogen ( abla H_2}) gases to produce ammonia ( abla NH_3}).
It's a foundational industrial reaction used to produce fertilizers, among other products.

The reaction is represented by:
  • abla N_2(g) + 3 abla H_2(g) ightarrow 2 abla NH_3(g)
This process is highly exothermic, releasing energy and usually performed under conditions favoring lower temperatures and higher pressures to optimize the output of ammonia.

However, according to the thermodynamic equation for Gibbs free energy, at higher temperatures:
  • The positive contribution of abla -T abla S^{ }} increases, making abla G^{ }} more positive, indicating a less favorable process.
Thus, while practical ammonia synthesis balances temperature and pressure, it highlights the complex interplay between temperature and thermodynamic properties such as enthalpy and entropy changes.

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Most popular questions from this chapter

In which of the following processes is \(\Delta S\) negative? (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{CO}_{2}(s) \rightleftarrows \mathrm{CO}_{2}(g)\) (c) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\)

Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the following reactions without referring to a table of thermodynamic data, and explain your predictions. (a) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(g)\) (Hint: The reaction is explosive under suitable conditions.) (b) \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{NaCl}(s)\) (Hint: Is table salt stable?) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)\) (Hint: The reaction occurs in the presence of a suitable catalyst.) (d) \(2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s)\) $$\rightleftarrows 2 \mathrm{CuO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ (Hint: The reaction occurs spontaneously to produce \(\mathrm{CuO}\) when the starting material is heated.)

Industrial chemicals can be made from coal by a process that starts with the reaction between red-hot coal and steam to form a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). This mixture, which is known as synthesis gas, can be converted to methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in the presence of a ruthenium/cobalt catalyst. The methanol produced in the reaction can then be converted into a host of other products, ranging from acetic acid to gasoline. The partial pressure of methanol when a mixture of 1.00 atm \(\mathrm{CO}\) and \(2.00 \mathrm{~atm} \mathrm{H}_{2}\) comes to equilibrium at \(65^{\circ} \mathrm{C}\) is 0.98 atm. Calculate the standard free energy of the reaction at \(65^{\circ} \mathrm{C}\). $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}(g)$$

Which of the following processes should have the most positive value of \(\Delta S ?\) (a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) (b) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{6}(l)\) (c) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\)

For some processes, even though the entropy change is favorable, the process doesn't occur. Explain why.

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